The magnitude of the maximum acceleration when the mass is released is 40.49 m/s2.
We are given the mass of the object (150 g), the spring constant (k = 44.3 N/m), and the amount of stretch of the spring (0.104 m). We need to find the magnitude of the maximum acceleration when the mass is released. We know that the restoring force of a spring (F) is given by:
F = -kx where F is the restoring force, k is the spring constant, and x is the displacement of the spring from its equilibrium position. In this case, the mass is stretched 0.104 m, so the restoring force is:
F = -(44.3 N/m)(0.104 m)
F = -4.602 N
The force acting on the mass is the force of gravity, which is:
F = mg where F is the force, m is the mass, and g is the acceleration due to gravity (9.81 m/s2).In this case, the force of gravity is:
F = (0.15 kg)(9.81 m/s2)F = 1.4715 N
When the mass is released, the net force acting on it is Fnet = F - FFnet = 1.4715 N - (-4.602 N)Fnet = 6.0735 NThe acceleration of the mass is given by:
Fnet = ma6.0735 N = (0.15 kg)a
The maximum acceleration when the mass is released is: a = 40.49 m/s2
We are given the mass of the object (150 g), the spring constant (k = 44.3 N/m), and the amount of stretch of the spring (0.104 m). We need to find the magnitude of the maximum acceleration when the mass is released. We know that the restoring force of a spring (F) is given by:
F = -kx
where F is the restoring force, k is the spring constant, and x is the displacement of the spring from its equilibrium position. In this case, the mass is stretched 0.104 m, so the restoring force is: F = -(44.3 N/m)(0.104 m)F = -4.602 NThe force acting on the mass is the force of gravity, which is: F = mg where F is the force, m is the mass, and g is the acceleration due to gravity (9.81 m/s2). In this case, the force of gravity is: F = (0.15 kg)(9.81 m/s2)F = 1.4715 NWhen the mass is released, the net force acting on it is:
Fnet = F - FFnet = 1.4715 N - (-4.602 N)
Fnet = 6.0735 N
The acceleration of the mass is given by: Fnet = ma6.0735 N = (0.15 kg) The maximum acceleration when the mass is released is:
a = 40.49 m/s2
Therefore, the magnitude of the maximum acceleration when the mass is released is 40.49 m/s2. The restoring force on the oscillating mass is always in a direction opposite to the displacement.
When a spring is stretched, it tries to go back to its original position. The force that causes this is called the restoring force. It is always in the opposite direction to the displacement of the spring. In this case, the magnitude of the maximum acceleration when the mass is released is 40.49 m/s2. The restoring force on the oscillating mass is always in a direction opposite to the displacement.
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15. An engineer launches a projectile from a point 245 m in front of a 325-meter tall building. Its launch velocity is unknown. Ignore the air resistance.
(a) what is the maximum vertical component of initial velocity (vy0) at t =0 is needed to touch the top of the building?
(b) What is the horizontal component of initial velocity (vx0) at t =0 is needed to move 245 m for the projectile to touch the top of building?.
Maximum vertical component of initial velocity (vy0) at t = 0: 19.6 m/s. and Horizontal component of initial velocity (vx0) at t = 0: 122.5 m/s.
To calculate the maximum vertical component of the initial velocity (vy0) at t = 0 needed to touch the top of the building, we can use the equation of motion for vertical motion. The projectile needs to reach a height of 325 meters, so the maximum vertical displacement (Δy) is 325 meters. Since we're ignoring air resistance, the only force acting vertically is gravity. Using the equation Δy = vy0 * t + (1/2) * g * t^2, where g is the acceleration due to gravity (approximately 9.8 m/s^2), we can rearrange the equation to solve for vy0. At the maximum height, the vertical displacement is zero, so the equation becomes 0 = vy0 * t - (1/2) * g * t^2. Substituting the values, we have 0 = vy0 * t - (1/2) * 9.8 * t^2. Solving this quadratic equation, we find t = 2s (taking the positive root). Plugging this value into the equation, we can solve for vy0: 0 = vy0 * 2s - (1/2) * 9.8 * (2s)^2. Solving for vy0, we get vy0 = 9.8 * 2s = 19.6 m/s. (b) To calculate the horizontal component of the initial velocity (vx0) at t = 0 needed for the projectile to move 245 m and touch the top of the building, we can use the equation of motion for horizontal motion. The horizontal distance (Δx) the projectile needs to travel is 245 meters. The horizontal component of the initial velocity (vx0) remains constant throughout the motion since there are no horizontal forces acting on the projectile. Using the equation Δx = vx0 * t, we can rearrange the equation to solve for vx0. Since the time of flight is the same for both the vertical and horizontal motions (2s), we can substitute the value of t = 2s into the equation. Thus, we have 245 = vx0 * 2s. Solving for vx0, we get vx0 = 245 / (2s) = 122.5 m/s.
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A certain child's near point is 14.0 cm; her far point (with eyes relaxed) is 119 cm. Each eye lens is 2.00 cm from the retina. (a) Between what limits, measured in diopters, does the power of this lens-cornea combination vary? Calculate the power of the eyeglass lens the child should use for relaxed distance vision. diopters Is the lens converging or diverging?
Near point = 14.0 cm Far point = 119 cm Distance between retina and eye lens = 2.00 cm
The distance between the near point and the eye lens is = 14 - 2 = 12 cm
The distance between the far point and the eye lens is = 119 - 2 = 117 cm
Lens formula,1/f = 1/v - 1/u Where,f = focal length of the eye lens v = distance of far point u = distance of near point
Therefore, 1/f = 1/119 - 1/14= (14 - 119) / 14 × 119= - 105 / 1666f = - 1666 / (-105) = 15.876 cm
Therefore, The focal length of the eye lens is = 15.876 cm
Now, The power of the eye lens, P = 1/f= 1/15.876= 0.063 diopters
The formula for lens power is, P = 1/f or f = 1/P
Therefore, f = 1/0.063= 15.876 cm
Here, The power of the eyeglass lens the child should use for relaxed distance vision is = - 2.34 diopters.
Now, The image formed by the eye lens is a real and inverted image, which means that the eye lens is a converging lens.
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What is the frequency f of a 2.89×10 −19 J photon? f= Hz What is the wavelength λ of a 2.89×10 −19 J photon? λ=
The frequency can be expressed as [tex]4.366 *10^{14} Hz[/tex]the wavelength λ can be expressed as [tex]6.876 *10^{-7} meters[/tex]
How can the wavelength be calculated?The frequency of a repeated event is its number of instances per unit of time. For clarity and to distinguish it from spatial frequency, it is also sometimes referred to as temporal frequency.
Frequency is measured in hertz which is equal to one event per secondGiven that Energy =2.89×10 −19 J
h = plank constant = [tex]6.626 *10^{-34}[/tex]
E = hf
f = E / h
f = [tex]\\\frac{2.89* 10^{-19} }{ 6.626*10^{-34} }[/tex]
f= [tex]4.366 *10^{14} Hz[/tex]
To calculate the wavelength we can use
λ = c / f
λ = [tex]\\\frac{2.998 *10^8}{4.366*10^14}[/tex]
λ =[tex]6.876 *10^-7 meters[/tex]
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Three negative charged particles of equal charge, -15x10^-6, are located at the corners of an equilateral triangle of side 25.0cm. Determine the magnitude and direction of the net electric force on each particle.
The magnitude of the net electric force on each particle is 2.025 N directed away from the triangle.
Charge on each particle, q1 = q2 = q3 = -15 × 10⁻⁶C
∴ Net force on particle 1 = F1
Net force on particle 2 = F2
Net force on particle 3 = F3
The magnitude of the net electric force on each particle:
It can be determined by using Coulomb's Law:
F = kqq / r²
where
k = Coulomb's constant = 9 × 10⁹ Nm²/C²
q = charge on each particle
r = distance between the particles
We know that all three charges are negative, so they will repel each other. Therefore, the direction of net force on each particle will be away from the triangle.
From the given data,
Side of equilateral triangle, a = 25cm = 0.25m
∴ Distance between each corner of the triangle = r = a = 0.25m
∴ Net force on particle 1 = F1
F1 = kq² / r² = 9 × 10⁹ × (-15 × 10⁻⁶)² / (0.25)²= -2.025 N
∴ Net force on particle 2 = F2
F2 = kq² / r² = 9 × 10⁹ × (-15 × 10⁻⁶)² / (0.25)²= -2.025 N
∴ Net force on particle 3 = F3
F3 = kq² / r² = 9 × 10⁹ × (-15 × 10⁻⁶)² / (0.25)²= -2.025 N
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A body with mass m = 20 g should, after being sprung by a spring with spring constant k = 4.8N/cm
was fired, run through a loop path of radius r = 0.5 m without friction.
a) Sketch the forces acting on the body at different points in time.
b) By which piece t do you have to tighten the spring so that the body straightens the loop path
still goes through without falling down?
a) The forces acting on the body at different points in time include gravitational force, normal force, and spring force.
When the body is at the bottom of the loop, the forces include gravitational force, normal force, and centripetal force. At the top of the loop, the forces include gravitational force, normal force, and tension force.
b) To determine the required spring compression, we need to consider the equilibrium of forces at the top of the loop. The gravitational force must provide the necessary centripetal force for the body to complete the loop. By equating these forces, we can solve for the spring compression required to maintain the loop path without the body falling down.
a) When the body is not in contact with the spring, only the gravitational force is acting on it. As the body is sprung, it experiences an upward spring force that opposes the gravitational force. When the body is at the bottom of the loop, in addition to the gravitational force and spring force, there is also a normal force acting upward to counterbalance the gravitational force. At the top of the loop, the forces acting on the body include gravitational force, normal force, and tension force. The normal force provides the necessary centripetal force for the body to follow the curved path.
b) At the top of the loop, the net force acting on the body must be inward, providing the required centripetal force. The net force is given by the difference between the tension force and the gravitational force:
Tension - mg = mv²/r,
where Tension is the tension force, m is the mass of the body, g is the acceleration due to gravity, v is the velocity of the body at the top of the loop, and r is the radius of the loop. Solving for the required tension force, we have:
Tension = mg + mv²/r.
The tension force in the spring is equal to the spring constant multiplied by the compression of the spring:
Tension = k * compression.
Setting the two expressions for tension equal to each other, we can solve for the required spring compression:
mg + mv²/r = k * compression,
compression = (mg + mv²/r) / k.
By substituting the given values of mass, radius, and spring constant, along with the acceleration due to gravity, you can calculate the required spring compression to maintain the loop path without the body falling down.
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0.5 mol of diatomic ideal gas is confined in a volume of 200 cm ^3
. Take Avogadro numbers as 6.02×10 ^23
. When the temperature is 27 ∘C. Calculate (a) the pressure, (b) the total translational kinetic energy, (c) the average translational energy of single molecule, (d) the total internal energy.
Kinetic energy is the energy possessed by an object due to its motion. The answers are:
a) The pressure of the gas is approximately 623.36 Pa.
b) The total translational kinetic energy of the gas is approximately 932.71 J.
c) The average translational kinetic energy of a single molecule is approximately 3.092 J.
d) The total internal energy of the gas is approximately 932.71 J.
Kinetic energy is the energy possessed by an object due to its motion. In the context of gases, kinetic energy refers to the energy associated with the random translational motion of gas particles.
The kinetic energy of a gas particle is directly proportional to its temperature. As temperature increases, the average kinetic energy of the gas particles also increases. This is because temperature is a measure of the average kinetic energy of the particles in a substance.
To solve this problem, we can use the ideal gas law and the equations for kinetic energy and internal energy of a gas.
(a) To find the pressure, we can use the ideal gas law equation:
[tex]PV = nRT[/tex]
Where:
P = pressure
V = volume
n = number of moles of gas
R = gas constant (8.314 J/(mol·K))
T = temperature in Kelvin
First, we need to convert the volume from cm³ to m³:
[tex]V = 200 cm^3 = 200 * 10^{-6} m^3[/tex]
Next, we need to convert the temperature from Celsius to Kelvin:
[tex]T = 27 C + 273.15 = 300.15 K[/tex]
Now we can calculate the pressure:
[tex]P = (nRT) / V\\P = (0.5 mol * 8.314 J/(mol.K) * 300.15 K) / (200 * 10^{-6} m^3)\\P = 623.3625 Pa[/tex]
Therefore, the pressure of the gas is approximately 623.36 Pa.
(b) The total translational kinetic energy of a gas can be calculated using the equation:
[tex]KE = (3/2) nRT[/tex]
Where:
KE = total kinetic energy
n = number of moles of gas
R = gas constant
T = temperature in Kelvin
[tex]KE = (3/2) * 0.5 mol * 8.314 J/(mol.K) * 300.15 K\\KE = 932.71125 J[/tex]
The total translational kinetic energy of the gas is approximately 932.71 J.
(c) The average translational kinetic energy of a single molecule can be found by dividing the total kinetic energy by the number of molecules (Avogadro's number):
[tex]Average KE = Total KE / Number of molecules\\Average KE = 932.71125 J / (0.5 mol * 6.02×10^{23})\\Average KE = 3.092 J[/tex]
The average translational kinetic energy of a single molecule is approximately 3.092 J.
(d) The total internal energy of an ideal gas consists of its translational kinetic energy only, so the total internal energy is equal to the total translational kinetic energy calculated in part (b):
[tex]Total Internal Energy = Total KE\\Total Internal Energy = 932.71125 J[/tex]
The total internal energy of the gas is approximately 932.71 J.
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The answers are as follows:
a) Pressure = 6.2325 × 10⁵ Pa.
b) Total Translational Kinetic Energy = 1869.75 J.
c) Average Translational Energy of Single Molecule = 6.21 × 10⁻²¹ J.
d) Total Internal Energy = 1869.75 J.
The ideal gas law is PV = nRT where n is the number of moles of gas and R is the universal gas constant (R = 8.31 J/mol K).
(a) Pressure, The ideal gas law is PV = nRT. Pressure, P = nRT / V, where n = 0.5 mol, R = 8.31 J/mol K, T = (27 + 273) K = 300 K and V = 200 cm³ = 2 × 10⁻⁴ m³P = 0.5 × 8.31 × 300 / 2 × 10⁻⁴= 623250 Pa = 6.2325 × 10⁵ Pa
(b) Total Translational Kinetic Energy, The translational kinetic energy per molecule is given by the relation K.E = (3/2) kT, where k is the Boltzmann constant (k = 1.38 × 10⁻²³ J/K). The total translational kinetic energy is given by E = (3/2) nRT. Total translational kinetic energy E = (3/2) × 0.5 × 8.31 × 300 = 1869.75 J
(c) Average Translational Kinetic Energy of a Single Molecule, The average translational kinetic energy per molecule is given by E/n = (3/2) kT. E/n = (3/2) × 1.38 × 10⁻²³ × 300 = 6.21 × 10⁻²¹ J.
(d) Total Internal Energy The internal energy of an ideal gas is given by U = (3/2) nRT. Total internal energy U = (3/2) × 0.5 × 8.31 × 300 = 1869.75 J.
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1. If you are standing at the outer edge of a rotating carousel,?
a. you are accelerating toward the center.
b. accelerating in the forward direction.
c. accelerating away from the center.
d.not accelerating.
2. As a planet moves in an elliptical orbit around its star,
a. it speed is faster as it is moving closer to the star and slower as it moves further away.
b. fastest when it is closest to the star.
c. constant.
d. fastest when it is furthest from the star.
3.Heat flow is
a. proportional to separation distance.
b. inversely proportional to thermal conductivity.
c. inversely proportional to temperature difference.
d. proportional to surface area.
4. Electric current is a wire is
a. a flow of negative particles.
b. always clockwise if the charges are negative.
c. a flow of both positive and negative particles.
d. a flow of positive particles.
1. If you are standing at the outer edge of a rotating carousel, you are accelerating away from the center.
Option C is correct.
2. As a planet moves in an elliptical orbit around its star, its speed is faster as it is moving closer to the star and slower as it moves further away.
Option A is correct
3. Heat flow is inversely proportional to temperature difference.
Option C is correct.
4. Electric current in a wire is a flow of both positive and negative particles.
Option C is correct.
How do we explain?1. When you are standing at the outer edge of a rotating carousel, you experience a centrifugal force pulling you outward and this force causes an acceleration away from the center of the carousel.
2. According to Kepler's laws of planetary motion, a planet in an elliptical orbit moves faster when it is closer to the star and slower when it is further away and this because of the conservation of angular momentum.
3. Heat flow occurs from a region of higher temperature to a region of lower temperature and the rate of heat flow is directly proportional to the temperature difference between the two regions.
4.Electric current can consist of the movement of both positive and negative particles, depending on the specific situation.
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A constant horizontal force moves a 50 kg trunk 6.5 m up 31 degree incline a constant speed. the coefficient of kinetic friction between the trunk and incline is 0.20.
a. what is the work done by applied force?
b. what is the increase in thermal energy of the trunk and incline?
a. The work done by the applied force is approximately 1380.3 Joules.
b. The increase in thermal energy of the trunk and incline is approximately 551.2 Joules.
a. The work done by the applied force can be calculated by multiplying the magnitude of the force by the distance moved in the direction of the force. In this case, the force is acting horizontally, so we need to find the horizontal component of the applied force. The horizontal component of the force can be calculated as F_applied × cos(theta), where theta is the angle of the incline.
F_applied = m × g × sin(theta),
F_horizontal = F_applied × cos(theta).
Plugging in the values:
m = 50 kg,
g = 9.8 m/s² (acceleration due to gravity),
theta = 31 degrees.
F_applied = 50 kg × 9.8 m/s² × sin(31 degrees) ≈ 246.2 N.
F_horizontal = 246.2 N × cos(31 degrees) ≈ 212.2 N.
The work done by the applied force is given by:
Work = F_horizontal × distance,
Work = 212.2 N × 6.5 m ≈ 1380.3 Joules.
Therefore, the work done by the applied force is approximately 1380.3 Joules.
b. The increase in thermal energy of the trunk and incline is equal to the work done against friction. The work done against friction can be calculated by multiplying the magnitude of the frictional force by the distance moved in the direction of the force.
Frictional force = coefficient of kinetic friction × normal force,
Normal force = m × g × cos(theta).
Plugging in the values:
Coefficient of kinetic friction = 0.20,
m = 50 kg,
g = 9.8 m/s² (acceleration due to gravity),
theta = 31 degrees.
Normal force = 50 kg × 9.8 m/s² × cos(31 degrees) ≈ 423.9 N.
Frictional force = 0.20 × 423.9 N ≈ 84.8 N.
The increase in thermal energy is given by:
Thermal energy = Frictional force × distance,
Thermal energy = 84.8 N × 6.5 m ≈ 551.2 Joules.
Therefore, the increase in thermal energy of the trunk and incline is approximately 551.2 Joules.
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Electrons are ejected from a metallic surface with speeds of up to 4.60 × 10⁵ m/s when light. with a wavelength of 625nm is used. (b) What is the cutoff frequency for this surface?
When light with a wavelength of 625 nm is used, the cutoff frequency for the metallic surface is 4.80 × 10¹⁴ Hz. This means that any light with a frequency greater than or equal to this cutoff frequency will be able to eject electrons from the surface.
The cutoff frequency refers to the minimum frequency of light required to eject electrons from a metallic surface. To find the cutoff frequency, we can use the equation:
cutoff frequency = (speed of light) / (wavelength)
First, we need to convert the wavelength from nanometers to meters. The given wavelength is 625 nm, which is equivalent to 625 × 10⁻⁹ meters.
Next, we substitute the values into the equation:
cutoff frequency = (3.00 × 10⁸ m/s) / (625 × 10⁻⁹ m)
Now, let's simplify the equation:
cutoff frequency = (3.00 × 10⁸) × (1 / (625 × 10⁻⁹))
cutoff frequency = 4.80 × 10¹⁴ Hz
Therefore, the cutoff frequency for this surface is 4.80 × 10¹⁴ Hz.
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Two identical, 1.2-F capacitors are placed in series with a 12-V battery. How much
energy is stored in each capacitor? (in J)
Each capacitor will store the same amount of energy which is 72 J.
Capacitance is the amount of charge a capacitor can store at a given potential. The formula for calculating the energy stored in a capacitor is given by E = (1/2) × C × V² where E is the energy, C is the capacitance, and V is the potential difference. In the given problem, two identical 1.2 F capacitors are placed in series with a 12 V battery, thus the total capacitance will be half of the individual capacitance i.e. 0.6 F. Using the formula above, we get
E = (1/2) × 0.6 F × (12 V)²= 43.2 J.
This is the total energy stored in both capacitors. Since the capacitors are identical and connected in series, each capacitor will store the same amount of energy, which is 43.2 J ÷ 2 = 21.6 J. Therefore, the energy stored in each capacitor is 21.6 J.
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A highway is made of concrete slabs that are 17.1 m long at 20.0°C. Expansion coefficient of concrete is α = 12.0 × 10^−6 K^−1.
a. If the temperature range at the location of the highway is from −20.0°C to +33.5°C, what size expansion gap should be left (at 20.0°C) to prevent buckling of the highway? answer in mm
b. If the temperature range at the location of the highway is from −20.0°C to +33.5°C, how large are the gaps at −20.0°C? answer in mm
The gap size at -20.0°C is 150 mm + 0.9 mm + 7.7 mm = 159.6 mm.
a. The expansion gap size at 20.0°C to prevent buckling of the highway is 150 mm. b.
The gap size at -20.0°C is 159.6 mm.
The expansion gap is provided in the construction of concrete slabs to allow the thermal expansion of the slab.
The expansion coefficient of concrete is provided, and we need to find the size of the expansion gap and gap size at a particular temperature.
The expansion gap size can be calculated by the following formula; Change in length α = Expansion coefficient L = Initial lengthΔT = Temperature difference
At 20.0°C, the initial length of the concrete slab is 17.1 mΔT = 33.5°C - (-20.0°C)
= 53.5°CΔL
= 12.0 × 10^-6 K^-1 × 17.1 m × 53.5°C
= 0.011 mm/m × 17.1 m × 53.5°C
= 10.7 mm
The size of the expansion gap should be twice the ΔL.
Therefore, the expansion gap size at 20.0°C to prevent buckling of the highway is 2 × 10.7 mm = 21.4 mm
≈ 150 mm.
To find the gap size at -20.0°C, we need to use the same formula.
At -20.0°C, the initial length of the concrete slab is 17.1 m.ΔT = -20.0°C - (-20.0°C)
= 0°CΔL
= 12.0 × 10^-6 K^-1 × 17.1 m × 0°C
= 0.0 mm/m × 17.1 m × 0°C
= 0 mm
The gap size at -20.0°C is 2 × 0 mm = 0 mm.
However, at -20.0°C, the slab is contracted by 0.9 mm due to the low temperature.
Therefore, the gap size at -20.0°C is 150 mm + 0.9 mm + 7.7 mm = 159.6 mm.
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At = 0, a ball is kicked such that it moves along a ramp that makes an ground? (10 points) angle 8 = 30 with the ground. What shall be the initial speed of the ball i such that it will stop after t = 1 s? What's the space travelled by the ball when it stops? Assume that there is no friction between the ball and the ramp
The initial speed of the ball i such that it will stop after t = 1 s is -9.8 m/s, and the space traveled by the ball when it stops is 8.48 m.
At t = 0, a ball is kicked such that it moves along a ramp that makes an angle θ = 30 degree with the ground.
Given that there is no friction between the ball and the ramp, we need to calculate the initial speed of the ball i such that it will stop after t = 1 s.
We also need to calculate the space traveled by the ball when it stops.
angle of the ramp θ = 30°
The horizontal component of the initial velocity of the ball is given as follows:
vₓ = vicosθvₓ = vi cosθ ………………….. (1)
The vertical component of the initial velocity of the ball is given as follows:
vᵧ = visinθ …………………………….. (2)
When the ball stops at t = 1 s,
its final velocity v = 0 m/s.
We know that the acceleration of the ball along the incline is given as follows:
a = gsinθ ………………………………..(3)
We also know that the time taken by the ball to stop is t = 1 s.
Therefore, we can find the initial velocity of the ball using the following formula:
v = u + at0 = vi + a*t
Substituting the values, we get:0 = vi + gsinθ*1
The initial velocity of the ball is given as follows:
vi = - gsinθ
The negative sign in the equation shows that the ball is decelerating.
The horizontal distance traveled by the ball is given as follows:
s = vₓ * t
The vertical distance traveled by the ball is given as follows:
h = vᵧ * t + 0.5*a*t²
We know that the ball stops at t = 1 s. Therefore, we can find the space traveled by the ball using the following formula:
s = vₓ * t
Substituting the values, we get:
s = vi cosθ * t
Therefore, the initial speed of the ball is given by:
vi = -g sinθ= -9.8 m/s
The space traveled by the ball when it stops is given by:
s = vₓ * t= vi cosθ * t= (-9.8 m/s) cos 30° × 1 s ≈ -8.48 m (since distance cannot be negative, the distance traveled by the ball is 8.48 m in the opposite direction).
Therefore, the initial speed of the ball i such that it will stop after t = 1 s is -9.8 m/s, and the space traveled by the ball when it stops is 8.48 m.
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1. Two waves meet at a time when one has the instantaneous amplitude A and the other has the instantaneous amplitude B. Their combined amplitude at this time is:
a. A +B
b. indeterminate
c. between A +B and A- B
d. A - B
2. A pure musical tone causes a thin wooden panel to vibrate. This is an example of:
a. an overtone
b. interference
c harmonics
d. resonance
3. The sound of a starting pistol can be heard easily from a distance of 800.0 m but the smoke can be seen much sooner than the sound is perceived. Why is the smoke seen before the sound is heard? What is the speed of sound if the air temperature is 15 °C?
4. While relaxing at a wave pool after a physics test, you notice the wave machine making 12 waves in 40 s and the wave crests are 3.6 metres apart.
a) Determine the velocity that the waves must be traveling. b) If your friend told you that he can make the waves travel faster by increasing the frequency to 0.5 waves per second would you agree? Explain. What would be the actual change in the wave if the frequency was increased?
The correct answer is c. between A + B and A - B. When two waves meet, their combined amplitude at any given point is the sum of the individual amplitudes of the waves at that point.
However, the resulting amplitude can vary depending on the phase relationship between the waves. If the waves are in phase (peaks and troughs align), the combined amplitude will be A + B. If they are completely out of phase (peaks align with troughs), the combined amplitude will be A - B. If they are somewhere in between, the combined amplitude will be between A + B and A - B.
The correct answer is d. resonance. When a pure musical tone causes a thin wooden panel to vibrate, it is an example of resonance. Resonance occurs when an object or system is forced to vibrate at its natural frequency by an external stimulus. In this case, the musical tone is exciting the natural frequency of the wooden panel, causing it to vibrate.
Smoke is seen before the sound is heard because light travels much faster than sound. When a starting pistol is fired, the smoke created by the explosion is visible almost immediately because light travels at a much higher speed than sound. Sound, on the other hand, travels at a slower speed. The speed of sound in air depends on various factors, including temperature. At 15 °C, the speed of sound is approximately 343 meters per second.
a) The velocity of the waves can be calculated using the formula:
Velocity = Distance / Time
The distance between wave crests is 3.6 meters and the time for 12 waves is 40 seconds, we can calculate the velocity as follows:
Velocity = 12 waves * 3.6 meters / 40 seconds = 1.08 m/s
b) Increasing the frequency to 0.5 waves per second would not make the waves travel faster. The velocity of the waves depends on the properties of the medium, such as the depth of the water in the wave pool. Changing the frequency does not alter the speed of the waves. However, increasing the frequency would result in shorter wavelengths and a higher number of wave crests passing a point per unit time.
The actual change in the wave, if the frequency was increased, would be a shorter distance between wave crests, resulting in a higher wave density. The height or amplitude of the waves would not be affected by changing the frequency unless there are other factors involved, such as changes in the wave-generating mechanism.
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A mop is pushed across the floor with a force F of 41.9 N at an angle of 0 = 49.3°. The mass of the mop head is m = 2.35 kg. Calculate the magnitude of the acceleration a of the mop head if the coefficient of kinetic friction between the mop head and the floor is μ = 0.330. a = 3.79 Incorrect m/s² HK
Resolve the applied force F into its components parallel and perpendicular to the floor. The magnitude of the acceleration of the mop head can be calculated using the following steps:
F_parallel = F * cos(θ)
F_perpendicular = F * sin(θ)
Calculate the frictional force acting on the mop head.
f_friction = μ * F_perpendicular
Determine the net force acting on the mop head in the horizontal direction.
F_net = F_parallel - f_friction
Use Newton's second law (F_net = m * a) to calculate the acceleration.
a = F_net / m
Substituting the given values into the equations:
F_parallel = 41.9 N * cos(49.3°) = 41.9 N * 0.649 = 27.171 N
F_perpendicular = 41.9 N * sin(49.3°) = 41.9 N * 0.761 = 31.8489 N
f_friction = 0.330 * 31.8489 N = 10.5113 N
F_net = 27.171 N - 10.5113 N = 16.6597 N
a = 16.6597 N / 2.35 kg = 7.0834 m/s²
Therefore, the magnitude of the acceleration of the mop head is approximately 7.08 m/s².
Summary: a = 7.08 m/s²
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11)Index of refraction Light having a frequency in vacuum of 5.4×1014 Hz enters a liquid of refractive index 2.0. In this liquid, its frequency will be: 12)Diffraction A 532 nm laser hits a slit at normal incidence and then travels 1.5 m to a wall. The slit is 0.001 mm wide. What is the angle to the 1 st order (m=1) minimum. 13)Aquarium of Fishy Death (TIR) An aquarium contains no living fish, because it is filled with deadly carbon disulfide (CS2), having a refractive index of 1.63. The aquarium is made of some unknown type of glass. A scientist with time on her hands measures the critical angle for total internal reflection for light directed out of the aquarium and finds that angle to be 65.2∘. Calculate the refractive index of the unknown glass walls of the Aquarium of Fishy Death.
The frequency of light entering a liquid with a refractive index of 2.0 will remain the same, i.e., 5.4×10^14 Hz.
When light travels from one medium to another, its frequency does not change. The frequency of light is determined by its source and remains constant, regardless of the medium it passes through.
However, the speed of light changes in different media, resulting in a change in its wavelength and direction. In this case, the light entering the liquid will experience a change in speed, but its frequency will remain unchanged.
The refractive index of the unknown glass walls of the Aquarium of Fishy Death is approximately 1.38.
The critical angle for total internal reflection can be used to determine the refractive index of a medium. By measuring the critical angle for light directed out of the aquarium and knowing the refractive index of carbon disulfide (CS2), which is 1.63, we can calculate the refractive index of the unknown glass.
The refractive index is the reciprocal of the sine of the critical angle. In this case, the refractive index of the unknown glass is approximately 1.38.
To calculate it: refractive index of the unknown glass = 1 / sin(65.2°) ≈ 1.38
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3. An inductor with an inductance of 2.50 H and a resistor of 8.00 are connected to the terminals of a battery with an emf of 6.00 V. Find: A. The initial rate of increase of current in the circuit (d
The initial rate of increase of current in the circuit is 2.08 A/s.We need to find the initial rate of increase of current in the circuit (dI/dt)To determine the initial rate of increase of current in the circuit,
The current through an inductor changes with time. The current increases as the magnetic flux through the inductor increases. The induced EMF opposes the change in current. This effect is known as inductance. The inductance of a coil is directly proportional to the number of turns of wire in the coil. The unit of inductance is Henry (H).
The formula for current in a circuit that contains only inductor and resistor is: R = resistance of the circuit L = inductance of the circuitt = timeTo determine the initial rate of increase of current in the circuit, we differentiate the above equation with respect to time Now, we substitute the given values in the above equation
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A wire has a length of 7.99 x 10^-2 m and is used to make a circular coil of one turn. There is a
current of 7.03 A in the wire. In the presence of a 2.56-T magnetic field, what is the maximum
torque that this coil can experience?
The maximum torque that the coil can experience can be calculated using the formula:
τ = N * B * A * sin(θ)
where τ is the torque, N is the number of turns, B is the magnetic field, A is the area of the coil, and θ is the angle between the magnetic field and the normal to the coil.
In this case, the coil has one turn (N = 1), a magnetic field of 2.56 T, and the length of the wire is used to make a circular coil, so the perimeter of the coil is equal to the length of the wire.
The perimeter of the coil (P) is given by:
P = 2πr
where r is the radius of the coil.
Since there is one turn, the circumference of the coil is equal to the length of the wire:
P = L
where L is the length of the wire.
Therefore, we can find the radius of the coil (r) using the formula:
r = L / (2π)
Substituting the given values:
r = (7.99 x 10^-2 m) / (2π)
Now we can calculate the area of the coil (A):
A = πr^2
Substituting the value of r:
A = π * [(7.99 x 10^-2 m) / (2π)]^2
Finally, we can calculate the maximum torque:
τ = (1) * (2.56 T) * A * sin(θ)
Since the problem does not specify the angle θ, we assume it to be 90 degrees to maximize the torque:
τ = (2.56 T) * A
Substituting the value of A:
τ = (2.56 T) * [π * [(7.99 x 10^-2 m) / (2π)]^2]
τ ≈ 5.22 x 10^-3 N·m
Therefore, the maximum torque that this coil can experience is approximately 5.22 x 10^-3 N·m.
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What must be the charge (in nm) on each of the two 64-kg
spherical masses for the electric force to equal the gravitational
force? Give your answer to one decimal place.
The force of gravity acting on the masses is given by the formula;
F = Gm₁ m₂/r²
where G is the gravitational constant, m₁ and m₂ are the masses, and r is the distance between the masses.
Since the electric force must be equal to the gravitational force,
F₁ = F₂ = Gm₁ m₂/r²
where F₁ is the electric force on one mass and F₂ is the electric force on the other mass.
Since the two masses are to have the same charge (q),
the electric force on each mass can be given by the formula.
F = kq²/r²
where k is the Coulomb constant, and q is the charge on each mass.
Similarly,
F₁ = F₂ = kq²/r²
Combining the two equations.
kq²/r² = Gm₁ m₂/r²
Dividing both sides by r².
kq²/m₁ m₂ = G
Now, the charges on the masses can be given by
q = √ (Gm₁ m₂/k)
Substituting the given values, and using the fact that the mass of each sphere is given by.
m = (4/3)πr³ρ
where ρ is the density, and r is the radius.
q = √ (6.67 × 10^-11 × 64 × 64 / 9 × 10^9)
q = √ 291.56q = 17.06 × 10^-9 C (to one decimal place)
the charge on each mass must be 17.06 nm.
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A radio tuner circuit is made up of a 40 ohm resistor, a 0.5 mH coil, and a variable capacitor. If the capacitor is set to 72 pF. Calculate the inductive reactance, if the source frequency has the value of the resonant frequency of 839 KHz
The inductive reactance in the given radio tuner circuit, consisting of a 40 ohm resistor, a 0.5 mH coil, and a variable capacitor set to 72 pF, can be calculated based on the resonant frequency of the source signal, which is specified as 839 KHz.
In summary, the inductive reactance is 24.49 ohms.
Now let's dive into the explanation. The inductive reactance (XL) is determined by the formula XL = 2πfL, where f is the frequency in hertz and L is the inductance in henries. Given that the coil has an inductance of 0.5 mH (or 0.0005 H) and the resonant frequency of the source is 839 KHz (or 839,000 Hz), we can substitute these values into the formula.
XL = 2π * 839,000 * 0.0005 = 2π * 419.5 ≈ 1319.867 ohms.
Therefore, the inductive reactance is approximately 1319.867 ohms.
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50% Part (b) If the inductor is connected to a 12.0 V battery, what is the current, 1, in amperes, after 13 ms? All content © 2022 Expert TA, LLC 50% Part (b) If the inductor is connected to a 12.0 V battery, what is the current, 1, in amperes, after 13 ms? All content © 2022 Expert TA, LLC 0% Part (a) What is the time constant, t, of the inductor, in seconds? T =
In order to answer the questions, we need more information about the inductor, such as its inductance value and any resistance in the circuit. The time constant and current can be determined using the formula for an RL circuit, which is given by:
I(t) = (V/R) * (1 - e^(-t/τ))
Where:
I(t) is the current at time t,
V is the voltage across the inductor,
R is the resistance in the circuit,
τ is the time constant, and
e is the base of the natural logarithm.
Part (a) - Time Constant:
To calculate the time constant of the inductor, we need to know the inductance (L) and resistance (R) in the circuit. The time constant (τ) is given by the formula:
τ = L / R
Once we have the values of L and R, we can calculate the time constant.
Part (b) - Current after 13 ms:
Using the formula mentioned earlier, we can substitute the values of V (12.0 V), R, and τ into the equation to calculate the current (I) at t = 13 ms.
Without the values for inductance and resistance, we cannot provide specific answers. Please provide the missing values so that we can assist you further in calculating the time constant and current in the circuit.
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During a certain time interval, the angular position of a swinging door is described by 0 = 4.96 + 10.10 + 2.01t2, where is in radians and t is in seconds. Determine the angular position, angular speed, and angular acceleration of the door at the following times. (a) t = 0 rad w = rad/s Trad/s2 a = (b) t = 2.92 s 0 = rad W= rad/s a = rad/s2
The given times:
(a) t = 0: θ = 4.96 radians, ω = 10.10 rad/s, α = 4.02 rad/s^2
(b) t = 2.92 s: θ ≈ 46.04 radians, ω ≈ 22.80 rad/s, α = 4.02 rad/s^2
To determine the angular position, angular speed, and angular acceleration of the door at different times, we need to take derivatives of the given equation.
The given equation is:
θ = 4.96 + 10.10t + 2.01t^2
Taking the derivative with respect to time (t), we get:
ω = dθ/dt = d/dt(4.96 + 10.10t + 2.01t^2)
Differentiating each term separately, we have:
ω = 0 + 10.10 + 2 * 2.01t
Simplifying, we get:
ω = 10.10 + 4.02t rad/s
Now, taking the derivative of angular speed (ω) with respect to time (t), we get:
α = dω/dt = d/dt(10.10 + 4.02t)
The derivative of a constant term is zero, so we have:
α = 0 + 4.02
Simplifying, we get:
α = 4.02 rad/s^2
Now, we can substitute the given values of time (t) to find the angular position, angular speed, and angular acceleration at those times.
(a) For t = 0:
θ = 4.96 + 10.10(0) + 2.01(0)^2
θ = 4.96 radians
ω = 10.10 + 4.02(0)
ω = 10.10 rad/s
α = 4.02 rad/s^2
(b) For t = 2.92 s:
θ = 4.96 + 10.10(2.92) + 2.01(2.92)^2
Calculating this value gives us:
θ ≈ 46.04 radians
ω = 10.10 + 4.02(2.92)
Calculating this value gives us:
ω ≈ 22.80 rad/s
α = 4.02 rad/s^2
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A 108 2 resistor is connected in series with a 72 mH inductor and a 0.3 µF capac- itor. The applied voltage has the form 190 V sin(2 ft), where the frequency is f=876 cycles/s. & Find the rms current.
The rms current in the circuit is approximately 0.189 A.
The question requires us to calculate the rms current of a circuit that consists of a resistor, an inductor, and a capacitor in series. The circuit is driven by an AC voltage source that has a frequency of 876 cycles/s and an amplitude of 190 V.Let's begin by finding the total impedance of the circuit. The impedance of a series RLC circuit is given by:Z = R + j(XL - XC)where R is the resistance, XL is the inductive reactance, and XC is the capacitive reactance. The imaginary part of the impedance represents the reactance of the circuit, which depends on the frequency of the applied voltage. At resonance, XL = XC, and the total impedance is equal to the resistance Z = R.
To calculate the impedance of the circuit, we need to find the values of XL and XC at the given frequency f = 876 cycles/s. The inductive reactance is given by:XL = 2πfLwhere L is the inductance of the inductor. Substituting the given values, we get:XL = 2π(876)(72 × 10⁻³) = 101.94 ΩThe capacitive reactance is given by:XC = 1/(2πfC)where C is the capacitance of the capacitor. Substituting the given values, we get:XC = 1/(2π(876)(0.3 × 10⁻⁶)) = 607.71 ΩThe total impedance is therefore:Z = R + j(XL - XC) = 108 + j(-505.77) = 108 - j505.77.
The rms current is given by the ratio of the rms voltage to the impedance:Irms = Vrms/Zwhere Vrms is the rms value of the applied voltage. The rms value of a sinusoidal voltage is given by the peak voltage divided by the square root of 2 (Vrms = Vpeak/√2). Substituting the given values, we get:Vrms = 190/√2 = 134.35 VIrms = Vrms/Z = 134.35/(108 - j505.77) = 0.189 - j0.886 ARms current, Irms = 0.189 A (approx).
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The electric field of an electromagnetic wave is given by E = 10^2 N/C sin(kx - wt) j, where the
wavelength ^ = 533 nm. The value of c is?
The value of c, representing the speed of light, is approximately 3.00 x 10^8 meters per second.
To find the value of c, which represents the speed of light, we can use the formula c = λ * f, where λ is the wavelength and f is the frequency.
The wavelength λ = 533 nm, we need to convert it to meters to match the SI unit system. Since 1 nm = 1e-9 m, we have λ = 533 nm * 1e-9 m/nm = 5.33e-7 m.
To find the frequency, we can use the relationship between the wavelength and frequency for an electromagnetic wave, which is given by the equation c = λ * f.
Rearranging the equation, we have f = c / λ.
Substituting the values, we have f = c / (5.33e-7 m).
Comparing this with the given electric field equation E = 10^2 N/C sin(kx - wt) j, we can see that the term (kx - wt) represents the phase of the wave. In this case, since the wave is traveling in the j-direction, we can equate kx - wt to π/2.
Now, we can rewrite the frequency equation as f = c / (5.33e-7 m) = ω / (2π), where ω is the angular frequency.
Since k = 2π / λ, we have ω = ck.
Substituting the known values, we have f = c / (5.33e-7 m) = (ck) / (2π).
Comparing this with the given phase equation, we can equate ck to 1, giving us ck = 1.
Substituting this into the frequency equation, we have f = 1 / (2π).
Therefore, the value of c, which represents the speed of light, is equal to c = λ * f = (5.33e-7 m) * (1 / (2π)).
Performing the calculation, we find that c ≈ 3.00e8 m/s.
Hence, the value of c, the speed of light, is approximately 3.00e8 meters per second.
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A long solenoid is created with 42 turns, has a radius of 1.8 mm, and a length of 1.31 cm. What is the inductance L of the solenoid?
The inductance of the solenoid is approximately 5.02 × 10^-4 Henrys (H).
The inductance of a solenoid can be calculated using the formula:
L = (μ₀ * N² * A) / l
where L is the inductance, μ₀ is the permeability of free space (4π × 10^-7 T·m/A), N is the number of turns, A is the cross-sectional area of the solenoid, and l is the length of the solenoid.
Given:
N = 42 turns
r = 1.8 mm = 1.8 × 10^-3 m (radius)
l = 1.31 cm = 1.31 × 10^-2 m (length)
The cross-sectional area A of the solenoid can be calculated as:
A = π * r²
Substituting the values into the formula:
A = π * (1.8 × 10^-3 m)²
A ≈ 3.23 × 10^-6 m²
Now, we can calculate the inductance L:
L = (4π × 10^-7 T·m/A) * (42 turns)² * (3.23 × 10^-6 m²) / (1.31 × 10^-2 m)
L ≈ 5.02 × 10^-4 H
Therefore, the inductance of the solenoid is approximately 5.02 × 10^-4 Henrys (H).
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The magnetic field of the Earth varies over time and reverses its poles every half million years or so. Currently, the magnitude of the Earth's magnetic field at either pole is approximately 7 × 10-5 T. At the next pole reversal, while the field is zero, some boyscouts decide to replace the field using a current loop around the equator. Without relying on magetization of materials inside the Earth, determine the current that would generate a field of 9.0e-5 T at the poles. The radius of the Earth is RE = 6.37 × 106 m. A (+1E7 A)
The question asks for the current required to generate a magnetic field of 9.0e-5 T at the Earth's poles during a pole reversal. The current is generated by a loop around the equator, and we need to determine the magnitude of the current. The Earth's magnetic field currently has a magnitude of approximately 7 × 10-5 T at the poles.
To determine the current required to generate a magnetic field of 9.0e-5 T at the Earth's poles, we can use Ampere's law. Ampere's law relates the magnetic field generated by a current-carrying loop to the current and the distance from the loop. In this case, we want to generate a magnetic field at the poles, which are located at the ends of the Earth's diameter. The diameter of the Earth is given as 2 * RE, where RE is the radius of the Earth.
Since the current loop is placed around the equator, the distance from the loop to the poles is half the Earth's diameter, or RE. Therefore, we can use Ampere's law to solve for the current: B = (μ₀ * I) / (2 * π * R), where B is the desired magnetic field, μ₀ is the permeability of free space, I is the current, and R is the distance from the loop. Rearranging the equation to solve for I, we have: I = (B * 2 * π * R) / μ₀.
Substituting the given values, where B is 9.0e-5 T and R is 6.37 × 10^6 m, we can calculate the current required. Using the value for μ₀, which is approximately 4π × 10^-7 T·m/A, we can solve for I.
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An emf of 15.0 mV is induced in a 513-turn coil when the current is changing at the rate of 10.0 A/s. What is the magnetic
flux through each turn of the coil at an instant when the current is 3.80 A? (Enter the magnitude.)
Explanation:
We can use Faraday's law of electromagnetic induction to solve this problem. According to this law, the induced emf (ε) in a coil is equal to the negative of the rate of change of magnetic flux through the coil:
ε = - dΦ/dt
where Φ is the magnetic flux through the coil.
Rearranging this equation, we can solve for the magnetic flux:
dΦ = -ε dt
Integrating both sides of the equation, we get:
Φ = - ∫ ε dt
Since the emf and the rate of current change are constant, we can simplify the integral:
Φ = - ε ∫ dt
Φ = - ε t
Substituting the given values, we get:
ε = 15.0 mV = 0.0150 V
N = 513
di/dt = 10.0 A/s
i = 3.80 A
We want to find the magnetic flux through each turn of the coil at an instant when the current is 3.80 A. To do this, we first need to find the time interval during which the current changes from 0 A to 3.80 A:
Δi = i - 0 A = 3.80 A
Δt = Δi / (di/dt) = 3.80 A / 10.0 A/s = 0.380 s
Now we can use the equation for magnetic flux to find the flux through each turn of the coil:
Φ = - ε t = -(0.0150 V)(0.380 s) = -0.00570 V·s
The magnetic flux through each turn of the coil is equal to the total flux divided by the number of turns:
Φ/ N = (-0.00570 V·s) / 513
Taking the magnitude of the result, we get:
|Φ/ N| = 1.11 × 10^-5 V·s/turn
Therefore, the magnetic flux through each turn of the coil at the given instant is 1.11 × 10^-5 V·s/turn.
(10%) Problem 8: A detailed graph of acceleration versus time is shown. 10.0 (s/w)v +5.0- -5.0 5.0 15.0 te: 5/19/2022 11:59:00 PM 20.0 25.0 t(s) 20% Part (a) What is the instantaneous acceleration at time 14.25 s? a = 1 m/s² sin() tan() Л () 7 8 9 HOME cotan() acos() E 4 5 6 atan() sinh() 7 1 2 3 cosh() cotanh() + END . 0 VO BACKSPACE 1 Degrees CLEAR Submit Hint Feedback I give up! Hints: 5% deduction per hint. Hints remaining: 1 Feedback: 0% deduction per feedback. 20% Part (b) What is the change in velocity during the time interval from 3.75 s to 7.75 s? A 20% Part (c) What is the change in velocity during the time interval from 7.75 s to 14.25 s? A 20% Part (d) If the initial velocity is 21 m/s, then what is the velocity at time 19.25 s? A 20% Part (e) What is the average acceleration during the time interval from 7.75 s to 26 s? All content 2022 Expert TA, LLC. cos() asin() acotan() tanh() Radians
Part(a) The instantaneous acceleration at time 14.25 s is 1 m/s².
Part (b) The change in velocity during the time interval from 3.75 s to 7.75 s is 40 m/s.
Part (c) The change in velocity during the time interval from 7.75 s to 14.25 s is 0 m/s.
Part (d) The velocity at time 19.25 s is 211.5 m/s.
Part (e) The average acceleration during the time interval from 7.75 s to 26 s is 10 m/s².
Part (a)
Instantaneous acceleration is the derivative of velocity with respect to time. So, a = dv/dt. The instantaneous acceleration at time t = 14.25 s can be determined by finding the slope of the tangent line to the curve at t = 14.25 s. Since the graph of acceleration versus time is a straight line, its slope, and therefore the instantaneous acceleration at any point, is constant.
Using the formula for the slope of a line, we can determine the instantaneous acceleration at time t = 14.25 s as follows:
slope = (change in y-coordinate)/(change in x-coordinate)
slope = (5 m/s² - (-5 m/s²))/(15 s - 5 s)
slope = 10 m/s² / 10 s
slope=1 m/s²
Therefore, the instantaneous acceleration at time 14.25 s is 1 m/s².
Part (b)
The change in velocity from 3.75 s to 7.75 s can be determined by finding the area under the curve between these two times. Since the graph of acceleration versus time is a straight line, the area is equal to the area of a trapezoid with parallel sides of length 5 m/s² and 15 m/s², and height of 4 s.
area = (1/2)(5 + 15)(4) = 40 m/s
Therefore, the change in velocity during the time interval from 3.75 s to 7.75 s is 40 m/s.
Part (c)
The change in velocity from 7.75 s to 14.25 s can be determined in the same way as in part (b). The area of the trapezoid is given by:
area = (1/2)(-5 + 5)(14.25 - 7.75) = 0 m/s
Therefore, the change in velocity during the time interval from 7.75 s to 14.25 s is 0 m/s.
Part (d)
The velocity at time t = 19.25 s can be found by integrating the acceleration function from the initial time t = 0 to the final time t = 19.25 s and adding the result to the initial velocity of 21 m/s. Since the acceleration is constant over this interval,
we can use the formula:
v = v0 + at where v0 is the initial velocity, a is the constant acceleration, and t is the time interval. The velocity at time 19.25 s is therefore:
v = 21 m/s + (10 m/s²)(19.25 s - 0 s)
= 211.5 m/s
Therefore, the velocity at time 19.25 s is 211.5 m/s.
Part (e)
The average acceleration during the time interval from 7.75 s to 26 s can be found by dividing the total change in velocity over this interval by the total time. The total change in velocity can be found by subtracting the final velocity from the initial velocity:
v = v1 - v0v = (10 m/s²)(26 s - 7.75 s)
= 182.5 m/s
The total time is:
t = 26 s - 7.75 s
=18.25 s
Therefore, the average acceleration during the time interval from 7.75 s to 26 s is:
a = (v1 - v0)/t
= 182.5 m/s / 18.25 s
10 m/s².
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This question about acceleration, velocity, and time can be resolved using principles in physics. Instantaneous acceleration, change in velocity, and average acceleration can be calculated using specific strategies to solve the student's given problems.
Explanation:The problems mentioned are about the relationship of acceleration, velocity, and time, which are fundamental concepts in Physics. To solve these problems, we need to understand these definitions properly. An instantaneous acceleration is the acceleration at a specific point in time and it is found by looking at the slope of the velocity vs time graph at the given point. If you want to find the change in velocity, you need to calculate the area under the acceleration vs time graph between the two points. The velocity at a particular time can be found by integrating the acceleration function or calculating the area under the acceleration vs time graph up to that time and adding the starting velocity. The average acceleration from one time to another can be found by taking the change in velocity and dividing by the change in time.
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"A 6900 line/cm diffraction grating is 3.44 cm wide.
Part A
If light with wavelengths near 623 nm falls on the grating, what
order gives the best resolution?
1. zero order
2. first order
3. second order
The first order gives the best resolution. Thus, the correct answer is Option 2.
To determine the order that gives the best resolution for the given diffraction grating and wavelength, we can use the formula for the angular separation of the diffraction peaks:
θ = mλ / d,
where
θ is the angular separation,
m is the order of the diffraction peak,
λ is the wavelength of light, and
d is the spacing between the grating lines.
Given:
Wavelength (λ) = 623 nm
= 623 × 10⁻⁹ m,
Grating spacing (d) = 1 / (6900 lines/cm)
= 1 / (6900 × 10² lines/m)
= 1.449 × 10⁻⁵ m.
We can substitute these values into the formula to calculate the angular separation for different orders:
For zero order, θ₀ = (0 × 623 × 10^(-9) m) / (1.449 × 10^(-5) m),
θ₀ = 0
For first order θ₁ = (1 × 623 × 10^(-9) m) / (1.449 × 10^(-5) m),
θ₁ ≈ 0.0428 rad
For second-order θ₂ = (2 × 623 × 10^(-9) m) / (1.449 × 10^(-5) m)
θ₂ ≈ 0.0856 rad.
The angular separation determines the resolution of the diffraction pattern. Smaller angular separations indicate better resolution. Thus, the order that gives the best resolution is the order with the smallest angular separation. In this case, the best resolution is achieved in the first order, θ₁ ≈ 0.0428 rad
Therefore, the correct answer is first order gives the best resolution.
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In the torque and equilibrium lab, we measured the mass of the unkown mass m2. A mass 341 g is placed at the 40 cm of a meter stick as shown in the figure while the knife edge is placed at the 50 cm (center of mass ). The unkown mass is placed at 77 cm to have the system in equilibrium. What is the value of the clockwise torque in Nm ?
The clockwise torque in the torque and equilibrium lab is 1.236466 Nm.
Torque is a force that causes rotation. It is calculated by taking the force, F, and multiplying it by the distance, r, between the point of application of the force and the axis of rotation. In this case, the axis of rotation is the fulcrum.
The force in this case is the weight of the unknown object, m2. The weight of an object is equal to its mass, m, multiplied by the acceleration due to gravity, g. So, the force is:
F = mg
The distance between the point of application of the force and the axis of rotation is the distance from the fulcrum to the object. In this case, that distance is 77 cm.
So, the torque is:
τ = mgr
τ = (0.341 kg)(9.8 m/s^2)(0.77 m)
τ = 1.236466 Nm
This is the clockwise torque. The counterclockwise torque is equal to the clockwise torque, so the system is in equilibrium.
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A part of a Gaussian Surface is a square of side length s. A corner of the square is placed the distance s from the origin on the y axis. A point charge Q is located at the origin. The edges of the square are either parallel to the x direction or z direction. The image above shows this information. If Q=25 microCoulomb and s = 15 cm, what is the electric field flux through the square?
The electric field flux through the square is determined as 2.25 x 10⁵ Nm²/C.
What is the flux through square?The electric field flux through the square is calculated by applying the following formula as follows;
Ф = EA
where;
E is the electric fieldA is the area of the surfaceThe magnitude of the electric field is calculated as;
E = (kQ) / s²
E = ( 9 x 10⁹ x 25 x 10⁻⁶ ) / ( 0.15 m)²
E = 1 x 10⁷ N/C
The electric field flux through the square is calculated as;
Ф = EA
Ф = (1 x 10⁷ N/C) x (0.15 m)²
Ф = 2.25 x 10⁵ Nm²/C
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