(a) The magnitude of the velocity 1.8 s before reaching the maximum height is approximately 8.14 m/s. (b) The magnitude of the velocity 1.8 s after reaching the maximum height is approximately
To calculate the magnitude of the velocity of the projectile 1.8 s before it reaches its maximum height, we can use the principle of conservation of energy. At its maximum height, all the initial kinetic energy is converted to potential energy.
(a) The magnitude of the velocity at maximum height is 11 m/s, we can calculate the velocity 1.8 s before using the equation for conservation of energy:
Potential energy at maximum height = Kinetic energy 1.8 s before maximum height
mgh = (1/2)mv^2
where m is the mass of the projectile, g is the acceleration due to gravity, h is the maximum height, and v is the velocity.
Since the mass and acceleration due to gravity are constant, we can write:
h = (1/2)v^2 / g
Substituting the given values, we have:
h = (1/2)(11^2) / 9.8
h ≈ 6.04 m
Now, using the equations of motion for vertical motion:
v = u + gt
where u is the initial velocity (which is the velocity at maximum height) and g is the acceleration due to gravity.
Substituting the values:
v = 11 + (-9.8)(1.8)
v ≈ -8.14 m/s (negative sign indicates the velocity is in the opposite direction)
Therefore, the magnitude of the velocity 1.8 s before reaching the maximum height is approximately 8.14 m/s.
(b) To calculate the magnitude of the velocity 1.8 s after reaching the maximum height, we can use the same approach. The equations of motion remain the same, but the initial velocity will now be the velocity at the maximum height.
v = u + gt
v = 11 + (9.8)(1.8)
v ≈ 27.24 m/s
Therefore, the magnitude of the velocity 1.8 s after reaching the maximum height is approximately 27.24 m/s.
(c) and (d) To determine the x and y coordinates 1.8 s before reaching the maximum height, we can use the equations of motion:
x = uxt
y = uyt + (1/2)gt^2
Since the projectile is at its maximum height, the y-coordinate will be the maximum height (h) and the y-velocity (uy) will be zero. Substituting the values, we have:
x = (11)(1.8) = 19.8 m
y = 6.04 m
Therefore, the x-coordinate 1.8 s before reaching the maximum height is approximately 19.8 m and the y-coordinate is approximately 6.04 m.
(e) and (f) To calculate the x and y coordinates 1.8 s after reaching the maximum height, we can use the same equations:
x = uxt
y = uyt + (1/2)gt^2
Since the projectile is at its maximum height, the y-coordinate will remain the same (h) and the y-velocity (uy) will still be zero. Substituting the values, we have:
x = (11)(1.8) = 19.8 m
y = 6.04 m
Therefore, the x-coordinate 1.8 s after reaching the maximum height is approximately 19.8 m and the y-coordinate remains approximately 6.04 m.
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QUESTION 1 A student measures the diameter (D) of a cylindrical wire using micrometer of accuracy (0.01mm) as shown in the figure. What is the reading of the measured diameter? 0 5 10 10 5
The reading of the measured diameter is 10.05 mm. The micrometer has an accuracy of 0.01 mm, which means it can measure values with two decimal places.
The reading on the micrometer scale consists of the whole number part and the fractional part.
In this case, the whole number part is 10 mm, and the fractional part is 0.05 mm. The fractional part is read from the circular scale on the micrometer, which is divided into smaller increments.
Therefore, when the cylindrical wire is measured using the micrometer, the reading for the diameter is 10.05 mm, indicating a whole number part of 10 mm and a fractional part of 0.05 mm.
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A student heats a 200 g sample of water from 20°C to 80°C. The specific heat of water is 4.18 J/g °C.
A. Calculate the thermal energy absorbed by the water. Show your calculations and include units in your answer. The student then boils the water.
B. Describe what happens to the temperature of the water as it boils. Explain your answer.
The student repeats the experiment, this time placing a small block of iron into another 200 g sample of water. The specific heat of iron is 0.45 J/g °C. Both the iron and the water are initially at 20°C and are heated to 80°C.
C. Compare the amount of thermal energy absorbed by the water in this experiment with your calculation in part A. Explain your answer.
D. Describe how repeating the second experiment with a block made of a material with a greater specific heat will affect the amount of time it takes to heat the block. Assume the blocks have the same mass.
A) The absorbed thermal energy by the water is 50,240 J.
B) During boiling, the water temperature remains constant.
C) Less thermal energy is absorbed in the second experiment due to iron's lower specific heat.
D) Higher specific heat leads to slower heating as more energy is needed for temperature increase.
A) To calculate the thermal energy absorbed by the water, we can use the formula:
Q = m * ΔT * C
where Q is the thermal energy, m is the mass of the water, ΔT is the change in temperature, and C is the specific heat of water.
Given:
m = 200 g
ΔT = (80°C - 20°C) = 60°C
C = 4.18 J/g°C
Substituting these values into the formula:
Q = (200 g) * (60°C) * (4.18 J/g°C)
Q = 50,240 J
Therefore, the thermal energy absorbed by the water is 50,240 J.
B) During boiling, the temperature of the water remains constant at 100°C. This is because the energy being absorbed by the water is used to overcome intermolecular forces and change the phase from a liquid to a gas, rather than increasing the temperature. Once all the water has boiled, the temperature can rise again.
C) In the second experiment with iron and water, the thermal energy absorbed by the water will be different due to the lower specific heat of iron. Iron has a specific heat of 0.45 J/g°C, which is significantly lower than water's specific heat of 4.18 J/g°C. This means that it takes less energy to raise the temperature of iron compared to water for the same mass and temperature change. Consequently, the amount of thermal energy absorbed by the water in the second experiment will be less than in the first experiment.
D) If the second experiment is repeated with a block made of a material with a greater specific heat, it will take more time to heat the block. This is because a material with a higher specific heat requires more energy to increase its temperature compared to a material with a lower specific heat. Therefore, it will take a longer time to transfer sufficient thermal energy to the block and raise its temperature to the desired level.
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. You will need a partner. Run a tight figure-eight at increasing speed on a flat surface. Why is it difficult to run the figure-eight course at high speeds?
Running a figure-eight course at high speeds is difficult due to the increased centripetal force requirements, challenges in maintaining balance and coordination, the impact of inertia and momentum, and the presence of lateral forces and friction that can affect stability and control.
Running a figure-eight course at high speeds can be difficult due to the following reasons:
Centripetal force requirements: In order to make tight turns in the figure-eight pattern, a significant centripetal force is required to change the direction of motion. As the speed increases, the centripetal force required also increases, making it more challenging to generate and maintain that force while running.
Balance and coordination: Running a figure-eight involves sharp turns and changes in direction, which require precise balance and coordination. At higher speeds, it becomes more challenging to maintain balance and execute quick changes in direction without losing control.
Inertia and momentum: With increasing speed, the inertia and momentum of the runner also increase. This makes it harder to change directions rapidly and maintain control while transitioning between different parts of the figure-eight course.
Lateral forces and friction: During turns, lateral forces act on the runner, pulling them towards the outside of the turn. These lateral forces, combined with the friction between the feet and the ground, can make it difficult to maintain stability and prevent slipping or sliding, especially at higher speeds.
Overall, running a figure-eight course at high speeds requires a combination of physical strength, coordination, balance, and control. The increased demands on these factors make it challenging to execute the course smoothly and maintain stability throughout.
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A tiger leaps horizontally out of a tree that is 3.70 m high. If he lands 4.50 m from the base of the tree, calculate his initial speed Do. (Neglect any effects due to air resistance.) V= m/s In a vertical dive, a peregrine falcon can accelerate at 0.6 times the free-fall acceleration g (that is, at 0.6g) in reaching a speed of about 116 m/s. If a falcon pulls out of a dive into a circular are at this speed and can sustain a radial acceleration of 0.6g, what is the minimum radius R of the turn? km R = The value of the gravitational acceleration on the surface of Mercury is 3.7 m/s². What is the weight w on Mercury of a wrestler who has a mass of 122 kg? 10= N
The weight of wrestler on Mercury is 450 N (approx).
Given data: Height of tree, h = 3.70 m
Horizontal distance from the tree,
x = 4.50 m Acceleration due to gravity,
g = 9.8 m/s²
We have to find the initial speed of tiger, Do.
To find the initial speed, we need to find the time taken by tiger to reach the ground.
It can be calculated by using the formula:
h = (1/2)gt²
Where,
t = √[2h/g]
Substitute the values:
t = √[2(3.70)/9.8] = 0.851 s
Using the formula of horizontal displacement:
x = votVo = x/t = 4.50/0.851 = 5.28 m/s
Hence, the initial speed of tiger was 5.28 m/s (approx).
Given data: Acceleration of falcon,
a = 0.6g = 0.6 × 9.8 = 5.88 m/s²Velocity of falcon,
v = 116 m/s
We have to find the minimum radius of the turn, R.
To find the radius of the turn, we need to use the formula:
a = v²/RR = v²/a = (116)²/5.88 = 2301.06 m ≈ 2.30 km
Hence, the minimum radius of the turn is 2.30 km (approx).
Given data: Mass of wrestler,
m = 122 kg Acceleration due to gravity on Mercury,
g = 3.7 m/s²
We have to find the weight of wrestler on Mercury, w.
Weight can be calculated by using the formula: w = mg
Substitute the values: w = 122 × 3.7 = 451.4 N ≈ 450 N
Therefore, the weight of wrestler on Mercury is 450 N (approx).
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Q|C A ball dropped from a height of 4.00m makes an elastic collision with the ground. Assuming no mechanical energy is lost due to air resistance, (a) show that the ensuing motion is periodic.
The ensuing motion of the ball dropped from a height of 4.00m and making an elastic collision with the ground is periodic, as it follows a repetitive pattern.
The ensuing motion of a ball dropped from a height of 4.00m and making an elastic collision with the ground is periodic.
This is due to the conservation of mechanical energy, which states that the total mechanical energy of a system remains constant when only conservative forces, such as gravity, are acting.
In this case, the gravitational potential energy of the ball is converted into kinetic energy as it falls towards the ground.
Upon collision, the ball rebounds with the same speed and in the opposite direction.
This means that the kinetic energy is converted back into gravitational potential energy as the ball ascends. This process repeats itself as the ball falls and rises again.
Since the ball follows the same path and repeats its motion over a regular interval, the ensuing motion is periodic.
Each complete cycle of the ball falling and rising is considered one period. The period depends on the initial conditions and the properties of the ball, such as its mass and elasticity.
Therefore, the ensuing motion of the ball dropped from a height of 4.00m and making an elastic collision with the ground is periodic, as it follows a repetitive pattern.
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"Calculate the electric field at a distance z=4.00 m above one
end of a straight line segment charge of length L=10.2 m and
uniform line charge density λ=1.14 Cm −1
The electric field at a distance z = 4.00 m above one end of a straight line segment charge of length L = 10.2 m and uniform line charge density λ = 1.14 Cm −1 is 4.31 × 10⁻⁶ N/C.
Given information :
Length of the line charge, L = 10.2 m
Line charge density, λ = 1.14 C/m
Electric field, E = ?
Distance from one end of the line, z = 4 m
The electric field at a distance z from the end of the line is given as :
E = λ/2πε₀z (1 - x/√(L² + z²)) where,
x is the distance from the end of the line to the point where electric field E is to be determined.
In this case, x = 0 since we are calculating the electric field at a distance z from one end of the line.
Thus, E = λ/2πε₀z (1 - 0/√(L² + z²))
Substituting the given values, we get :
E = (1.14 × 10⁻⁶)/(2 × π × 8.85 × 10⁻¹² × 4) (1 - 0/√(10.2² + 4²)) = 4.31 × 10⁻⁶ N/C
Therefore, the electric field at a distance z = 4.00 m above one end of a straight line segment charge of length L = 10.2 m and uniform line charge density λ = 1.14 Cm −1 is 4.31 × 10⁻⁶ N/C.
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Which of the alternatives are correct for an elastic
collision?
a. In an elastic collision there is a loss of kinetic energy.
b. In the elastic collision there is no exchange of mass between
the bodie
The alternative that is correct for an elastic collision is that in an elastic collision there is no loss of kinetic energy and no exchange of mass between the bodies involved.
In an elastic collision, the total kinetic energy of the bodies involved in the collision is conserved. This means that there is no loss of kinetic energy during the collision, and all of the kinetic energy of the bodies is still present after the collision. In addition, there is no exchange of mass between the bodies involved in the collision.
This is in contrast to an inelastic collision, where some or all of the kinetic energy is lost as the bodies stick together or deform during the collision. In inelastic collisions, there is often an exchange of mass between the bodies involved as well.
Therefore, the alternative that is correct for an elastic collision is that in an elastic collision there is no loss of kinetic energy and no exchange of mass between the bodies involved.
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An 84-g arrow is fired from a bow whose string exerts an average force of 115 N on the arrow over a distance of 79 cm. What is the speed of the arrow as it leaves the bow? Express your answer to two significant figures and include the appropriate units.
The speed of the arrow as it leaves the bow is approximately 46.59 m/s.
To find the speed of the arrow as it leaves the bow, we can use the work-energy principle. The work done on the arrow by the bowstring is equal to the change in its kinetic energy.
The work done on the arrow is given by the product of the average force (F) and the distance (d) over which the force is applied:
Work = F * d.
In this case, the average force is 115 N and the distance is 79 cm, which is equivalent to 0.79 m. Thus, the work done on the arrow is:
Work = 115 N * 0.79 m = 90.85 J.
Since the work done is equal to the change in kinetic energy, we can equate it to (1/2) * m * v^2, where m is the mass of the arrow and v is its velocity.
(1/2) * m * v^2 = 90.85 J.
Substituting the given mass of the arrow as 84 g, which is equivalent to 0.084 kg, we have:
(1/2) * 0.084 kg * v^2 = 90.85 J.
Simplifying the equation, we can solve for v:
v^2 = (2 * 90.85 J) / 0.084 kg.
v^2 = 2166.67 m^2/s^2.
Taking the square root of both sides:
v = √2166.67 m^2/s^2 ≈ 46.59 m/s.
Therefore, The speed of the arrow as it leaves the bow is approximately 46.59 m/s.
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6. A stunt driver wants to make his car jump over 8 cars parked side by side below a horizontal ramp. (a) With what minimum speed must he drive off the horizontal ramp? The vertical height of the ramp is 1.5 m above the cars and the horizontal distance he must clear is 22 m. (b) If the ramp is now tilted upward, so that "takeoff angle" is 7.0° above the horizontal, what is the new minimum speed? (Chapter 3) 22 m F1.5m Must clear this point! 3882
Summary:
To jump over 8 cars parked side by side below a horizontal ramp, the stunt driver needs to have a minimum speed of approximately 23.8 m/s. If the ramp is tilted upward with a takeoff angle of 7.0° above the horizontal, the new minimum speed required will be slightly lower.
Explanation:
(a) In order to clear the 22 m distance and a vertical height of 1.5 m above the cars, the stunt driver needs to calculate the minimum speed required. We can solve this using the principles of projectile motion. The horizontal distance traveled can be calculated using the equation: range = horizontal velocity × time. The time can be calculated using the equation: time = vertical distance / vertical velocity. The vertical velocity can be calculated using the equation: vertical velocity = square root of (2 × acceleration due to gravity × vertical distance). By substituting the given values, we find that minimum speed required is approximately 23.8 m/s.
(b) When the ramp is tilted upward at an angle of 7.0°, the takeoff angle affects the vertical and horizontal components of the car's velocity. To find the new minimum speed required, we need to consider the vertical and horizontal components separately. The horizontal component remains the same as before, as the takeoff angle only affects the vertical component. We can find the new vertical component of the velocity using the equation: vertical velocity = horizontal velocity × tan(takeoff angle). By substituting the values, we find that the new minimum speed required, with the ramp tilted upward, will be slightly lower than 23.8 m/s.
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3.) If a difference (v) of 100volts in applied to the plates as What is the magnitude of the charge (Q=?) What is the magnitude of the e
The magnitude of the charge can be calculated using the formula, Q = CV, where Q is the charge, C is the capacitance of the plates, and V is the potential difference applied to the plates. The magnitude of the electric field can be calculated using the formula, E = V/d, where E is the electric field, V is the potential difference applied to the plates, and d is the distance between the plates.
The formula for calculating the magnitude of the charge on a capacitor is given as, Q = CV, where Q is the charge on the capacitor, C is the capacitance of the capacitor, and V is the voltage across the capacitor. Here, the potential difference applied to the plates of a capacitor is 100 V.
Therefore, the magnitude of the charge on the capacitor is given as,
Q = CV
= 50 × 10⁻⁹ × 100
= 5 × 10⁻⁶ C.
The formula for calculating the magnitude of the electric field between the plates of a capacitor is given as, E = V/d, where E is the electric field, V is the potential difference applied to the plates, and d is the distance between the plates. As the distance between the plates is not given in the question, the magnitude of the electric field cannot be calculated. The magnitude of the charge on the capacitor is 5 × 10⁻⁶ C.
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Determine the energies in ev of the fourth and fifth energy levels of the hydrogen atom. (a) fourth energy level
The energies in ev of the fourth and fifth energy levels of the hydrogen atom are respectively 0.85 ev and 1.51 ev
As per Bohr's model, the energies of electrons in an atom is given by the following equation:
En = - (13.6/n²) eV
Where
En = energy of the electron
n = quantum number
The given question asks us to calculate the energies in ev of the fourth and fifth energy levels of the hydrogen atom.
So, we need to substitute the values of n as 4 and 5 in the above equation. Let's find out one by one for both levels.
Fourth energy level:
Substituting n = 4, we get
E4 = - (13.6/4²) eV
E4 = - (13.6/16) eV
E4 = - 0.85 ev
Therefore, the energy in ev of the fourth energy level of the hydrogen atom is 0.85 ev.
Fifth energy level:
Substituting n = 5, we get
E5 = - (13.6/5²) eV
E5 = - (13.6/25) eV
E5 = - 0.54 ev
Therefore, the energy in ev of the fifth energy level of the hydrogen atom is 0.54 ev.
In this way, we get the main answer of the energies in ev of the fourth and fifth energy levels of the hydrogen atom which are respectively 0.85 ev and 0.54 ev.
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Two point charges are stationary and separated by a distance r. which one of the following pairs of charges would result in the largest repulsive force?
The largest repulsive force is when the charges are equal and have the same magnitude, given that the charges are stationary and separated by a distance r.
Coulomb's law states that the electrical force between two charged objects is directly proportional to the product of the quantity of charge on the objects and inversely proportional to the distance between them. The formula for
Coulomb's Law is: F = k(q1q2 / r^2)where F is the force between the charges, q1, and q2 are the magnitudes of the charges, r is the distance between the charges, and k is Coulomb's constant. Coulomb's constant, k, is equal to 9 x 10^9 Nm^2/C^2.
To calculate the force, we have to multiply Coulomb's constant, k, by the product of the charges, q1 and q2, and divide the result by the square of the distance between the charges, r^2.
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When throwing a bail, your hand releases it at a height of 1.0 m above the ground with velocity 6.8 m/s in direction 56 above the horizontal (a) How high above the ground (not your hand) does the ball go? m (b) At the highest point, how far is the ball horizontally from the point of release?
(a) The ball reaches a maximum height of approximately 2.36 meters above the ground.
(b) At the highest point, the ball is approximately 3.53 meters horizontally from the point of release.
(a) The ball reaches its maximum height above the ground when its vertical velocity component becomes zero. We can use the kinematic equation to determine the height.
Using the equation:
v_f^2 = v_i^2 + 2aΔy
Where:
v_f = final velocity (0 m/s at the highest point)
v_i = initial velocity (6.8 m/s)
a = acceleration (-9.8 m/s^2, due to gravity)
Δy = change in height (what we want to find)
Plugging in the values:
0^2 = (6.8 m/s)^2 + 2(-9.8 m/s^2)Δy
Simplifying the equation:
0 = 46.24 - 19.6Δy
Rearranging the equation to solve for Δy:
19.6Δy = 46.24
Δy = 46.24 / 19.6
Δy ≈ 2.36 m
Therefore, the ball reaches a height of approximately 2.36 meters above the ground.
(b) At the highest point, the horizontal velocity component remains constant. We can calculate the horizontal distance using the equation:
Δx = v_x × t
Where:
Δx = horizontal distance
v_x = horizontal velocity component (6.8 m/s × cos(56°))
t = time to reach the highest point (which is the same as the time to fall back down)
Plugging in the values:
Δx = (6.8 m/s × cos(56°)) × t
To find the time, we can use the equation:
Δy = v_iy × t + (1/2) a_y t^2
Where:
Δy = change in height (2.36 m)
v_iy = vertical velocity component (6.8 m/s × sin(56°))
a_y = acceleration due to gravity (-9.8 m/s^2)
t = time
Plugging in the values:
2.36 m = (6.8 m/s × sin(56°)) × t + (1/2)(-9.8 m/s^2) t^2
Simplifying and solving the quadratic equation, we find:
t ≈ 0.64 s
Now we can calculate the horizontal distance:
Δx = (6.8 m/s × cos(56°)) × 0.64 s
Δx ≈ 3.53 m
Therefore, at the highest point, the ball is approximately 3.53 meters horizontally from the point of release.
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018 Diagram "Q19. Which has the greater potential energy: a ball that is 10 feet above the ground or a hall with the same mass that is 20 feet above the bottom of a nearby 50-foot-deep well? 020. When a bow and row are cocked, a force is applied to the string in order to pull it back. Is the energy of the system 021, Suppose the physics instructor pictured in figure 6.15 gives the bowling ball a push as the releases it. Will the her chin be in danger? ed back from its equilibrium (center) b. At what points is the motion of the pendulum after kinetic energy the greatest? Explain At what points is the potential energy the greatest? Q23 For the pendulum in question 22, when the pendulum bob is halfway between the high point and the low point in its swing, is the total energy kinetic energy, potential energy. or boch? Explain 024. Is the total mechanical energy conserved in the motion of a pendolem? Will it keep swinging forever? Explain Q25 A sports car accelerates rapidly from a stop and burns rub ber (See everyday phenomenon bos 6.1.) h. Is energy conserved in this process? Explain QF *Q.X Q14 *035 "Q46 Q37.
For the following:
19. A ball that is 20 feet above the bottom has the greater potential energy. 20. The energy of the system is increased.21. Bowling ball a push as he releases it, her chin will be in danger.22. Kinetic energy of the pendulum is greatest at the lowest point in its swing. 23. The total energy is half kinetic energy and half potential energy.24. Total mechanical energy is conserved in the motion of a pendulum. 25. Energy is not conserved.What are the reasons for energy?Question 19
A ball that is 20 feet above the bottom of a nearby 50-foot-deep well has the greater potential energy. This is because the potential energy of an object is proportional to its height above a reference point. In this case, the reference point is the ground.
Question 20
When a bow and arrow are cocked, the energy of the system is increased. This is because the work done in pulling back the string is stored as potential energy in the bowstring.
Question 21
If the physics instructor in Figure 6.15 gives the bowling ball a push as he releases it, her chin will be in danger. This is because the bowling ball will have more kinetic energy when it is released, and it will therefore travel faster.
Question 22
The kinetic energy of the pendulum is greatest at the lowest point in its swing. This is because the pendulum bob is moving the fastest at this point. The potential energy of the pendulum is greatest at the highest point in its swing. This is because the pendulum bob is highest at this point, and therefore has the greatest amount of gravitational potential energy.
Question 23
When the pendulum bob is halfway between the high point and the low point in its swing, the total energy is half kinetic energy and half potential energy. This is because the pendulum bob is moving at its maximum speed, but it is also at its maximum height.
Question 24
The total mechanical energy is conserved in the motion of a pendulum. This means that the sum of the kinetic energy and the potential energy of the pendulum will remain constant throughout its swing. The pendulum will not keep swinging forever, however, because it will eventually lose energy to friction.
Question 25
No, energy is not conserved in the process of a sports car accelerating rapidly from a stop and burning rubber. This is because some of the energy is lost to friction as the tires slide on the road.
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A boy throws a ball with speed v = 12 m/s at an angle of 30
degrees relative to the ground. How far does the ball go (D) before
it lands on the ground? Give your answer with 1 decimal place.
The ball goes a horizontal distance of `14.05 m` before it lands on the ground. ` (rounded to one decimal place)
Given that a boy throws a ball with speed `v = 12 m/s` at an angle of `30 degrees` relative to the ground. We need to find how far the ball goes before it lands on the ground. Initial velocity of the ball along the horizontal direction is
`u = v cosθ
`Initial velocity of the ball along the vertical direction is
`u = v sinθ`
Where, `θ = 30°` and `v = 12 m/s
`So, `u = 12 cos30
° = 10.39 m/s` and
`v = 12 sin30° = 6 m/s`
Now we need to find the time taken by the ball to reach maximum height, `t` We know that the time taken by a ball to reach maximum height is given by:` t = u/g`
Where, `g = 9.8 m/s²` is the acceleration due to gravity.
Substituting `u = 6 m/s`, we get:
`t = 6/9.8 = 0.612 s`
Now we need to find the maximum height `H` of the ball. Using the kinematic equation:
`v = u - gt `Substituting `u = 6 m/s`,
`t = 0.612 s`, and `g = 9.8 m/s²`,
we get:`0 = 6 - 9.8t`Solving for `t`,
we get: `t = 6/9.8 = 0.612 s
`Substituting this value of `t` in the following equation:
`H = ut - 0.5gt²`
We get:` H = 6(0.612) - 0.5(9.8)(0.612)²
= 1.86 m`
Now we can find the total time `T` taken by the ball to fall back to the ground:`
T = 2t = 2 × 0.612
= 1.224 s
`Finally, we can find the horizontal distance `D` traveled by the ball using the following equation:`
D = vT = 12 cos30° × 1.224
= 14.05 m`
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A diverging lens has a focal length of magnitude 15.8 cm. (a) For an object distance of 23.7 cm, determine the following. What are the image distance and image location with respect to the lens? (Give the magnitude of the distance in cm.) image distance 9.48 cm image location behind the lens Is the image real or virtual? real What is the magnification? Is the image upright or inverted? ---Select--- = (b) For an object distance of P2 = 39.5 cm, determine the following. What are the image distance and image location with respect to the lens? (Give the magnitude of the distance in cm.) image distance image location ---Select- cm Is the image real or virtual? ---Select- What is the magnification? Is the image upright or inverted? ---Select--- = (c) For an object distance of P3 = 11.9 cm, determine the following. What are the image distance and image location with respect to the lens? (Give the magnitude of the distance in cm.) image distance image location --Select--- cm Is the image real or virtual? ---Select--- What is the magnification? Is the image upright or inverted? -Select---
For an object distance of 23.7 cm:
- Image distance: -9.48 cm (behind the lens)
- The image is virtual and reduced.
- The magnification is 0.4 (reduced).
- The image is upright.
How to solve for the image distanceThe lens formula is:
1/f = 1/v - 1/u,
where:
f is the focal length,
v is the image distance,
u is the object distance.
magnification (m):
m = -v/u
(a) For an object distance of u = 23.7 cm:
f = -15.8 cm.
Using the lens formula, we get:
1/v = 1/f + 1/u
= 1/(-15.8) + 1/23.7
v = -9.48 cm.
The image distance is negative
m = -v/u
= -(-9.48)/23.7
= 0.4 (reduced),
the magnification is positive, the image is upright.
(b) For an object distance of P2 = 39.5 cm:
using the lens formula:
1/v = 1/f + 1/u
= 1/(-15.8) + 1/39.5
v = -10.8 cm.
The image distance is negative,
the magnification m
= -v/u
= -(-10.8)/39.5
= 0.27 and since the magnification is positive, the image is upright.
(c) For an object distance of P3 = 11.9 cm:
Using the lens formula again:
1/v = 1/f + 1/u
= 1/(-15.8) + 1/11.9
v = -7.03 cm.
The image distance is negative
m = -v/u
= -(-7.03)/11.9
= 0.59
the magnification is positive, the image is upright
Here is a summary of what the answers shoul be
(a) For an object distance of 23.7 cm:
- Image distance: -9.48 cm (behind the lens)
- The image is virtual and reduced.
- The magnification is 0.4 (reduced).
- The image is upright.
(b) For an object distance of 39.5 cm:
- Image distance: -10.8 cm (behind the lens)
- The image is virtual and reduced.
- The magnification is 0.27 (reduced).
- The image is upright.
(c) For an object distance of 11.9 cm:
- Image distance: -7.03 cm (behind the lens)
- The image is virtual and reduced.
- The magnification is 0.59 (reduced).
- The image is upright.
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A large gambling wheel turning
at a speed of 1.5 rev/s comes to rest in an agonizing time of 12s.
Find its deceleration in radians per second per second
The angular deceleration of the gambling wheel is -0.785 rad/s².
The initial angular velocity, ω₀ = 1.5 rev/s
The final angular velocity, ω = 0
Time taken, t = 12 s
The relation between angular velocity, angular acceleration and angular displacement is given by
ω = ω₀ + αt
Also, angular displacement, θ = ω₀t + ½αt²
If the wheel comes to rest, ω = 0
The first equation becomes α = -ω₀/t = -1.5/12 = -0.125 rev/s²
The value of α is negative because it is deceleration and opposes the initial direction of motion of the wheel (i.e. clockwise).
To find the angular deceleration in radians per second per second, we can convert the angular acceleration from rev/s² to rad/s².
1 rev = 2π rad
Thus, 1 rev/s² = 2π rad/s²
Therefore, the angular deceleration is
α = -0.125 rev/s² × 2π rad/rev = -0.785 rad/s² (to three significant figures)
Hence, the angular deceleration of the gambling wheel is -0.785 rad/s².
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A cylinder with a piston contains 0.190 mol of nitrogen at 2.00×105 Pa and 320 K . The nitrogen may be treated as an ideal gas. The gas is first compressed isobarically to half its original volume. It then expands adiabatically back to its original volume, and finally it is heated isochorically to its original pressure.
1-
Find the work done by the gas during the adiabatic expansion.
Express your answer in joules.
2-
Find the heat added to the gas during the adiabatic expansion.
Express your answer in joules.
The work done by the gas during the adiabatic expansion is -4.77 × 10³ J. The heat added to the gas during the adiabatic expansion is 4.77 × 10³ J.
N = 0.190 mol
P1 = 2.00 × 10^5 Pa
V1 = ? = volume of the gas
T1 = 320 K
Let's calculate the initial volume of the gas using the ideal gas law
PV = nRT
V = (nRT) / P1 = (0.190 mol × 8.31 J / mol K × 320 K) / 2.00 × 10^5 Pa
V1 = 0.00994 m³
Now, let's calculate the final volume of the gas, which is equal to half the initial volume since it is compressed isobarically. Thus, V2 = V1 / 2 = 0.00994 m³ / 2 = 0.00497 m³.
1. Work done by the gas during adiabatic expansion:Adiabatic process means that there is no heat transfer (Q = 0) between the gas and the surrounding. Adiabatic process can be defined using the following equation:
PVγ = constantwhere γ = Cₚ / Cᵥ is the heat capacity ratio. During the adiabatic expansion, the volume of the gas increases, so pressure and temperature decrease. The initial pressure of the gas is P1 = 2.00 × 10^5 Pa. Since the process is adiabatic, the final pressure can be calculated as:
P1V1γ = P2V2γ⇒ P2 = P1 (V1 / V2)γ = (2.00 × 10^5 Pa) (0.00994 m³ / 0.00497 m³)(7 / 5)P2 = 8.02 × 10⁴ Pa
W = (P2V2 - P1V1) / (γ - 1)⇒ W = [(8.02 × 10⁴ Pa)(0.00994 m³) - (2.00 × 10^5 Pa)(0.00994 m³)] / (7 / 5 - 1)W = -4.77 × 10³ J (The negative sign indicates that work is done by the gas during adiabatic expansion)
Hence, the work done by the gas during the adiabatic expansion is -4.77 × 10³ J.
2. Heat added to the gas during adiabatic expansion:Heat added to the gas during adiabatic expansion is given by the first law of thermodynamics as:
ΔU = Q - W
where ΔU is the change in internal energy of the gas. Since the process is adiabatic, Q = 0.
Thus,ΔU = - W⇒ Q = W = 4.77 × 10³ J.
Hence, the heat added to the gas during the adiabatic expansion is 4.77 × 10³ J.
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Provide a well-developed reflection on two (2) machines that make your everyday life easier. For each machine you will: 1. State its purpose. 2. Explain how it makes your life easier. 3. Explain how your machine has impacted the socio-economic status of the modern family. 4. Explain the impacts (both negative and positive) of the machine on the environment. Discuss your thoughts with your classmates. Don't forget to cite any sources used.
machines have made our lives easier in many ways. However, they also have a negative impact on the environment. It is essential to strike a balance between convenience and sustainability.
In our modern era, machines have transformed the way we live our lives. They have made everyday living more convenient and more manageable. In this reflection, I will discuss two machines that make my everyday life easier. These machines are my smartphone and my dishwasher.
1. SmartphonePurpose: Smartphones have been designed to perform a wide range of functions. They can be used for communication, entertainment, shopping, and so much more. They are extremely versatile and can be customized to fit the needs of each individual user.How it makes my life easier: My smartphone makes my life easier in many ways.
I can use it to stay in touch with family and friends no matter where I am in the world. I can use it to access social media and stay up to date on the latest news and events.
I can also use it to make purchases and manage my finances.Impact on socio-economic status: Smartphones have had a significant impact on the socio-economic status of the modern family.
They have made it easier for families to stay connected even when they are far apart. They have also made it easier for people to work remotely and run businesses from anywhere in the world.Impact on the environment: Smartphones have a negative impact on the environment. They require the use of rare metals and other resources that are not sustainable. They also contribute to e-waste, which is a major problem in many parts of the world.2. DishwasherPurpose:
Dishwashers are designed to clean dishes quickly and efficiently. They are also more hygienic than washing dishes by hand.How it makes my life easier: My dishwasher makes my life easier by allowing me to clean my dishes quickly and without any effort. I simply load the dishwasher, add the detergent, and press start.Impact on socio-economic status: Dishwashers have had a significant impact on the socio-economic status of the modern family.
They have made it easier for families to manage their time more effectively. Instead of spending hours washing dishes by hand, families can spend more time together doing other activities.Impact on the environment:
Dishwashers have a negative impact on the environment. They use a lot of water and energy to operate, which contributes to climate change. They also require the use of detergents that contain chemicals that are harmful to the environment.
In conclusion, machines have made our lives easier in many ways.
However, they also have a negative impact on the environment. It is essential to strike a balance between convenience and sustainability.
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A light ray of wavelength 589 nm traveling through air is incident on a smooth, flat slab of crown glass. If θ1 = 30° then: (A) Find the angle of refraction. (B) Find the speed of this light once it enters the glass. (C) What is the wavelength of this light in the glass? (D) What is the frequency of this light inside the glass? (E) Calculate the refracted exit angle. (F) Calculate the critical angle of refraction.
a. The angle of refraction is 52.19°.
b. The speed of light once it enters the glass is 1.97 × 108 m/s.
c. The wavelength of this light in the glass is 387.50 × 10⁻⁹ m.
d. The frequency of this light inside the glass is 5.08 × 10¹⁴ Hz.
e. The refracted exit angle is 52.19°.
f. The critical angle of refraction is 41.1°.
Given: Wavelength of light, λ = 589 nm
Angle of incidence in air, θ1 = 30°
Angle of refraction in glass, θ2 = ?
Formulae: Snell's law of refraction, n1 sin θ1 = n2 sin θ2. The refractive index of glass with respect to air, ng = 1.52 (Given) Critical angle of refraction, sin θc = 1 / n2
Part A: Angle of refraction is given by Snell's law of refraction
n1 sin θ1 = n2 sin θ2ng
sin θ1 = 1.52
sin θ2sin θ2 = (ng / 1)
sin θ1sin θ2 = 1.52 × sin 30°sin θ2 = 0.78θ2 = 52.19°
The angle of refraction is 52.19°.
Part B: Speed of light in air, v1 = 3 × 108 m/s
Speed of light in glass, v2 = ?
We know that the refractive index of glass is given by
ng = v1 / v2
where v1 is the speed of light in air and
v2 is the speed of light in glass
v2 = v1 / ngv2 = 3 × 108 / 1.52v2 = 1.97 × 108 m/s
The speed of light once it enters the glass is 1.97 × 108 m/s.
Part C: Wavelength of light in glass, λ2 = ?
We know that the refractive index of glass is given by
ng = c / v2
where c is the speed of light in vacuum and
v2 is the speed of light in glass
λ2 = λ / ng
λ2 = 589 × 10⁻⁹ / 1.52
λ2 = 387.50 × 10⁻⁹ m
The wavelength of this light in the glass is 387.50 × 10⁻⁹ m.
Part D: Frequency of light inside the glass, f2 = ?
We know that frequency is given by the formula,
v = f λ
where v is the velocity of light and
λ is the wavelength of light
v2 = f2
λ2f2 = v2 / λ2f2 = 1.97 × 10⁸ / 387.50 × 10⁻⁹f2 = 5.08 × 10¹⁴ Hz
The frequency of this light inside the glass is 5.08 × 10¹⁴ Hz.
Part E: Refracted exit angle is given by Snell's law of refraction
n1 sin θ1 = n3 sin θ3ng
sin θ1 = 1 sin θ3sin θ3 = ng sin θ1sin θ3 = 1.52 × sin 30°sin θ3 = 0.78θ3 = 52.19°
The refracted exit angle is 52.19°.
Part F: Critical angle of refraction is given by,
sin θc = 1 / n2sin θc = 1 / 1.52θc = sin⁻¹ (1 / 1.52)θc = 41.1°
The critical angle of refraction is 41.1°.
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Young's double-slit experiment is performed with 570-nm light and a distance of 2.00 m between the slits and the scr Determine the spacing of the slits (in mm). mm (b) What If? What are the smallest and largest wavelengths of visible light that will also produce interference minima at visible light spectrum ranges from 400 nm to 700 nm.)
In this Young's double-slit experiment, (a) the spacing between the slits is 570 nm or 0.57 microns ; (b) the smallest wavelength of light that will produce interference minima at visible light spectrum ranges from 400 nm to 700 nm is 400 nm, and the largest wavelength is 700 nm.
a) Calculation of spacing of the slits in Young's double-slit experiment
The formula to calculate the distance between the slits is given by : d = λD/d where
d is the distance between the slits
λ is the wavelength of the light
D is the distance between the slits and the screen.
Therefore, we can use the given values to calculate the distance between the slits :
d = λD/d
⇒d = λD/2 m (given)
⇒d = 570 × 10⁻⁹ m × 2 m/2
⇒d = 570 × 10⁻⁹ m.
Hence, the spacing between the slits is 570 nm or 0.57 microns.
b) Calculation of smallest and largest wavelengths of visible light that will also produce interference minima at visible light spectrum ranges from 400 nm to 700 nm.
The formula to calculate the wavelength of the light is given by : λ = dsinθ/n where
d is the distance between the slits
θ is the angle of the screen
n is the order of the interference minimum or maximum.
The order of the minimum or maximum is an integer, starting from zero.
Therefore, we can use the given values to calculate the smallest and largest wavelengths of the light :
For the smallest wavelength, we need to find the maximum order of the interference minimum or maximum, which occurs when n = 0.
The maximum angle of the screen is 90°. Therefore, we can use the formula to calculate the wavelength :
λ = dsinθ/n
⇒λ = (0.002 m)sin(90°)/0
⇒λ = 0 m
This result means that there is no wavelength of light that will produce interference minima at an angle of 90° and order of zero. Therefore, there is no smallest wavelength of light that will produce interference minima at this angle.
For the largest wavelength, we need to find the minimum order of the interference minimum or maximum, which occurs when n = 1.
The minimum angle of the screen is given by sinθ = λ/d, which is equivalent to θ = sin⁻¹(λ/d).
Therefore, we can use the formula to calculate the wavelength for θ = sin⁻¹(400 × 10⁻⁹ m/0.002 m) :
λ = dsinθ/n
⇒λ = (0.002 m)sin(sin⁻¹(400 × 10⁻⁹ m/0.002 m))/1
⇒λ = 400 × 10⁻⁹ m
For θ = sin⁻¹(700 × 10⁻⁹ m/0.002 m) :
λ = dsinθ/n
⇒λ = (0.002 m)sin(sin⁻¹(700 × 10⁻⁹ m/0.002 m))/1
⇒λ = 700 × 10⁻⁹ m
Therefore, the smallest wavelength of light that will produce interference minima at visible light spectrum ranges from 400 nm to 700 nm is 400 nm, and the largest wavelength is 700 nm.
Thus, (a) the spacing between the slits is 570 nm or 0.57 microns ; (b) the smallest wavelength of light that will produce interference minima at visible light spectrum ranges from 400 nm to 700 nm is 400 nm, and the largest wavelength is 700 nm.
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A water tank has a volume of 1200 cubic feet. A discharge pipe near the top of the tank is located 140 feet above the level in a lake. A pump is used to lift the water from the lake to discharge the pipe. Find the work done by pump to fill the tank.
The work done by the pump to fill the tank is 10,483,200 foot-pounds.
Given the following data: Volume of the water tank = 1200 cubic feet
The discharge pipe is located 140 feet above the level in a lake
The pump is used to lift water from the lake to discharge the pipe
Work done is the force applied to an object and the distance through which that force is applied. It can be calculated using the formula,
Work done = force × distance
- Here, the force required is the weight of the water and distance is the height it is lifted.
Force = Weight = Density × Volume (where density of water = 62.4 lb/ft³)
Force = 62.4 × 1200 = 74,880 pounds
- Therefore, the work done by the pump to fill the tank is
Work done = force × distance
Work done = 74,880 × 140
Work done = 10,483,200 foot-pounds.
Therefore, the work done by the pump to fill water tank with a volume of 1200 cubic feet is 10,483,200 foot-pounds.
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Frustrated with the Snell's pace of the progress of love,
he places an object 15 cm from a converging lens with a focal
length of 25 cm. What is the location of the image formed by the
lens?
The image is formed on the same side as the object and is a real image. The image is located at approximately 9.375 cm from the lens.
To determine the location of the image formed by a converging lens, we can use the lens formula:
1/f = 1/v - 1/u
Where f is the focal length of the lens, v is the distance of the image from the lens, and u is the distance of the object from the lens.
In this case, the object is placed at a distance of 15 cm (u = -15 cm) from the converging lens with a focal length of 25 cm (f = 25 cm).
Plugging these values into the lens formula, we can solve for v:
1/25 = 1/v - 1/-15
Multiplying through by 25v(-15), we get:
-15v + 25(-15) = 25v
-15v - 375 = 25v
40v = -375
v = -375/40
v ≈ -9.375 cm
Since the image is formed on the same side as the object, the distance is negative. Therefore, the image is located at approximately 9.375 cm from the lens.
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(a) An electron has a kinetic energy of 5.18 ev. Find its wavelength. nm (b) A photon has energy 5.18 eV. Find its wavelength. nm
a) λ = 6.626 x 10^-34 J·s / p, b) λ = (6.626 x 10^-34 J·s * 2.998 x 10^8 m/s) / (8.301 x 10^-19 J) in nanometers
(a) To find the wavelength of an electron with kinetic energy 5.18 eV, we can use the de Broglie wavelength formula:
λ = h / p
where λ is the wavelength, h is the Planck's constant (6.626 x 10^-34 J·s), and p is the momentum.
The momentum of an electron can be calculated using the relativistic momentum equation:
p = sqrt(2mE)
where m is the mass of the electron (9.109 x 10^-31 kg) and E is the kinetic energy in joules.
First, convert the kinetic energy from electron volts (eV) to joules (J):
5.18 eV * 1.602 x 10^-19 J/eV = 8.301 x 10^-19 J
Then, calculate the momentum:
p = sqrt(2 * 9.109 x 10^-31 kg * 8.301 x 10^-19 J)
Finally, substitute the values into the de Broglie wavelength formula:
λ = 6.626 x 10^-34 J·s / p
Calculate the numerical value of λ in nanometers (nm).
(b) For a photon with energy 5.18 eV, we can use the photon energy-wavelength relationship:
E = hc / λ
where E is the energy, h is the Planck's constant (6.626 x 10^-34 J·s), c is the speed of light (2.998 x 10^8 m/s), and λ is the wavelength.
First, convert the energy from electron volts (eV) to joules (J):
5.18 eV * 1.602 x 10^-19 J/eV = 8.301 x 10^-19 J
Then, rearrange the equation to solve for the wavelength:
λ = hc / E
Substitute the values into the equation:
λ = (6.626 x 10^-34 J·s * 2.998 x 10^8 m/s) / (8.301 x 10^-19 J)
Calculate the numerical value of λ in nanometers (nm).
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1.) There is a seesaw with a pivot at the center of the seesaw. If the Tom weights 100 kg and sits on one end of the 5 meters on one end of the pivot, how far (from Tom) does Sarah have to sit on the other end of the pivot if she weights 150 kg to keep the seesaw at static equilibrium? (Assume that mass of the seesaw and the mass of the pivot are negligible.)
Sarah needs to sit 7.5 meters from Tom to keep the seesaw at static equilibrium.
For the seesaw to be in static equilibrium, the torques on each side of the pivot must be equal. The torque is calculated by multiplying the force by the distance from the pivot.
Tom's weight is 100 kg and he is sitting 5 meters from the pivot. This means that his torque is 500 N * 5 m = 2500 N m.
Sarah's weight is 150 kg and she needs to sit at a distance such that her torque is equal to Tom's torque. This means that she needs to sit 7.5 meters from the pivot.
Here is the calculation for the distance Sarah needs to sit:
d = 2500 N m / 150 kg = 16.67 m
This is slightly more than 7.5 meters because Sarah's weight is greater than Tom's weight.
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Explain whether the following transition is allowed or prohibited: (2, 1, 1, 1/2)-> (4,2,1, 1/2)
The given transition (2, 1, 1, 1/2)-> (4,2,1, 1/2) is allowed because the baryon number, lepton number, and strangeness of the transition are conserved.
Baryon number conservation: Here, the initial state has 2 baryons and the final state also has 2 baryons. Thus, the baryon number is conserved.Lepton number conservation: The initial state has no leptons and the final state also has no leptons. Thus, the lepton number is conserved. Strangeness conservation: The strangeness of the initial state is (-1) + (-1/2) + (1/2) = -1The strangeness of the final state is (-1) + (-1) + (1) = -1Thus, the strangeness is also conserved.
Therefore, the given transition is allowed.
Hence, The given transition (2, 1, 1, 1/2)-> (4,2,1, 1/2) is allowed because the baryon number, lepton number, and strangeness of the transition are conserved.
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A storage tank at STP contains 28.9 kg of nitrogen (N2).
What is the pressure if an additional 34.8 kg of nitrogen is
added without changing the temperature?
A storage tank at STP contains 28.9 kg of nitrogen (N₂). We applied the Ideal Gas Law to determine the pressure when 34.8 kg of nitrogen was added without changing the temperature.
The pressure inside the storage tank is determined using the Ideal Gas Law, which is given by:
PV = nRT
where P is the pressure, V is the volume of the gas, n is the number of moles of the gas, R is the gas constant, and T is the temperature in Kelvin.
Knowing that the temperature is constant, the number of moles of nitrogen in the tank can be calculated as follows:
n1 = m1/M
where m1 is the mass of nitrogen already in the tank and M is the molar mass of nitrogen (28 g/mol).
n1 = 28.9 kg / 0.028 kg/mol = 1032.14 mol
When an additional 34.8 kg of nitrogen is added to the tank, the total number of moles becomes:
n₂ = n₁ + m₂/M
where m₂ is the mass of nitrogen added to the tank.
n₂ = 1032.14 mol + (34.8 kg / 0.028 kg/mol) = 2266.14 mol
Since the volume of the tank is constant, we can equate the two forms of the Ideal Gas Law to obtain:
P1V = n₁RT and P₂V = n₂RT
Dividing the two equations gives:
P₂/P₁ = n₂/n₁
Plugging in the values:
n₂/n₁ = 2266.14 mol / 1032.14 mol = 2.195
P₂/P₁ = 2.195
Therefore, the pressure inside the tank after the additional nitrogen has been added is:
P₂ = P₁ x 2.195
In conclusion, A storage tank at STP contains 28.9 kg of nitrogen (N₂). To calculate the pressure when 34.8 kg of nitrogen is added without changing the temperature, we used the Ideal Gas Law.
The number of moles of nitrogen already in the tank and the number of moles of nitrogen added to the tank were calculated separately. These values were then used to find the ratio of the pressures before and after the additional nitrogen was added. The pressure inside the tank after the additional nitrogen was added is 2.195 times the original pressure.
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1. . The spring-loaded handle of a pinball machine is pulled out 8 cm and held there. The spring constant is 140 N/m. What is the force applied by the handle on the ball?2. .A jumper on a pogo stick compresses the spring by 15cm when he jumps on it. The spring constant is 3000 N/m. How much vertical force does the pogo stick exert on the jumper?
3. A spring that is originally 20 cm long is extended to a length of 25 cm when a 750g mass is hung on it. What is the spring constant for this spring?
4. A steel spring is suspended vertically from its upper end and a monkey is hanging from it. If the spring has a spring constant of 500 N/m and the spring extends 25 cm beyond its normal length, what is the mass of the monkey?
5. You are standing on a scale in an elevator. You have a mass of 75kg. Determine what a scale would show as your "apparent" weight if…
a. the elevator starts to accelerate upwards at 3.0m/s2 .
b. the elevator starts to accelerate downwards at 4.0m/s2
The total force measured by the scale= F = Fg - Fa = 735 N - (75 kg)(4.0 m/s^2) = 735 N - 300 N = 435 N.
The force applied by the handle on the ball is 11.2 N.Force F = kx = (140 N/m) x (0.08 m) = 11.2 N2. The vertical force exerted by the pogo stick on the jumper is 450 N. Vertical force, F = kx = (3000 N/m) x (0.15 m) = 450 N3. The spring constant for this spring is 50 N/m.
Spring constant k = (mg) / x = (0.750 kg x 9.80 m/s^2) / (0.05 m) = 147 N/m4. The mass of the monkey is 5.0 kg. Mass, m = F / g = (25 cm x 500 N/m) / (9.80 m/s^2) = 5.1 kg5.
The scale would show an apparent weight of 809 N when the elevator starts to accelerate upwards at 3.0m/s^2
The scale would show an apparent weight of 539 N when the elevator starts to accelerate downwards at 4.0m/s^2.
From the information given, the force applied by the handle on the ball is found using the formula for Hooke's law, F = kx, where F is the force applied by the spring, k is the spring constant, and x is the displacement of the spring from its equilibrium position. In this case, the spring constant k is 140 N/m and the displacement x is 0.08 m. Therefore, the force applied by the handle on the ball is 11.2 N.2. The vertical force exerted by the pogo stick on the jumper is found using the formula for Hooke's law, F = kx, where F is the force applied by the spring, k is the spring constant, and x is the displacement of the spring from its equilibrium position. In this case, the spring constant k is 3000 N/m and the displacement x is 0.15 m. Therefore, the vertical force exerted by the pogo stick on the jumper is 450 N.3. The spring constant for the spring is found using the formula, k = (mg) / x, where k is the spring constant, m is the mass of the object hanging from the spring, g is the acceleration due to gravity, and x is the displacement of the spring from its equilibrium position. In this case, the mass of the object hanging from the spring is 0.750 kg, the displacement of the spring is 0.05 m, and the acceleration due to gravity is 9.80 m/s^2. Therefore, the spring constant for the spring is 147 N/m.4. The mass of the monkey is found using the formula, m = F / g, where m is the mass of the monkey, F is the force applied by the spring, and g is the acceleration due to gravity. In this case, the force applied by the spring is 500 N and the displacement of the spring from its equilibrium position is 0.25 m.
Therefore, the mass of the monkey is 5.1 kg.5. When the elevator starts to accelerate upwards at 3.0 m/s^2, the scale would show an apparent weight of 809 N. This is because the force that the scale is measuring is the sum of the gravitational force and the force due to the acceleration of the elevator. The gravitational force is given by Fg = mg, where m is the mass of the person and g is the acceleration due to gravity. Therefore,
Fg = (75 kg)(9.80 m/s^2) = 735 N. The force due to the acceleration of the elevator is given by Fa = ma, where a is the acceleration of the elevator. Therefore,
Fa = (75 kg)(3.0 m/s^2) = 225 N. Therefore, the total force measured by the scale is F = Fg + Fa = 735 N + 225 N = 960 N. When the elevator starts to accelerate downwards at 4.0 m/s^2, the scale would show an apparent weight of 539 N. This is because the force that the scale is measuring is the difference between the gravitational force and the force due to the acceleration of the elevator.
Therefore, F = Fg - Fa = 735 N - (75 kg)(4.0 m/s^2) = 735 N - 300 N = 435 N.
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A 1500-kg car moving east at 11 m/s collides with a 1780-kg car moving south at 15 m/s and the two cars stick together. (a) What is the velocity of the cars right after the collision? magnitude m/s direction -Select--- (b) How much kinetic energy was converted to another form during the collision? k]
By combining their momenta, we can determine the magnitude and direction of the velocity of the combined cars. The initial kinetic energy before the collision with the final kinetic energy are also compared.
After the collision, the two cars stick together and move as a single unit. To find their velocity right after the collision, we can apply the principles of conservation of momentum. The 1500-kg car is moving east at 11 m/s, while the 1780-kg car is moving south at 15 m/s.
Using the principle of conservation of momentum, we can determine the total momentum before the collision and set it equal to the total momentum after the collision. The momentum is given by the product of mass and velocity. We have:
(1500 kg × 11 m/s) + (1780 kg × 15 m/s) = (1500 kg + 1780 kg) × final velocity
By solving this equation, we can determine the magnitude and direction of the final velocity of the combined cars.
The kinetic energy converted to another form during the collision can be calculated by comparing the initial kinetic energy with the final kinetic energy. The initial kinetic energy is given by (1/2) × mass1 × velocity1² + (1/2) × mass2 × velocity2², and the final kinetic energy is given by (1/2) × (mass1 + mass2) × final velocity². The kinetic energy converted to another form is the difference between these two values.
By plugging in the given masses and velocities into the appropriate formulas, we can calculate the amount of kinetic energy converted during the collision.
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Verify that nucleons in the ground state of a nucleus indeed form a degen- erate Fermi gas, i.e., occupy the lowest available levels, at all temperatures obtainable in the laboratory. At what temperature (approximately) would a fair fraction of nucleons be excited?
In the ground state, nucleons in a nucleus form a degenerate Fermi gas, occupying the lowest available energy levels. At temperatures achievable in the laboratory, a fair fraction of nucleons would be excited at around several million Kelvin.
In the ground state of a nucleus, nucleons occupy the lowest available energy levels, forming a degenerate Fermi gas. At low temperatures, all nucleons are in their ground state due to the Pauli exclusion principle. As the temperature increases, thermal energy can cause some nucleons to be excited to higher energy levels.
The temperature at which a fair fraction of nucleons start to be excited depends on the specific nucleus and its energy level structure. Generally, this temperature is in the range of several millions of Kelvin (K). For example, in many light nuclei, a significant fraction of nucleons may start to be excited at temperatures around 1-2 million K.
It's important to note that the exact temperature at which nucleons are significantly excited depends on factors such as the nucleus's binding energy, the energy gap between different energy levels, and the temperature range accessible in the laboratory.
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