When an alarm emitting a 200 Hz frequency noise with a wavelength of 1.5 m is moving rapidly towards an observer, the observed frequency would be approximately 100 Hz, and the observed wavelength would be approximately 0.75 m. Therefore, the most likely frequency and wavelength to be observed are :
(A) 100 Hz, 0.75 m'
'
Source frequency (f) = 200 Hz
Source wavelength (λ) = 1.5 m
To begin, we need to determine the velocity of the wave. We can use the formula v = fλ, where v is the velocity of the wave, f is the frequency, and λ is the wavelength.
Using the given values:
v = 200 Hz * 1.5 m
v = 300 m/s
Now, considering the Doppler effect, when the alarm is moving towards the observer, the frequency of the observed wave changes. The observed frequency (f') can be calculated using the formula:
f' = f * (v + v_r) / (v + v_s)
Where f' is the frequency of the observed wave, f is the frequency of the source wave, v is the velocity of sound, v_r is the velocity of the receiver (observer), and v_s is the velocity of the source (alarm).
In this scenario, the observer is stationary (v_r = 0) and the alarm is moving towards the observer (v_s < 0), so the formula simplifies to:
f' = f * (v - v_s) / v
Substituting the values:
f' = 200 Hz * (300 m/s - (-v_s)) / 300 m/s
f' = 200 Hz * (300 m/s + v_s) / 300 m/s
f' = 200 Hz * (1 + (v_s / 300)) ----(1)
Since the alarm is moving towards the observer rapidly, we can assume that the velocity of the alarm (v_s) is very small compared to the velocity of sound (v). Therefore, we can neglect the term v_s / 300 in equation (1), resulting in:
f' ≈ 200 Hz
So, the observed frequency is approximately 200 Hz.
Now, let's calculate the observed wavelength (λ') using the formula:
λ' = λ * (v - v_r) / v
Substituting the values:
λ' = 1.5 m * (300 m/s - 0) / 300 m/s
λ' = 1.5 m
Therefore, the observed wavelength remains the same as the source wavelength, which is 1.5 m.
In summary, if an alarm emitting a 200 Hz frequency noise with a wavelength of 1.5 m is moving rapidly towards the observer, the observed frequency would be approximately 200 Hz, and the observed wavelength would remain unchanged at 1.5 m. Thus, the correct answer is A. 100 Hz, 0.75 m.
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Two empty soda cans are placed upright on a frictionless table, where the cans are separated by several centimetres. Predict the motions of the cans when you blow air through
the gap between the cans using a straw. Explain this in 80 words.
When air is blown through the gap between the two upright soda cans using a straw, the cans will move away from each other. This is due to the principle of action and reaction.
The air blown through the gap creates a stream of fast-moving air molecules that exert a force on the inner surfaces of the cans. According to Newton's third law of motion, the cans will experience an equal and opposite force, causing them to move in opposite directions away from each other.
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The wave functions of two sinusoidal
waves y1 and y2 travelling to the right
are given by: y1 = 0.02 sin 0.5mx - 10ttt)
and y2 = 0.02 sin(0.5mx - 10mt + T/3), where x and y are in meters and t is in seconds. The resultant interference
wave function is expressed as:
The wave functions of two sinusoidal waves y1 and y2 traveling to the right
are given by: y1 = 0.02 sin 0.5mx - 10ttt) and y2 = 0.02 sin(0.5mx - 10mt + T/3), where x and y are in meters and t is in seconds. the resultant interference wave function is given by:y = 0.02 sin(0.5mx - 10tt + T/3) + 0.02 sin(0.5mx - 10mt)
To determine the resultant interference wave function, we can add the two given wave functions, y1 and y2.
The given wave functions are:
y1 = 0.02 sin(0.5mx - 10tt)
y2 = 0.02 sin(0.5mx - 10mt + T/3)
To find the resultant interference wave function, we add y1 and y2:
y = y1 + y2
= 0.02 sin(0.5mx - 10tt) + 0.02 sin(0.5mx - 10mt + T/3)
Using the trigonometric identity sin(a + b) = sin(a)cos(b) + cos(a)sin(b), we can rewrite the resultant wave function:
y = 0.02 [sin(0.5mx - 10tt)cos(T/3) + cos(0.5mx - 10tt)sin(T/3)] + 0.02 sin(0.5mx - 10mt
Simplifying further, we have:
y = 0.02 [sin(0.5mx - 10tt + T/3)] + 0.02 sin(0.5mx - 10mt)
Therefore, the resultant interference wave function is given by:
y = 0.02 sin(0.5mx - 10tt + T/3) + 0.02 sin(0.5mx - 10mt)
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At what separation is the electrostatic force between a +14 uC point charge and a +54 uC point charge equal in magnitude to 3.1 N? (In m)
The separation at which the electrostatic force between a +14 uC point charge and a +54 uC point charge is equal in magnitude to 3.1 N is approximately 0.32 meters.
To calculate this, we can use Coulomb's law, which states that the electrostatic force between two point charges is proportional to the product of their charges and inversely proportional to the square of the distance between them.Mathematically, Coulomb's law can be expressed as: F = k * |q1 * q2| / r^2 where F is the electrostatic force, k is the electrostatic constant (k = 8.99 x 10^9 N*m^2/C^2), q1 and q2 are the charges of the two point charges, and r is the separation between them.
In this case, we have q1 = +14 uC = +14 x 10^-6 C and q2 = +54 uC = +54 x 10^-6 C. We are given that the magnitude of the electrostatic force is 3.1 N. By rearranging Coulomb's law, we can solve for the separation:
r = sqrt(k * |q1 * q2| / F)
Substituting the given values, we find:
r = sqrt((8.99 x 10^9 N*m^2/C^2) * |(14 x 10^-6 C) * (54 x 10^-6 C)| / (3.1 N))
Calculating this expression gives us a separation of approximately 0.32 meters.
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Gravity is an inverse-square force like electricity and magnetism. If lighter weight moose has a weight of 3640 N on Earth's surface (approximately 6.37 · 10^6 m from Earth's center), what will the moose's weight due to Earth in newtons be at the Moon's orbital radius (approximately 3.84 · 10^8 m from Earth's center) to two significant digits?
To two significant digits, the weight of the moose due to Earth at the Moon's orbital radius would be approximately 60 N.
To calculate the weight of the moose due to Earth at the Moon's orbital radius, we need to consider the inverse-square relationship of gravity and apply it to the given distances.
Given:
Weight of the moose on Earth's surface = 3640 N
Distance from Earth's center at Earth's surface (r1) = 6.37 × 10^6 m
Distance from Earth's center at Moon's orbital radius (r2) = 3.84 × 10^8 m
The gravitational force between two objects is given by the equation F = (G * m1 * m2) / r^2, where F is the force, G is the gravitational constant, m1 and m2 are the masses of the objects, and r is the distance between their centers.
To find the weight of the moose at the Moon's orbital radius, we need to calculate the force at that distance using the inverse-square relationship.
First, we calculate the ratio of the distances squared:
(r2/r1)^2 = (3.84 × 10^8 m / 6.37 × 10^6 m)^2
Next, we calculate the weight at the Moon's orbital radius:
Weight at Moon's orbital radius = Weight on Earth's surface * (r1^2 / r2^2)
Substituting the given values:
Weight at Moon's orbital radius ≈ 3640 N * (6.37 × 10^6 m)^2 / (3.84 × 10^8 m)^2
Calculating the weight at the Moon's orbital radius:
Weight at Moon's orbital radius ≈ 60 N
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A car of mass 1374 kg accelerates from rest to 15.2 m/s in 5.40 s. How much force was required to do this?
The force required to accelerate the car from rest to 15.2 m/s in 5.40 s is approximately 3858.5 N.
To calculate the force required to accelerate the car, we can use Newton's second law of motion, which states that the force acting on an object is equal to the product of its mass and acceleration:
F = m * a
Where:
F is the force (what we're trying to find)m is the mass of the car (1374 kg)a is the acceleration of the car (which can be calculated using the formula Δv / Δt, where Δv is the change in velocity and Δt is the change in time)Given that the car starts from rest (initial velocity, v₀ = 0) and reaches a final velocity of 15.2 m/s in 5.40 s, we can calculate the acceleration:
Δv = v - v₀ = 15.2 m/s - 0 m/s = 15.2 m/s
Δt = 5.40 s
a = Δv / Δt = 15.2 m/s / 5.40 s
Now, let's calculate the force:
F = (1374 kg) * (15.2 m/s / 5.40 s)
F ≈ 3858.5 N
Therefore, the force required to accelerate the car from rest to 15.2 m/s in 5.40 s is approximately 3858.5 Newtons.
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Find the total surface area of the washer, rounded to one
decimal place, for x = 14 mm and y = 24 mm. Hint: Think of the
washer as a cylinder through which a hole has been drilled.
The total surface area of the washer, considering the outer and inner cylinders, is approximately 1051.4 mm². The outer cylinder contributes to the surface area while the inner cylinder, representing the hole, does not affect it.
To find the total surface area of the washer, we need to calculate the surface area of the outer cylinder and subtract the surface area of the inner cylinder.
The surface area of a cylinder is given by the formula:
[tex]A_{cylinder[/tex]= 2πrh
where r is the radius of the cylinder's base and h is the height of the cylinder.
In this case, the washer can be seen as a cylinder with a hole drilled through it, so we need to calculate the surface areas of both the outer and inner cylinders.
Let's calculate the total surface area of the washer:
Calculate the surface area of the outer cylinder:
Given x = 14 mm, the radius of the outer cylinder ( [tex]r_{outer[/tex] ) is half of x, so [tex]r_{outer[/tex] = x/2 = 14/2 = 7 mm.
The height of the outer cylinder ([tex]h_{outer[/tex]) is y = 24 mm.
[tex]A_{outer_{cylinder[/tex] = 2π [tex]r_{outer[/tex][tex]h_{outer[/tex] = 2π(7)(24) ≈ 1051.4 mm² (rounded to one decimal place).
Calculate the surface area of the inner cylinder:
Given the inner radius (r_inner) is 7 mm less than the outer radius, so r_inner = r_outer - 7 = 7 - 7 = 0 mm (since the inner hole has no radius).
The height of the inner cylinder ([tex]h_{inner[/tex]) is the same as the outer cylinder, y = 24 mm.
[tex]A_{inner_{cylinder[/tex] = 2π [tex]r_{inner[/tex] [tex]h_{inner[/tex] = 2π(0)(24) = 0 mm².
Subtract the surface area of the inner cylinder from the surface area of the outer cylinder to get the total surface area of the washer:
Total surface area = [tex]A_{outer_{cylinder[/tex] - [tex]A_{inner_{cylinder[/tex] = 1051.4 - 0 = 1051.4 mm².
Therefore, the total surface area of the washer, rounded to one decimal place, is approximately 1051.4 mm².
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A 19 0-kg child descends a slide 1,80 m high and reaches the bottom with a speed of 1.25 m/s Part A How much thermal energy due to friction was generated in this process? Express your answer to three significant figures and include the appropriate units.
The thermal energy generated due to friction in this process is approximately 3,195 J.
To calculate the thermal energy generated due to friction, we need to consider the change in potential energy and kinetic energy of the child.
The change in potential energy (ΔPE) of the child can be calculated using the formula:
ΔPE = mgh
where:
m is the mass of the child (190 kg),
g is the acceleration due to gravity (approximately 9.8 m/s²),
and h is the height of the slide (1.80 m).
ΔPE = (190 kg) × (9.8 m/s²) × (1.80 m)
ΔPE ≈ 3,343.2 J
The change in kinetic energy (ΔKE) of the child can be calculated using the formula:
ΔKE = (1/2)mv²
where:
m is the mass of the child (190 kg),
and v is the final velocity of the child (1.25 m/s).
ΔKE = (1/2) × (190 kg) × (1.25 m/s)²
ΔKE ≈ 148.4 J
The thermal energy due to friction can be calculated by subtracting the change in kinetic energy from the change in potential energy:
Thermal energy = ΔPE - ΔKE
Thermal energy = 3,343.2 J - 148.4 J
Thermal energy ≈ 3,194.8 J
Therefore, the thermal energy generated due to friction in this process is approximately 3,194.8 Joules (J).
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A 957-g empty iron kettle is put on a stove. How much heat. in joules. must it absorb to raise its temperature from 15.0°C to 37.0°C? (The specific heat for iron is 113 cal/kg•C°, 1 cal = 4.190 J) 10,900 J 9950 J 2380 J 16,700 J A monatomic ideal gas undergoes an isothermal expansion at 300 K, as the volume increased from 0.02 m2 to 0.14 m3. The final pressure of the gas is 140 kPa. The ideal gas constant is R = 8.314 J/mol · K. The change in the internal (thermal) energy of the gas is closest to 0.00 kJ. -38 kJ. -19 kJ. 19 kJ. 38 kJ.
1. The heat absorbed by the iron kettle is approximately 10,900 J.
2. The change in the internal energy of the gas is closest to 0.00 kJ.
1. To calculate the heat absorbed by the iron kettle, we can use the formula:
Q = m × c × ΔT
where Q is the heat, m is the mass of the iron kettle, c is the specific heat of iron, and ΔT is the change in temperature.
Given:
m = 957 g = 0.957 kg (converting to kilograms)
c = 113 cal/kg·°C = 113 × 4.190 J/kg·°C (converting to joules)
ΔT = (37.0°C - 15.0°C)
Substituting the values into the formula:
Q = 0.957 kg × (113 × 4.190 J/kg·°C) × (37.0°C - 15.0°C)
Q ≈ 10900 J
Therefore, the heat absorbed by the iron kettle is approximately 10900 J.
2. For an isothermal process, the change in internal (thermal) energy of the gas is zero. Therefore, the change in internal energy is closest to 0.00 kJ.
Therefore, the answer is 0.00 kJ.
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An electron moves north at a velocity of 9.8 x 104 m/s and has a
magnetic force of 5.6x10 -18 N west exerted on it. If the magnetic
field points upward, what is the magnitude of the magnetic
field.
i
The magnitude of the magnetic field is 3.5x[tex]10^-5[/tex] Tesla. To determine the magnitude of the magnetic field, we can use the formula for the magnetic force experienced by a moving charged particle in a magnetic field:
F = qvB sin(θ)
where F is the magnetic force, q is the charge of the particle, v is the velocity of the particle, B is the magnetic field, and θ is the angle between the velocity vector and the magnetic field vector.
In this case, we are given the magnetic force (F = 5.6x10^-18 N), the velocity of the electron (v = 9.8x10^4 m/s), and the direction of the magnetic force (west). We need to find the magnitude of the magnetic field (B).
Since the force is perpendicular to the velocity, the angle θ between the velocity vector and the magnetic field vector is 90 degrees. Therefore, sin(θ) = 1.
B = F / (qv)
B = (5.6x[tex]10^-18[/tex]N) / (1.6x1[tex]0^-19[/tex] C x 9.8x[tex]10^4[/tex] m/s)
B = 3.5x[tex]10^-5[/tex] T
Therefore, the magnitude of the magnetic field is 3.5x[tex]10^-5[/tex]Tesla.
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7. A bar magnet passes through a circular loop of wire as shown below. At the instant that the middle of the magnet passes through the loop, the induced current (as seen from the coil # 2) is A counte
The passage of a bar magnet through a circular loop of wire induces a current in the wire.
At the instant that the middle of the magnet passes through the loop, the induced current is a counterclockwise direction in coil #2.
This phenomenon is known as electromagnetic induction and is described by Faraday's Law. When a magnetic field changes in intensity or moves relative to a conductor (such as a wire), it induces an electromotive force (EMF) in the conductor, which in turn creates an electric current. In this case, as the bar magnet passes through the circular loop of wire, the magnetic field changes, which induces a current in the wire.
This induced current follows Lenz's Law, which states that the direction of the induced current is always in such a direction as to oppose the change that produced it. In this case, as the north pole of the bar magnet enters the loop, it creates a magnetic field pointing upwards through the loop. Therefore, the induced current creates a magnetic field in the opposite direction (downwards) to oppose the change. This corresponds to a counterclockwise induced current in coil #2.
As the bar magnet continues to pass through the loop, the magnetic field changes again, and the induced current will change accordingly. Once the bar magnet exits the loop, the induced current will stop. This phenomenon has numerous applications in everyday life, including electromagnetic induction used in power plants to generate electricity
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An advertisement claims that a particular automobile can "stop on a dime". What net force would be necessary to stop a 850 kg automobile travelling initially at 57.0 km/h in a distance equal to the diameter of a dime, 1.8 cm? Express your answer with the appropriate units.
R=
Given data: Mass of automobile, m = 850 kg, Initial velocity, v = 57.0 km/h = 15.83 m/s, Distance travelled to stop the car, d = Diameter of a dime = 1.8 cm = 0.018 m. Using the kinematic equation of motion,v² = u² + 2adBy applying the above formula, we can determine the distance travelled by the automobile to come at rest by a force F as:0 = v² + 2ad ⇒ d = -v² / 2a. Neglecting the negative sign as we need only magnitude of acceleration,
a. Force required to stop the automobile can be calculated by Newton's second law of motion, F = ma. Now, acceleration of automobile is given by ,a = (v²) / (2d). Putting the given values, we geta = (15.83 m/s)² / [2 × 0.018 m] = 11,062.5 m/s². Thus, the net force required to stop the automobile travelling initially at 57.0 km/h in a distance equal to the diameter of a dime is F = ma = 850 kg × 11,062.5 m/s² = 9,403,125 N.
Hence, the required net force is 9,403,125 N.
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A assumptive radioactive sample's half-life is unknown. In an initial sample of 9.9×1010 radioactive nuclei, the initial activity is 6.3187×107 Bq(1 Bq=1 decay/s). Part A - What is the decay constant in s−1 ? Use scientific notations, format 1.234∗10n. Part B - What is the half-life in Minutes? 1 min=60 s Enter a regular number with 2 digits after the decimal point, in minutes. Part C - What is the decay constant in min−1 ? Enter a regular number with 5 digits after the decimal polnt. Part D - After 7.60 minutes since the initial sample is prepared, what will be the number of radioactive nuclei that remain in the sample? Use scientific notations, format 1,234∗10n - Part E - How many minutes after the initial sample is prepared will the number of radioactive nucloi remaining in the sample reach 6.214×1010 ? Enter a regular number with 2 digits after the decimal point, in minutes.
Part A. The decay constant is λ = 6.3838383838383838e-04, Part B. The half-life in Minutes is 18.0759 min, Part C. The decay constant in min−1 is 0.038303 min^(-1) Part D. The number of radioactive nuclei that remain in the sample is 4.971874 and Part E. the initial sample is prepared will the number of radioactive nucloi remaining in the sample reach 6.214×1010 in 8.5334 min.
Part A: To find the decay constant, we can use the formula,
λ = (ln(2)) / (T1/2)
where λ is the decay constant and T1/2 is the half-life.
In this case, the initial activity (A0) is given as 6.3187×10^7 Bq.
The decay constant can be calculated as: λ = A0 / N0
Where N0 is the initial number of radioactive nuclei.
Given N0 = 9.9×10^10, we can substitute the values,
λ = (6.3187×10^7) / (9.9×10^10)
Simplifying, we get,
λ = 6.3838383838383838e-04 s^(-1) (scientific notation)
Part B: The half-life (T1/2) can be calculated using the formula: T1/2 = (ln(2)) / λ
Substituting the value of λ from Part A, we have: T1/2 = (ln(2)) / (6.3838383838383838e-04)
Calculating, we find,
T1/2 = 1084.5605336763952 s
Converting to minutes: T1/2 = 1084.5605336763952 / 60 = 18.0759 min
Part C: To convert the decay constant to min^(-1), we can use the conversion factor,
1 min^(-1) = 60 s^(-1)
Therefore, the decay constant in min^(-1) is: λ_min = λ * 60 = 6.3838383838383838e-04 * 60
Calculating, we get: λ_min = 0.038303 min^(-1)
Part D: After a time of 7.60 minutes, we can use the radioactive decay equation: N(t) = N0 * exp(-λ * t)
where N(t) is the number of radioactive nuclei at time t.
Substituting the values,
N(7.60) = (9.9×10^10) * exp(-6.3838383838383838e-04 * 7.60)
Calculating, we find,
N(7.60) = 4.971874330204165e10 (scientific notation)
Part E: To find the time it takes for the number of radioactive nuclei to reach 6.214×10^10, we can rearrange the radioactive decay equation: t = -(1/λ) * ln(N(t) / N0)
Substituting the values: t = -(1/6.3838383838383838e-04) * ln((6.214×10^10) / (9.9×10^10))
Calculating, we get,
t ≈ 8.5334 min (regular number with 2 digits after the decimal point)
Therefore, approximately 8.53 minutes after the initial sample is prepared, the number of radioactive nuclei remaining in the sample will reach 6.214×10^10.
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Numerical Response #2 A 400 g mass is hung vertically from the lower end of a spring. The spring stretches 0.200 m. The value of the spring constant is _____N/m.6. A node is where two or more waves produce A. destructive interference with no displacement B. destructive interference with maximum amplitude C. constructive interference with maximum amplitude D. constructive interference with no displacement
The value of the spring constant is determined by the mass and the amount the spring stretches. By rearranging the equation, the spring constant is found to be approximately 20 N/m.
The spring constant, denoted by k, is a measure of the stiffness of a spring and is determined by the material properties of the spring itself. It represents the amount of force required to stretch or compress the spring by a certain distance. Hooke's Law relates the force exerted by the spring (F) to the displacement of the spring (x) from its equilibrium position:
F = kx
In this scenario, a 400 g mass is hung vertically from the lower end of the spring, causing it to stretch by 0.200 m. To determine the spring constant, we need to convert the mass to kilograms by dividing it by 1000:
mass = 400 g = 0.400 kg
Now we can rearrange Hooke's Law to solve for the spring constant:
k = F / x
Substituting the values we have:
k = (0.400 kg * 9.8 m/s^2) / 0.200 m
Calculating this expression gives us:
k ≈ 19.6 N/m
Rounding to the nearest significant figure, we can say that the value of the spring constant is approximately 20 N/m.
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Question 4 A book of mass m is taken to a heighth with a constant speed. A rock of mass 2m is taken to the same height also at a constant speed. The rock rises to this height twice as fast as the book. The work the gravitational force does on the rock is one quarter of the the work done on the book one half of the work done on the book twice the work done on the book four times the work done on the book the same as the work done on the book
The work done by the gravitational force on the rock is four times the work done on the book.
The work done by the gravitational force is given by the equation W = mgh, where W is the work done, m is the mass of the object, g is the acceleration due to gravity, and h is the height. Since both the book and the rock are lifted to the same height with constant speed, the gravitational potential energy gained by each object is the same.
Let's assume the work done on the book is W_book. According to the problem, the rock rises to the same height twice as fast as the book. Since work done is directly proportional to the time taken, the work done on the rock, W_rock, is twice the work done on the book (2 * W_book).
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Using your knowledge of kinetic molecular theory and methods of transfer of heat, explain what happens when a person puts their hand down on a very hot stove top. Also explain how they may have had a warning that the stovetop would be hot before their hand touched the stove.
When a person puts their hand down on a very hot stove top, the heat energy is transferred from the stove top to the person's hand. Kinetic molecular theory explains that the temperature of a substance is related to the average kinetic energy of the particles that make up that substance. In the case of the stove top, the heat causes the particles to vibrate faster and move farther apart, which results in an increase in temperature.
The transfer of heat occurs by three methods, namely conduction, convection, and radiation. In this case, the heat is transferred through conduction. Conduction is the transfer of heat energy through a substance or between substances that are in contact. When the person's hand touches the stove top, the heat energy is transferred from the stove top to the person's hand through conduction.
Before touching the stove, the person may have had a warning that the stove top would be hot. This is because of the transfer of heat through radiation. Radiation is the transfer of heat energy through electromagnetic waves. The stove top, which is at a higher temperature than the surrounding air, emits heat energy in the form of radiation. The person may have felt the heat radiating from the stove top, indicating that the stove top was hot and that it should not be touched.
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A 17 kg curling stone traveling at 12.5 m/s [E] collides with another stone of the same mass. The collision is a glancing one. If the second stone is now traveling at 1.5 m/s [E25°N], calculate the velocity of the first stone after collision. 13 m/s [E6.8°S] 26 m/s [E18°N] 7.0 m/s [E1.4°N] 11 m/s [E3.3°S]
To solve this problem, we can analyze the conservation of momentum and the conservation of kinetic energy during the collision.
Let's break down the initial and final velocities of the stones into their x and y components for easier calculations. For the initial velocity of the first stone, we have:
Initial velocity of stone 1: v1 = 12.5 m/s [E]
Initial velocity of stone 2: v2 = 0 m/s [E]
The final velocity of the second stone is given as:
Final velocity of stone 2: vf2 = 1.5 m/s [E25°N]
To find the final velocity of the first stone (vf1), we need to calculate its x and y components separately. Let's assume the final velocity of the first stone has components vx1 and vy1.
Using the conservation of momentum, we know that the total momentum before the collision is equal to the total momentum after the collision. Since the masses of the stones are the same, we can write the equation:
(m1 * v1) + (m2 * v2) = (m1 * vx1) + (m2 * vf2)
Substituting the known values, we have:
(17 kg * 12.5 m/s) + (17 kg * 0 m/s) = (17 kg * vx1) + (17 kg * 1.5 m/s)
Simplifying the equation:
212.5 kg·m/s = 17 kg * vx1 + 25.5 kg·m/s
212.5 kg·m/s - 25.5 kg·m/s = 17 kg * vx1
187 kg·m/s = 17 kg * vx1
Dividing both sides by 17 kg:
vx1 = 11 m/s [E]
Now, we can use the conservation of kinetic energy to find the y-component of the final velocity of the first stone. Since the collision is glancing, the kinetic energy in the y-direction is conserved. We have:
(1/2) * m1 * v1^2 = (1/2) * m1 * vy1^2
Substituting the values:
(1/2) * 17 kg * (12.5 m/s)^2 = (1/2) * 17 kg * vy1^2
156.25 J = 8.5 kg * vy1^2
Dividing both sides by 8.5 kg:
vy1^2 = 18.3824
Taking the square root:
vy1 ≈ 4.286 m/s
Now we have the x and y components of the final velocity of the first stone. We can calculate the magnitude and direction using trigonometry:
Magnitude of vf1 = sqrt(vx1^2 + vy1^2) ≈ sqrt((11 m/s)^2 + (4.286 m/s)^2) ≈ 11.952 m/s
Direction of vf1 = atan(vy1 / vx1) ≈ atan(4.286 m/s / 11 m/s) ≈ atan(0.3896) ≈ 21.8°
The final velocity of the first stone after the collision is approximately 11.952 m/s [E21.8°N].
Among the given options, the closest value is 11 m/s [E3.3°S].
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A boy kicks a soccer ball from the ground, giving it an initial velocity of 34 m/s at some unknown angle. The ball reaches a maximum height of 19m above the ground. Use energy to determine the velocity?
the velocity of the soccer ball is approximately 27.29 m/s.To determine the velocity of the soccer ball, The total energy is the sum of the kinetic energy (0.5mv²) and the potential energy (mgh). Since the initial kinetic energy is zero, we can equate the potential energy at the maximum height to the total energy at the ground level. Solving for v, we get:
0.5mv² + mgh = mgh
0.5v² = 2gh
v² = 4gh
v = √(4gh)
Given that g is approximately 9.8 m/s² and h is 19m, we can substitute these values:
v = √(4 * 9.8 * 19) = √(745.6) ≈ 27.29 m/s
Therefore, the velocity of the soccer ball is approximately 27.29 m/s.
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Two cars collide at an icy intersection and stick together afterward. The first car has a mass of 1,200 kg and is approaching at 7.74 m/s due south. The second car has a mass of 805 kg and is
approaching at 15.7 m/s due west.
Calculate the final velocity (magnitude and direction) of the cars.
The final velocity of the two cars, after colliding at an icy intersection, is 6.51 m/s at an angle of 309 degrees from the south.
When two cars collide and stick together, their masses and velocities determine their final velocity.
In this case, using the law of conservation of momentum, we can calculate the final velocity of the two cars.
The initial momentum of the first car is (1200 kg)(7.74 m/s) = 9292.8 kgm/s south.
The initial momentum of the second car is (805 kg)(15.7 m/s) = 12648.5 kgm/s west.
After the collision, the total momentum of the two cars is conserved and is equal to (1200 + 805)*(final velocity).
Solving for the final velocity, we get a magnitude of 6.51 m/s.
The direction of the final velocity can be found using trigonometry, where the tangent of the angle between the final velocity and the south direction is equal to -15.7/7.74.
This gives us an angle of 309 degrees from the south.
Therefore, the final velocity of the two cars is 6.51 m/s at an angle of 309 degrees from the south.
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The position of a 51 gg oscillating mass is given by
x(t)=(1.5cm)cos11t, where t is in seconds. Determine the
amplitude.
The given position equation is x(t) = (1.5 cm)cos(11t). In this equation, the coefficient of the cosine function represents the amplitude of the oscillation.
To determine the amplitude of the oscillating mass, we can observe that the equation for position, x(t), is given by:
x(t) = (1.5 cm) * cos(11t)
The amplitude of an oscillating mass is the maximum displacement from the equilibrium position. In this case, the maximum displacement is the maximum value of the cosine function.
The maximum value of the cosine function is 1, so the amplitude of the oscillating mass is equal to the coefficient in front of the cosine function, which is 1.5 cm.
Therefore, the amplitude of the oscillating mass is 1.5 cm.
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21 of 37 Attempt Many expressions in special relativity contain the gamma (Y) factor. 1 Y= V1 - 02/22 In the equation, v is the speed of some object or reference frame and c is the speed of light. Find the numerical value of y for each of the listed speeds. v = 0. incorrect V = 0.450c. Y incorrect 0.990, Y Incorrect retel renants how the value of y depends on speed.
The value of γ (gamma) increases as the speed (v) approaches the speed of light (c).
The correct expression for the gamma factor (γ) in special relativity is:
γ = 1 / √(1 - (v^2 / c^2))
For the given speeds:
1. v = 0: γ = 1 / √(1 - (0^2 / c^2)) = 1 / √(1 - 0) = 1
2. v = 0.450c: γ = 1 / √(1 - (0.450c)^2 / c^2) = 1 / √(1 - 0.2025) = 1 / √(0.7975) ≈ 1.112
The value of γ depends on the speed (v) relative to the speed of light (c). As the speed approaches the speed of light (c), the value of γ increases, indicating greater time dilation and relativistic effects.
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1. A ball is kicked horizontally at 8 m/s30 degrees above the horizontal. How far does the ball travel before hitting the ground? (2pts) 2. A shell is fired from a cliff horizontally with initial velocity of 800 m/s at a target on the ground 150 m below. How far away is the target? (2 pts) 3. You are standing 50 feet from a building and throw a ball through a window that is 26 feet above the ground. Your release point is 6 feet off of the ground (hint: you are only concerned with Δy ). You throw the ball at 30ft/sec. At what angle from the horizontal should you throw the ball? (hint: this is your launch angle) ( 2 pts) 4. A golfer drives a golf ball from the tee down the fairway in a high arcing shot. When the ball is at the highest point during the flight: ( 1pt) a. The velocity and acceleration are both zero b. The x-velocity is zero and the y-velocity is zero c. The x-velocity is non-zero but the y-velocity is zero d. The velocity is non-zero but the acceleration is zero
1) Distance = 9.23 m ; 2) Horizontal distance = 24,481.7 m ; 3) θ = 33.2 degrees ; 4) When the ball is at the highest point during the flight, a) the velocity and acceleration are both zero and hence option a) is the correct answer.
1. The horizontal component of the ball's velocity is 8cos30, and the vertical component of its velocity is 8sin30. The ball's flight time can be determined using the vertical component of its velocity.
Using the formula v = u + at and assuming that the initial vertical velocity is 8sin30, the acceleration is 9.81 m/s² (acceleration due to gravity), and the final velocity is zero (because the ball is at its maximum height), the time taken to reach the maximum height can be calculated.
The ball will reach its maximum height after half of its flight time has elapsed, so double the time calculated previously to get the total time. Substitute the time calculated previously into the horizontal velocity formula to get the distance the ball travels horizontally before landing.
Distance = 8cos30 x 2 x [8sin30/9.81] = 9.23 m
Answer: 9.23 m
2. Using the formula v = u + gt, the time taken for the shell to hit the ground can be calculated by assuming that the initial vertical velocity is zero (since the shell is fired horizontally) and that the acceleration is 9.81 m/s². The calculated time can then be substituted into the horizontal distance formula to determine the distance the shell travels horizontally before hitting the ground.
Horizontal distance = 800 x [2 x 150/9.81]
= 24,481.7 m
Answer: 24,481.7 m³.
3) To determine the angle at which the ball should be thrown, the vertical displacement of the ball from the release point to the window can be used along with the initial velocity of the ball and the acceleration due to gravity.
Using the formula v² = u² + 2as and assuming that the initial vertical velocity is 30sinθ, the acceleration due to gravity is -32.2 ft/s² (because the acceleration due to gravity is downwards), the final vertical velocity is zero (because the ball reaches its highest point at the window), and the displacement is 20 feet (26-6), the angle θ can be calculated.
Angle θ = arc sin[g x (20/900 + 1/2)]/2, where g = 32.2 ft/s²
Answer: θ = 33.2 degrees
4. A golfer drives a golf ball from the tee down the fairway in a high arcing shot. When the ball is at the highest point during the flight, the velocity and acceleration are both zero. (1pt)
Answer: a. The velocity and acceleration are both zero. Thus, option a) is correct.
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A storage battery has an emf of 6.7 V and internal resistance of
0.50. Compute for the terminal voltage of the battery when it is
being charged with 2.0 A.
A storage battery with emf of 6.7 V and internal resistance of 0.50 is being charged with 2.0 A.
Let’s compute the terminal voltage of the battery
The expression for the terminal voltage of the battery while charging is given byV = emf - IR
Where, V is the terminal voltage of the battery
emf is the electromotive force of the battery
I is the charging current
R is the internal resistance of the battery
Given that,emf of the battery = 6.7 V
Internal resistance of the battery = 0.50 Ω
Charging current = 2.0 A
Therefore, the expression for terminal voltage of the battery isV = 6.7 - 2.0 × 0.50
= 6.7 - 1
= 5.7 V
So, the terminal voltage of the battery while charging with 2.0 A is 5.7 V.
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ii). Hence, what is the length of a meterstick measured by an observer travelling at α). 1610km/hr and β). 0.9c [c =3.0 x10 8m/s]
ii). Hence, a clock on a space rocket ticks off at a time interval of 1hour.what is the time elapse on earth if the space rocket is travelling at a speed α). 1610km/hr ? and β). 0.9c ? [c =3.0 x10 8m/s]
Length of a meterstick when measured by an observer at α). 1610km/hr is 0.9997 times its length at rest. Length of a meterstick when measured by an observer at β). 0.9c is 0.4359 times its length at rest.
i) The length of an object at rest can change depending on how fast it is moving. This phenomenon is known as length contraction. An observer travelling at a speed of 1610 km/hr would measure a meterstick to be slightly shorter than its actual length, that is, 0.9997 times its length at rest. Similarly, an observer travelling at a speed of 0.9c would measure the meterstick to be much shorter, only 0.4359 times its length at rest.
ii) Time dilation is another phenomenon associated with moving objects. As an object moves faster, time appears to slow down relative to a stationary observer. Thus, a clock on a space rocket travelling at 1610 km/hr would appear to tick off at a slower rate than a clock on earth. Therefore, if the space rocket travels for 1 hour, the time elapsed on earth would be slightly longer. If the space rocket is travelling at 0.9c, then time dilation is much more pronounced. The time elapsed on earth would be much longer than 1 hour due to the extreme time dilation.
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Question 5 of 7 > The largest tendon in the body, the Achilles tendon, connects the calf muscle to the heel bone of the foot. This tendon is typically 15.0 cm long, 5.00 mm in diameter, and has a Young's modulus of 1.65 x 10° Pa. If an athlete has stretched the tendon to a length of 16.1 cm. what is the tension T, in newtons, in the tendon?
Young's modulus is the constant that shows the ratio of stress to strain for a material that is being stretched or compressed. The formula for stress is.
The original length of the tendon is L1 = 15.0 cm The stretched length of the tendon is L2 = 16.1 cm The diameter of the tendon is d = 5.00 mm = 0.0050 m Young's modulus is Y = 1.65 x 1010 Pa To find the tension T in the tendon, we need to calculate the change in length and stress.
Change in length of tendonΔL[tex]= L2 - L1ΔL = 16.1 cm - 15.0 cmΔL = 1.1 \\[/tex]cm Now, we convert the change in length to meters,ΔL = 1.1 cm x 1 m/100 cmΔL = 0.011 m Stress on tendon Stress = Force/Area In this case, we are given the diameter of the tendon.
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T
The sound intensity 300.0 m from a wailing tornado siren is 0.10 W/m². What is the sound intensity level 50.0 m from the siren?
The sound intensity level at a distance of 50.0 m from the siren is approximately 1.33 W/m², calculated using the inverse square law for sound propagation and the formula for sound intensity level.
To calculate the sound intensity level at a distance of 50.0 m from the siren, we can start by using the inverse square law for sound propagation:
I₁/I₂ = (r₂/r₁)²
Where I₁ and I₂ are the sound intensities at distances r₁ and r₂, respectively. We are given that the sound intensity at a distance of 300.0 m is 0.10 W/m².
So, plugging in the values:
0.10 W/m² / I₂ = (50.0 m / 300.0 m)²
Simplifying:
I₂ = 0.10 W/m² / ((50.0 m / 300.0 m)²)
= 0.10 W/m² / (0.1667)²
= 0.10 W/m² / 0.02778
≈ 3.60 W/m²
Now, to determine the sound intensity level (L), we can use the formula:
L = 10 log₁₀ (I/I₀)
Where I is the sound intensity and I₀ is the reference intensity, typically 10^(-12) W/m².
Using the given sound intensity of 3.60 W/m²:
L = 10 log₁₀ (3.60 / 10^(-12))
= 10 log₁₀ (3.60) + 10 log₁₀ (10^12)
≈ 10 log₁₀ (3.60) + 120
≈ 10 (0.556) + 120
≈ 5.56 + 120
≈ 125.56 dB
Therefore, the sound intensity level at a distance of 50.0 m from the siren is approximately 125.56 dB.
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A skilled archer tries to hit an apple from the ground with an arrow. At the same time, as he fires, the apple falls freely. If the velocity of the arrow is <97,27,0> m/s, height of the apple is (200,100,0) and arrow starts from (-200,0,0)> m, calculate the time when the arrow will hit the apple and visualize it as a simulation. Create an arrow and apple as defined (you can add more objects as well). Define a loop to update the speeds and positions (for both apple and arrow) including the effect of gravity. Plot the graph (Position Vs Time)
The time when the arrow will hit the falling apple can be calculated by simulating the motion of both objects. Given the velocity of the arrow as <97,27,0> m/s, the initial position of the arrow as (-200,0,0) m, and the initial position of the apple as (200,100,0) m.
We can update the speeds and positions of both objects using a loop that incorporates the effect of gravity. By plotting the graph of position versus time, we can visually determine the time at which the arrow hits the apple.
To simulate the motion of the arrow and the falling apple, we need to update their speeds and positions over time. We can do this by incorporating the effect of gravity on both objects. Assuming the acceleration due to gravity is -9.8 m/s^2 (taking downward as the negative direction), we can use the following equations of motion:
Arrow:
Velocity of the arrow: v_arrow = <97, 27, 0> m/s
Initial position of the arrow: p_arrow = <-200, 0, 0> m
Apple:
Initial velocity of the apple: v_apple = <0, 0, 0> m/s
Initial position of the apple: p_apple = <200, 100, 0> m
Using a loop, we can update the positions and speeds of the arrow and the apple by considering the effect of gravity on their vertical components. The horizontal components of the velocities remain constant.
By tracking the positions of the arrow and the apple over time, we can plot a graph of their vertical positions versus time. The time at which the arrow and the apple intersect on the graph corresponds to the time when the arrow hits the apple.
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A swimming pool measures a length of 6.0 m, width 4.3 m , and depth 3.3 m. Compute the force exerted by the water against the bottom. Do not include the force due to air pressure. Express your answe
The force due to air pressure, is approximately 836,532 Newtons.
To compute the force exerted by the water against the bottom of the swimming pool, we need to consider the concept of pressure and the area of the pool's bottom.
The pressure exerted by a fluid at a certain depth is given by the formula P = ρgh, where P is the pressure, ρ is the density of the fluid, g is the acceleration due to gravity, and h is the depth.
In this case, the fluid is water, which has a density of approximately 1000 kg/m³, and the acceleration due to gravity is 9.8 m/s².
The depth of the pool is given as 3.3 m. Substituting these values into the formula, we can calculate the pressure at the bottom of the pool:
P = (1000 kg/m³)(9.8 m/s²)(3.3 m) = 32,340 Pa
To determine the force exerted by the water against the bottom, we need to multiply this pressure by the area of the pool's bottom. The area is calculated by multiplying the length and width of the pool:
Area = 6.0 m × 4.3 m = 25.8 m²
Now, we can calculate the force using the formula Force = Pressure × Area:
Force = (32,340 Pa)(25.8 m²) = 836,532 N
Therefore, the force exerted by the water against the bottom of the swimming pool, without considering the force due to air pressure, is approximately 836,532 Newtons.
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One
problematic property of light was how it was transmitted through
space from the sun to Earth. Explain how the properties of the
particle theory and wave theory each handled this
explanation?
The particle theory suggests that light is made up of tiny particles called photons, which travel in straight lines and interact with matter. On the other hand, the wave theory proposes that light is a form of electromagnetic radiation that propagates as waves, spreading out in all directions.
According to the particle theory of light, light is composed of discrete particles called photons. These photons are emitted by the sun and travel through space in straight lines until they encounter an object. When photons interact with matter, they can be absorbed, reflected, or transmitted depending on the properties of the material. This theory explains how light travels from the sun to Earth as a series of particle-like entities that move in specific paths.
On the other hand, the wave theory of light suggests that light is an electromagnetic wave that spreads out in all directions from its source, such as the sun. According to this theory, light is characterized by its wavelength, frequency, and amplitude. As an electromagnetic wave, light does not require a medium to propagate and can travel through the vacuum of space. The wave theory explains how light is transmitted as a continuous wave that fills the space between the sun and Earth, allowing it to reach our planet without the need for particles or a physical connection.
Both theories offer different perspectives on how light is transmitted through space. The particle theory focuses on the discrete nature of light as particles that interact with matter, while the wave theory emphasizes the wave-like properties of light as electromagnetic radiation that can propagate through a vacuum. Both theories have been supported by experimental evidence and are used to explain different phenomena related to light, highlighting the dual nature of light as both particles and waves
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A400 ohm resistor is connected in series with a 0.35 H inductor and AC-source. The potential difference across the resistor is VR-6.8 cos (680rad/s)t a) What is the circuit current at t-1.6s? t in Volts. b) Determine the inductive reactance of the inductor? c) What is the voltage across the inductor (V₁) at t=3.2s? J₁ = 2
In this series circuit, a 400-ohm resistor is connected with a 0.35 H inductor and an AC source.
The potential difference across the resistor is given by VR = 6.8 cos(680 rad/s)t. To solve the given questions, we need to determine the circuit current at t = 1.6 s, calculate the inductive reactance of the inductor, and find the voltage across the inductor (V₁) at t = 3.2 s.
a) To find the circuit current at t = 1.6 s, we can use Ohm's law. The potential difference across the resistor is VR = 6.8 cos(680 rad/s)(1.6 s). Since the resistor and inductor are in series, the current flowing through both components is the same. Therefore, the circuit current at t = 1.6 s is I = VR / R, where R is the resistance value of 400 ohms.
b) The inductive reactance of an inductor can be calculated using the formula XL = 2πfL, where f is the frequency and L is the inductance. In this case, the frequency is given by ω = 680 rad/s. Thus, the inductive reactance of the 0.35 H inductor is XL = 2π(680)(0.35).
c) To determine the voltage across the inductor (V₁) at t = 3.2 s, we need to consider the relationship between voltage and inductive reactance. The voltage across the inductor can be calculated using the formula V₁ = IXL, where I is the circuit current at t = 3.2 s, and XL is the inductive reactance determined in part (b).
By applying the necessary calculations, we can find the circuit current at t = 1.6 s, the inductive reactance of the inductor, and the voltage across the inductor at t = 3.2 s using the given information.
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A uniform plank of length 2.00 m and mass 29.2 kg is supported by three ropes. A 700 N person is a distance, d, of 0.44 m from the left end.
Part (a) Find the magnitude of the tension, T2, in the vertical rope on the left end. Give your answer in Newtons. Part (b) Find the magnitude of the tension, T1, in the rope on the right end. Give your answer in Newtons. Part (c) Find the magnitude of the tension, T3, in the horizontal rope on the left end. Give your answer in Newtons.
Ques 2: A uniform plank of length 2.00 m and mass 33.86 kg is supported by three ropes
If the tension, T1, cannot exceed 588 N of force without breaking, what is the maximum distance, d, the 700-N person can be from the left end? Be sure to answer in meters.
The maximum distance, d, the 700-N person can be from the left end is when T1 = 142.88 N which occurs when the person is at the very right end of the plank.
How to determine magnitude and distance?Part (a) To find the magnitude of tension, T2, in the vertical rope on the left end, consider the equilibrium of forces acting on the plank. The plank is in rotational equilibrium, which means the sum of the torques acting on the plank must be zero.
Since the person is located 0.44 m from the left end, the distance from the person to the left end is 2.00 m - 0.44 m = 1.56 m.
Denote the tensions in the ropes as T1, T2, and T3. The torques acting on the plank can be calculated as follows:
Torque due to T1: T1 × 2.00 m (clockwise torque)
Torque due to T2: T2 × 0.00 m (no torque since the rope is vertical)
Torque due to T3: T3 × 1.56 m (counter-clockwise torque)
Since the plank is in rotational equilibrium, the sum of the torques must be zero:
T1 × 2.00 m - T3 × 1.56 m = 0
The weight of the plank is acting at the center of the plank, which is at a distance of 1.00 m from either end. The weight can be calculated as:
Weight = mass × acceleration due to gravity
Weight = 29.2 kg × 9.8 m/s²
Weight = 285.76 N
The sum of the vertical forces must be zero:
T1 + T2 + T3 - 285.76 N = 0
The vertical forces must balance, so:
T1 + T2 + T3 = 285.76 N
Substitute the value of T2 = 0 (since there is no vertical tension) and solve for T1:
T1 + 0 + T3 = 285.76 N
T1 + T3 = 285.76 N
Part (b) To find the magnitude of tension, T1, in the rope on the right end, use the same equation as above:
T1 + T3 = 285.76 N
Part (c) To find the magnitude of tension, T3, in the horizontal rope on the left end, consider the horizontal forces acting on the plank. Since the plank is in horizontal equilibrium, the sum of the horizontal forces must be zero:
T3 = T1
So, T3 = T1
Ques 2: To find the maximum distance, d, the 700-N person can be from the left end, consider the maximum tension that the rope T1 can handle, which is 588 N.
Using the equation T1 + T3 = 285.76 N, we can substitute T3 = T1:
T1 + T1 = 285.76 N
2T1 = 285.76 N
T1 = 142.88 N
Since the person exerts a downward force of 700 N, the tension in T1 cannot exceed 588 N. Therefore, the maximum tension in T1 is 588 N.
Rearrange the equation T1 + T3 = 285.76 N to solve for T3:
T3 = 285.76 N - T1
T3 = 285.76 N - 588 N
T3 = -302.24 N
Since tension cannot be negative, T3 cannot be -302.24 N. Therefore, there is no valid solution for T3.
To find the maximum distance, d, rearrange the equation:
T1 + T3 = 285.76 N
142.88 N + T3 = 285.76 N
T3 = 285.76 N - 142.88 N
T3 = 142.88 N
Since T3 = T1, substitute T3 = T1:
142.88 N = T1
Therefore, the maximum distance, d, the 700-N person can be from the left end is when T1 = 142.88 N, which occurs when the person is at the very right end of the plank.
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