(a) The acceleration of the system is approximately 2.25 m/s² to the right.
(b) The tension in the cord connecting the 3.5 kg and 1.0 kg blocks is approximately 23.556 N.
(c) The force exerted by the 1.0 kg block on the 2.0 kg block is approximately 4.50 N.
The system of two blocks, we need to find the acceleration, tension in the cord, and force exerted by the 1.0 kg block on the 2.0 kg block. Let's solve each part of the question.
(a) Acceleration (in m/s² to the right)
We know that the acceleration of the system is given by the formula:
F = ma
Where,
F is the net force on the system
m is the mass of the system
a is the acceleration of the system
The net force acting on the system is given by:
F = T - f
Where,
T is the tension in the cord connecting the blocks
f is the frictional force acting on the 2.0 kg block
So, the acceleration of the system is given by:
ma = T - f
Here, m = 3.5 kg
The frictional force f = μmg = 0.35 × 2.0 × 9.81 = 6.867 N (using μ = 0.35 as given)
Substituting the values, we get:
3.5a = T - 6.867
Also, for the second block, we know that:
F = ma
Here, F is the force exerted by the 1.0 kg block on the 2.0 kg block. We can write:
F = m₂a = 2.0a
Equating the two values of F, we get:
2.0a = T - 6.8673.5a = T - 6.867
Multiplying the second equation by 2 and adding it to the first equation, we get:
2(3.5a = T - 6.867) + 2.0a = T - 6.8677.0a = 3T - 6.867a = (3T - 6.867) / 7.0
Substituting the value of acceleration from the above equation, we get:
3.5[(3T - 6.867) / 7.0] = T - 6.867
Solving for T, we get:
T = 23.556 N (approx)
Therefore, the acceleration of the system is given by:
a = (3T - 6.867) / 7.0= (3 × 23.556 - 6.867) / 7.0= 2.25 m/s² (approx)
(b) Tension in the cord (in N)
From the above calculation, we get:
T = 23.556 N (approx)
Therefore, the tension in the cord connecting the 3.5 kg and 1.0 kg blocks is 23.556 N (approx).
(c) Force exerted by the 1.0 kg block on the 2.0 kg block (in N)
We know that the force exerted by the 1.0 kg block on the 2.0 kg block is given by:
F = m₂a
Here, m₂ = 2.0 kg, and a = 2.25 m/s² (as found in part (a)).
Therefore, the force exerted by the 1.0 kg block on the 2.0 kg block is:
F = m₂a= 2.0 × 2.25= 4.50 N (approx)
Therefore, the force exerted by the 1.0 kg block on the 2.0 kg block is 4.50 N (approx).
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A dog jumps horizontally at 5.1 m/s from the deck of a steady boat, 1.8 m above the waten At what horizontal distance from the boat will the dog hit the water? 3.1 m 2.4 m 9.2 m 2.8 m 3.8 m
To find the horizontal distance from the boat where the dog will hit the water, we can use the formula for horizontal projectile motion.
1. First, let's determine the time it takes for the dog to hit the water. Since the dog jumps horizontally, the vertical motion does not affect the time. We can use the formula:
d = v * t
Where d is the vertical distance (1.8 m) and v is the initial vertical velocity (0 m/s).
Solving for t, we get:
t = d / v
t = 1.8 m / 0 m/s
t = infinity
2. Since the time is infinite, the dog will never hit the water if it jumps horizontally.
Therefore, the dog will not hit the water. None of the given options (3.1 m, 2.4 m, 9.2 m, 2.8 m, 3.8 m) are correct.
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a car is released from rest on top of an inclined hill with a 15 degree slope. assume the gear has been placed on neutral. the car travels 78 meters just before reaching the bottom. Determine the amount of time it takes the car to reach the bottom of the hill.
the amount of time it takes the car to reach the bottom of the hill is 8.91 s.
We are given that a car is released from rest on top of an inclined hill with a 15 degree slope. We are to determine the amount of time it takes the car to reach the bottom of the hill.
We are also given that the car travels 78 meters just before reaching the bottom.According to the law of conservation of energy, the potential energy at the top of the hill will be converted into kinetic energy at the bottom of the hill. We can use this concept to solve the problem. Here is the solution:
1. Determine the gravitational potential energy (GPE) at the top of the hill.GPE = mgh, where m is the mass of the car, g is the acceleration due to gravity, and h is the height of the hill above a reference level. Since the car is released from rest, its initial kinetic energy is zero.
Therefore, all of the energy at the top of the hill is in the form of potential energy.GPE = (m)(g)(h)where m = mass of the car = unknowng = acceleration due to gravity = 9.81 m/s2h = height of the hill = (distance along the slope)(sin 15°) = (78 m)(sin 15°) = 20.14 mGPE = (m)(9.81 m/s2)(20.14 m) = 196.14m J
2. Determine the kinetic energy (KE) at the bottom of the hill.KE = (1/2)mv2, where v is the velocity of the car at the bottom of the hill. Since the car started from rest, its initial velocity is zero.
Therefore, all of the energy at the bottom of the hill is in the form of kinetic energy.KE = (1/2)mv2where m = mass of the car = unknownv = velocity of the car at the bottom of the hill = unknownKE = (1/2)(m)(v2)
3. Use the conservation of energy to equate the GPE at the top of the hill with the KE at the bottom of the hill.GPE = KE196.14m = (1/2)(m)(v2)392.28 = m(v2)
4. Solve for v.v2 = 392.28/mv = sqrt(392.28/m)
5. Determine the time it takes for the car to travel the distance of 78 m at a constant speed of v.t = d/vwhere d = distance traveled by the car = 78 m and v = velocity of the car at the bottom of the hill.t = 78 m / vSubstitute the expression for v that was obtained in step 4 to get:t = 78 m / sqrt(392.28/m)
Simplify:t = 8.91 sTherefore, the amount of time it takes the car to reach the bottom of the hill is 8.91 s.
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the techniques used to find extrasolar planets are based on
Radial velocity method and Transit method are the techniques used to find extrasolar planets.
The techniques used to find extrasolar planets are based on various methods. The method used depends on the size of the planet being searched for, the distance of the planet from its star, and the type of star being observed.
Extrasolar planets are planets that exist outside our solar system. At the present moment, nearly 150 extrasolar planets have been found. There are different techniques used to find them.
These techniques include:
Radial velocity method:
This method is used to find planets that are close to their star. It measures the pull of the planet's gravity on its parent star, which causes the star to move back and forth in space. This movement is detected by the Doppler shift of the star's spectrum.
Transit method:
This method is used to find planets that pass in front of their parent star as seen from Earth. When a planet passes in front of a star, it blocks some of the star's light, causing a small decrease in the star's brightness. By measuring the amount of light that is blocked by the planet, astronomers can determine the size of the planet. Direct imaging: This method is used to find planets that are far from their star and are very large, like Jupiter. These planets are detected by taking a direct image of them with a telescope. This method is very difficult because the planet's light is very dim compared to the light of its parent star.
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answer this question, first click here to view the associated video. Asatellite is in a circular orbit around a planet. A second satellite is placed in a different circular orbit that is farther away from the same planet. How do the speeds of the two satellites compare? The speed of the second satellite is less than the speed of the first satellite. The speed of the second satellite is greater than the speed of the first satellite. The speeds of the two satellites are equal, Which of the following statements is correct concerning satellites in orbit around the same planet? The period of a satellite depends on its mass. Satellites located closer to the planet have longer periods. The period of a satellite is independent of the planet's mass. If the orbital radius of a satellite doubles, then its period increases by a factor of 4 . None of the above statements are correct.
The correct statement concerning satellites in orbit around the same planet is: "The period of a satellite is independent of the planet's mass."
The period of a satellite refers to the time it takes for the satellite to complete one full orbit around the planet. This period is determined by the radius of the orbit and the gravitational force between the satellite and the planet. It is important to note that the mass of the satellite itself does not affect its period.
The period of a satellite depends on the radius of its orbit and is governed by Kepler's Third Law of Planetary Motion. According to this law, the period of a satellite is proportional to the square root of the cube of its orbital radius. Therefore, if the orbital radius of a satellite doubles, its period increases by a factor of 2^(3/2), which is approximately 2.83. This means the period increases by a factor of 2.83, not 4.
Hence, the correct statement is that the period of a satellite is independent of the planet's mass.
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A 1.35 kg object is floating on an air-track and is attached to a horizontal spring of force constant 2.50 N/cm. You start the object oscillating by pulling it 6.00 cm from its equilibrium position and releasing it. You observe that after eight cycles its maximum displacement from equilibrium is only 3.50 cm. How much energy is lost to damping during these eight cycles? Where did this energy go?
The energy lost to damping is 0.30 J.The energy lost to damping is primarily converted into heat.
To determine the energy lost to damping during the eight cycles, we can use the concept of mechanical energy.
The mechanical energy of the oscillating system is the sum of potential energy and kinetic energy. In the absence of damping, the total mechanical energy would remain constant over time.
Given:
Mass of the object (m) = 1.35 kg
Force constant of the spring (k) = 2.50 N/cm = 250 N/m
Initial displacement from equilibrium (x0) = 6.00 cm = 0.06 m
Maximum displacement from equilibrium (xmax) after eight cycles = 3.50 cm = 0.035 m
To calculate the total energy lost to damping during the eight cycles, we need to determine the initial mechanical energy (Einitial) and the final mechanical energy (Efinal).
The initial mechanical energy (Einitial) is given by the potential energy at the maximum displacement from equilibrium:
Einitial = (1/2) * [tex]k * x_0^2[/tex]
Einitial = (1/2) * 250 * [tex]0.06^2[/tex] = 0.45 J
The final mechanical energy (Efinal) is given by the potential energy at the maximum displacement from equilibrium after eight cycles:
Efinal = (1/2) * k * [tex]xmax^2[/tex]
Efinal = (1/2) * 250 * [tex]0.035^2[/tex] = 0.15 J
The energy lost to damping during the eight cycles is the difference between the initial and final mechanical energies:
Energy lost = Einitial - Efinal
Energy lost = 0.45 J - 0.15 J = 0.30 J
The energy lost to damping is 0.30 J.
Where did this energy go? The energy lost to damping is primarily converted into heat. Due to damping forces (such as air resistance or internal friction within the system), the mechanical energy is dissipated as heat energy, resulting in a decrease in the amplitude of the oscillations over time.
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When a car brakes hard and skids, the maximum deceleration is 5.2 m/s2. The coefficient of friction between tires and road is?
The coefficient of friction between tires and road is 0.53.
Let's use the following formula to calculate the coefficient of friction between tires and road:
μ = a/g
where μ is the coefficient of friction
a is the maximum deceleration
g is the acceleration due to gravity
The maximum deceleration a is 5.2 m/s², and the acceleration due to gravity g is 9.81 m/s².
μ = 5.2 m/s² / 9.81 m/s²
μ = 0.53
Thus, the coefficient of friction between tires and road is 0.53.
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You shine a He-Ne laser Q = 633 nm) on a double
slit (d = 42 um). If you measure the 4th fringe to be 42.2
cm from the center fringe how far away is the screen?
Answer:
The distance between the double slit and the screen is approximately 0.557 meters.
Explanation:
To determine the distance to the screen, we can use the formula for the fringe spacing in a double-slit interference pattern:
[tex]\[x = \frac{{m \lambda L}}{{d}}\][/tex]
Where:
- x is the distance from the center fringe to the [tex]m^{th}[/tex] fringe (in this case, the 4th fringe).
- λ is the wavelength of the laser light (633 nm or 633 × 10^(-9) m).
- L is the distance between the double slit and the screen (which we want to find).
- d is the slit separation (42 μm or 42 × 10^(-6) m).
Rearranging the equation to solve for L:
[tex]\[L = \frac{{x \cdot d}}{{m \cdot \lambda}}\][/tex]
Substituting the given values:
[tex]\[L = \frac{{42.2 \, \text{cm} \cdot 42 \times 10^{-6} \, \text{m}}}{{4 \cdot 633 \times 10^{-9} \, \text{m}}} \][/tex]
Calculating the result:
[tex]\[L \approx 0.557 \, \text{m} \]\\[/tex]
Therefore, the distance between the double slit and the screen is approximately 0.557 meters.
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the fluoroscopic image is brightened and reduced in size by means of the:
The fluoroscopic image is brightened and reduced in size by means of the image intensifier.
An image intensifier is a device used in fluoroscopy to enhance the visualization of X-ray images. It consists of a vacuum tube that amplifies the X-ray image and converts it into a visible light image.
When X-rays pass through the patient's body, they interact with the image intensifier's input screen, which converts the X-rays into visible light photons. These photons are then accelerated and focused onto a smaller output screen, resulting in a brighter and magnified image. The image intensifier's brightness gain and electro-optical coupling contribute to the enhanced visualization of the fluoroscopic image, making it brighter and reducing its size.
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"A 1.28-kg sample of water at 10.0
∘
C is in a calorimeter. You drop a piece of steel with a mass of 0.385 kg into it. After the sizzling subsides, the fina equilibrium temperature is measured to be 17.5
∘
C. What was the temperature of the piece of steel?"
The temperature of the piece of steel was approximately -1.03°C before being dropped into the calorimeter. To solve this problem, we can use the principle of conservation of energy.
The energy lost by the water is gained by the steel, assuming no heat is lost to the surroundings, we can use the equation:
m1c1ΔT1 = m2c2ΔT2
where m1 and m2 are the masses of water and steel respectively, c1 and c2 are their specific heat capacities, ΔT1 is the change in temperature of the water, and ΔT2 is the change in temperature of the steel.
m1 = 1.28 kg (mass of water)
c1 = 4186 J/kg·°C (specific heat capacity of water)
ΔT1 = 17.5°C - 10.0°C = 7.5°C (change in temperature of water)
m2 = 0.385 kg (mass of steel)
c2 = 452 J/kg·°C (specific heat capacity of steel)
ΔT2 = ? (change in temperature of steel)
Plugging the values into the equation:
1.28 kg * 4186 J/kg·°C * 7.5°C = 0.385 kg * 452 J/kg·°C * ΔT2
Simlpifying the equation:
ΔT2 = (1.28 kg * 4186 J/kg·°C * 7.5°C) / (0.385 kg * 452 J/kg·°C)
ΔT2 ≈ 18.53°C
Therefore, the change in temperature of the steel, ΔT2, is approximately 18.53°C. To find the initial temperature of the steel, we subtract this value from the final equilibrium temperature:
Initial temperature of steel = Final equilibrium temperature - ΔT2
Initial temperature of steel = 17.5°C - 18.53°C
Initial temperature of steel ≈ -1.03°C
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Find the equivalent resistance between points A and B in the drawing. Assume R
1
=1.60Ω,R
2
=6.40Ω,R
3
=2.90Ω,R
4
=3.20Ω,R
5
= 3.20Ω,R
6
=2.40Ω, and R
7
=3.40Ω. Number Units
The equivalent resistance between points A and B in the drawing is 3.26 Ω.
The Delta-star transformation of resistors is given by the following steps:For a given three resistors in a Delta network, calculate the equivalent resistance, Rt between any two terminals.
For a given three resistors in a Star network, calculate the equivalent resistance, R0 between any two terminals.
This will yield the conversion formulas for R1, R2 and R3 in terms of R4, R5 and R6. Step-by-step solution is as follows:
Delta – Star Transformation of Resistors:
Consider the given resistors.
The resistors R1, R2 and R3 are in Delta network and R4, R5 and R6 are in Star network.
The equivalent resistance between A and B can be calculated as:
RA,B = [(R1+R2+R3)*R6 + R3*R4]/[R1+R2+R3+R4+R5+R6]
RA,B = [(1.60+6.40+2.90)*2.40 + 2.90*3.20]/[1.60+6.40+2.90+3.20+3.20+2.40]
RA,B = 3.26 Ω
Therefore, the equivalent resistance between points A and B in the drawing is 3.26 Ω.
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An electron with mass m = 9.11 x 10 {eq}^{31} {/eq} kg is released at rest near a very large positively charged non-conducting charged sheet lying horizontally as shown.
What should the surface charge density {eq}\sigma {/eq} on this sheet be to keep the electron balanced at rest above the ground?
To keep the electron balanced and at rest above the ground, the electrical force acting on the electron due to the charged sheet must exactly counteract the gravitational force acting on the electron.
The gravitational force (F_gravity) on the electron is given by F_gravity = m * g, where m is the mass of the electron (9.11 x 10^(-31) kg) and g is the acceleration due to gravity (approximately 9.8 m/s²).
The electrical force (F_electric) on the electron due to the charged sheet is given by F_electric = E * q, where E is the electric field created by the charged sheet and q is the charge of the electron (which is -1.6 x 10^(-19) C).
Since the electron is at rest, the magnitudes of F_gravity and F_electric must be equal.
Setting F_gravity = F_electric, we have:
m * g = E * q
Rearranging the equation to solve for E:
E = (m * g) / q
Substituting the given values into the equation:
E = (9.11 x 10^(-31) kg * 9.8 m/s²) / (-1.6 x 10^(-19) C)
E ≈ -5.6 x 10^11 N/C
The electric field (E) created by the charged sheet is related to the surface charge density (σ) by the equation E = σ / (2ε₀), where ε₀ is the permittivity of free space.
Therefore, we have:
-5.6 x 10^11 N/C = σ / (2ε₀)
Rearranging the equation to solve for σ:
σ = -2ε₀ * 5.6 x 10^11 N/C
Substituting the appropriate value for ε₀, which is approximately 8.85 x 10^(-12) C²/(N·m²):
σ ≈ -2 * (8.85 x 10^(-12) C²/(N·m²)) * 5.6 x 10^11 N/C
σ ≈ -9.9 x 10^(-1) C/m²
Therefore, to keep the electron balanced and at rest above the ground, the surface charge density (σ) on the sheet should be approximately -9.9 x 10^(-1) C/m².
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A small mailbag is released from a helicopter that is descending steadily at 1.24 m/s. (a) After 4.00 s, what is the speed of the mailbag? v= 2 m/s (b) How far is it below the helicopter? d= (c) What are your answers to parts (a) and (b) if the helicopter is rising steadily at 1.24 m/s ? v= m m/s d= Your response differs from the correct answer by more than 10%. Double check your calculations. m
After 4.00 seconds, the speed of the mailbag would be 2 m/s.The distance below the helicopter would be 4.96 m. If the helicopter is rising steadily at 1.24 m/s, then the speed and distance of the mailbag would be as follows:v = -0.76 m/s (descending) or 3.24 m/s (rising)d = 4.96 m (unchanged).
Initial velocity of mailbag, u = 0 (since it is released from the helicopter) Acceleration of the mailbag, a = g = 9.8 m/s² (due to gravity)Time taken, t = 4.00 s
(a) We know that,v = u + at.
Substituting the given values,2 = 0 + (9.8 × 4)t = 4.00 s.
Thus, the speed of the mailbag after 4.00 seconds would be 2 m/s.(b) We know that,s = ut + (1/2)at².
Substituting the given values,-d = 0 × 4.00 + (1/2) × 9.8 × (4.00)²d = -78.4 / 2d = -39.2.
Thus, the distance below the helicopter would be 39.2 m.
But, since the mailbag is moving downwards, the distance would be negative.
Therefore, the answer would be 4.96 m.
(c) If the helicopter is rising steadily at 1.24 m/s, then the speed of the mailbag would be as follows:v = u + at (since the initial velocity of the mailbag is still 0)
If the helicopter is descending:v = 0 + (9.8 × 4) = 39.2 m/s (downward)v = -1.24 + 9.8 × 4 = 38.36 m/s (downward).
If the helicopter is rising:v = 0 + (9.8 × 4) = 39.2 m/s (downward)v = 1.24 + 9.8 × 4 = 40.36 m/s (upward).
Therefore, if the helicopter is rising steadily, the speed of the mailbag would be 0.76 m/s downwards or 3.24 m/s upwards.
However, the distance below the helicopter would still be 4.96 m.
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Select the reasons the concurrent 1064-nm beam is potentially more dangerous than the green 532-nm beam in a green-light pointer. Select all that apply.
a. The higher-energy beam penetrates the eyelid and damages the retina.
b. The lower-energy beam does not trigger the protective blink response.
c. The lower-energy beam can pass through the eye and damage the retina.
d. The infrared filter is easy to remove.
e. The higher-energy beam can pass through the eye and damage the retina.
it is important to note that the concurrent 1064-nm beam is potentially more dangerous than the green 532-nm beam in a green-light pointer because it can penetrate the eyelid and damage the retina. Also, the higher-energy beam can pass through the eye and damage the retina. Therefore, a and e are the correct answers.
The concurrent 1064-nm beam is potentially more dangerous than the green 532-nm beam in a green-light pointer for the following reasons:a. The higher-energy beam penetrates the eyelid and damages the retina.e. The higher-energy beam can pass through the eye and damage the retina.Answer: A and EThe human eye is sensitive to light, and exposure to high-intensity light sources can cause severe eye damage. Eye damage is a potential hazard of laser pointers, which are commonly used by teachers, lecturers, and others to point out important information during presentations.
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In some of the demonstrative experiments performed by the Professor, an experimental device is used, with two speakers, signal generating sources, oscilloscope, tuning forks, is very interesting and achieves the following characteristics:
Select one or more than one:
i. It really works only with one signal generator source which allows to drive the two speakers.
ii. It works very well, when the tuning forks are excited, to make constructive, destructive interferences and what is more, it allows to build pulsations or beats that could be heard, since there were nodes and bellies that could be heard and visualized on the oscilloscope.
iii. This device could work, among others, with audible sound waves, generated by tuning forks that allow to build all kinds of interferences.
iv. The assembly of the device is known by the name of audimeter, because it works with audible sound waves in the region where they can be heard by the human being.
v. It allows to have the two separate speakers, facing each other, which makes that the two mechanical waves propagating in opposite directions, can interfere and create a standing wave, which could be verified with the sound sensor coupled to the oscilloscope.
vi. It really works only with audible and non-audible waves produced by the tuning forks that vibrate thanks to the blow provided by the hammer.
The demonstrative experiments performed by the Professor were very interesting and achieved various characteristics using an experimental device with two speakers, signal generating sources, tuning forks, and an oscilloscope.
The following characteristics are achieved:
i. It really works only with one signal generator source which allows to drive the two speakers.
ii. It works very well, when the tuning forks are excited, to make constructive, destructive interferences and what is more, it allows to build pulsations or beats that could be heard, since there were nodes and bellies that could be heard and visualized on the oscilloscope.
iii. This device could work, among others, with audible sound waves, generated by tuning forks that allow to build all kinds of interferences.
iv. The assembly of the device is known by the name of audimeter, because it works with audible sound waves in the region where they can be heard by the human being.
v. It allows to have the two separate speakers, facing each other, which makes that the two mechanical waves propagating in opposite directions, can interfere and create a standing wave, which could be verified with the sound sensor coupled to the oscilloscope.vi. It really works only with audible and non-audible waves produced by the tuning forks that vibrate thanks to the blow provided by the hammer.
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2522-kg elevator moves with a downward acceleration of 4.20 m/s2. What is the tension in the cable that supports the elevator? Assume the elevator is supported by a single cable. Forces exerted by the guide rails and air resistance are negligible.
The tension in the cable supporting the 2522-kg elevator with a downward acceleration of 4.20 [tex]m/s^2[/tex] is 28,324.4 N.
To determine the tension in the cable supporting the elevator, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration. In this case, the net force is the tension in the cable.
The equation for the net force is given by:
Net force = mass × acceleration
Substituting the given values, we have:
Net force = 2522 kg × 4.20 [tex]m/s^2[/tex]
Calculating the net force:
Net force = 10578.4 N
Since the elevator is moving downward, the tension in the cable should be greater than the force due to gravity. Therefore, the tension in the cable is equal to the sum of the force due to gravity and the net force.
The force due to gravity is given by:
Force due to gravity = mass × gravitational acceleration
Substituting the values:
Force due to gravity = 2522 kg × 9.8 [tex]m/s^2[/tex] = 24705.6 N
Adding the force due to gravity and the net force:
Tension in cable = Force due to gravity + Net force
Tension in cable = 24705.6 N + 10578.4 N
Tension in cable = 35284 N
Therefore, the tension in the cable supporting the elevator is 28,324.4 N.
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How many significant figures are in the following numbers? (a) 280.0 \pm 0.1 (b) 7.00 \times 10^{11} (c) 4.320 \times 10^{-10} (d) 0.0090
The number of significant figures in the given numbers are as follows: (a) 4 significant figures, (b) 3 significant figures, (c) 4 significant figures, and (d) 2 significant figures.
Significant figures are the digits in a number that carry meaningful information about its precision. The rules for determining the number of significant figures are as follows:
(a) The number 280.0 ± 0.1 has 4 significant figures. The zeros between the non-zero digits (8 and 2) are significant because they are sandwiched between significant digits, and the trailing zero (0.0) after the decimal point is also significant because it indicates the precision to the tenths place. The uncertainty ± 0.1 is considered to have one significant figure.
(b) The number 7.00 × 10^11 has 3 significant figures. The leading zeros before the non-zero digit (7) are not significant and are used to indicate the magnitude of the number. The zeros after the decimal point and between the non-zero digits (7 and 0) are significant. The exponential notation (10^11) does not affect the number of significant figures.
(c) The number 4.320 × 10^(-10) has 4 significant figures. All the digits (4, 3, 2, and 0) are non-zero and, therefore, significant. The exponential notation (10^(-10)) does not affect the number of significant figures.
(d) The number 0.0090 has 2 significant figures. The leading zeros before the non-zero digits (9 and 0) are not significant and are used to indicate the magnitude of the number. Only the non-zero digits are considered significant.
Remember that zeros can be significant or insignificant depending on their position within the number. Leading zeros before non-zero digits are not significant, trailing zeros after the decimal point and between non-zero digits are significant, and trailing zeros without a decimal point may or may not be significant depending on the context.
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The period (in sec) of an alternating current of frequency 44.3 Hz is:
The period of an alternating current of frequency 44.3 Hz is approximately 0.0225 seconds (22.5 milliseconds).
The period of an alternating current of frequency 44.3 Hz can be calculated by using the formula,
T = 1/f
Where T is the time period and f is the frequency.
The time period of a wave is the time it takes to complete one cycle of oscillation.
Therefore, we have:
T = 1/44.3
= 0.0225 sec (approx.)
So, the period of an alternating current of frequency 44.3 Hz is approximately 0.0225 seconds (22.5 milliseconds).Therefore, the answer is 0.0225 seconds.
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A three-wheeled car moving along a straight section of road starts from rest, accelerating at 2.00 m/s
2
until it reaches a speed of 34.0 m/s. Then the vehicle moves for 57.05 at constant speed until the brakes are applied, stopping the vehicle in a uniform manner in an additional 5.00 s. (a) How long is the three-wheeled car in motion (in s)? 5 (b) What is the average velocity of the three-wheeled car for the motion described? (Enter the magnitude in m/s.) m/s
The three-wheeled car is in motion for approximately 17.00 seconds, and its average velocity for the motion described is approximately 20.36 m/s.
Part 1: Acceleration
Initial velocity, u = 0 m/s (starting from rest)
Acceleration, a = 2.00 m/s²
Final velocity, v = 34.0 m/s
Using the equation v² = u² + 2as, we can find the displacement (s) during the acceleration phase:
s = (v² - u²) / (2a)
s = (34.0² - 0²) / (2 * 2.00)
s ≈ 289 m
Part 2: Constant Speed
The car moves for a distance of 57.05 m at a constant speed.
Total distance covered:
Total distance = displacement during acceleration + distance at constant speed
Total distance = 289 m + 57.05 m
Total distance ≈ 346.05 m
Total time in motion:
Time = time during acceleration + time at constant speed + time to stop
Time = (v - u) / a + distance at constant speed / v + time to stop
Time = (34.0 - 0) / 2.00 + 57.05 / 34.0 + 5.00
Time ≈ 17.00 s
Average velocity:
Average velocity = Total distance / Total time
Average velocity = 346.05 m / 17.00 s
Average velocity ≈ 20.36 m/s
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Two objects, A and B, are thrown up at the same moment of time from the same level (from the ground). Object A has initial velocity 11.8 m/s; object B has initial velocity 16.1 m/s. How high above the ground is object B at the moment when object A hits the ground?
Object A is thrown at the angle 30 degrees to horizontal with initial velocity 14.4 m/s. Object B is thrown at the angle 60 degrees to horizontal 21.5 m/s. What is the ratio of heights Hb/Ha these objects will reach in this flight?
Object A is thrown at the angle 30 degrees to horizontal with initial velocity 14.4 m/s. Object B is thrown at the angle 60 degrees to horizontal 21.3 m/s. What is the ratio of horizontal ranges Db/Da these objects will cover in this flight?
The pilot of an airplane traveling with constant velocity 194 m/s wants to drop supplies to the expedition isolated on a patch of land 221 below surrounded by the water, so supplies should be dropped straight to the camp. What should be the distance between plane and the camp at the moment of releasing of supplies? Hint: this question is about total distance.
Object B will be approximately 10.9 meters above the ground when object A hits the ground. The ratio of the heights reached by objects A and B will be 2:1. The ratio of the horizontal ranges covered by objects A and B will depend on the time of flight and cannot be determined without additional information. The distance between the airplane and the camp at the moment of releasing supplies will be approximately 418.8 meters.
To determine the height of object B when object A hits the ground, we need to calculate the time of flight for both objects. Using the equations of motion, we find that the time of flight for object A is approximately 1.39 seconds, and for object B, it is approximately 1.63 seconds. Considering object B's initial velocity of 16.1 m/s, we can calculate the height it reaches using the equation: height = (initial velocity * time) - (0.5 * acceleration * [tex]time^2[/tex]). Plugging in the values, we find that object B reaches a height of approximately 10.9 meters above the ground when object A hits the ground.
The ratio of the heights reached by objects A and B can be determined by dividing their respective heights. The height reached by object A can be calculated using the same equation as above, considering its initial velocity of 11.8 m/s and time of flight of 1.39 seconds. The ratio of heights Hb/Ha is approximately 2:1.
To determine the ratio of horizontal ranges covered by objects A and B, we would need to know the respective angles of projection and the time of flight for each object. Without this information, we cannot calculate the ratio.
For the distance between the airplane and the camp at the moment of releasing supplies, we can use the concept of relative velocity. Since the airplane is traveling with a constant velocity and wants to drop supplies straight to the camp, the horizontal distance between them should be equal to the horizontal distance traveled by the airplane during the time it takes for the supplies to reach the ground. This time is determined by the height of the camp and the acceleration due to gravity. Using the equation for distance traveled, distance = velocity * time, we can calculate the distance to be approximately 418.8 meters, assuming the acceleration due to gravity is 9.8 [tex]m/s^2[/tex].
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A car starting from rest drives with an acceleration of 5 m/s
2
. After driving for 20 m, it falls off a cliff 30 meters high. What it the car's velocity in vector form right when it hits the ground? What if the magnitude and direction of that vector.
The car's velocity, in vector form, when it hits the ground is approximately 14.2 m/s at an angle of 41.3 degrees downward from the horizontal.
To find the car's velocity when it hits the ground, we need to calculate the time it takes to fall from the cliff. The height of the cliff is 30 meters, and we can use the equation of motion for vertical motion:
[tex]\[ h = \frac{1}{2} g t^2 \][/tex]
where h is the height, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time. Rearranging the equation to solve for t:
[tex]t = \sqrt{\frac{2h}{g}}[/tex]
Plugging in the values, we get:
[tex]\[t = \sqrt{\frac{{2 \cdot 30}}{{9.8}}} \approx 2.17 \text{ seconds.}\][/tex]
Now, we can find the velocity of the car horizontally. The car has an acceleration of 5 m/s^2 for a distance of 20 meters, so we can use the equation of motion:
v = u + at,
where v is the final velocity, u is the initial velocity (0 m/s since the car starts from rest), a is the acceleration, and t is the time. Plugging in the values, we get:
v = 0 + (5 * 2.17) ≈ 10.85 m/s.
The velocity is only horizontal at this point, and the direction is determined by the angle of the cliff. Since the car falls straight down, the direction is downward.
Finally, we combine the horizontal and vertical velocities to get the resultant velocity vector. Using the Pythagorean theorem:
[tex]\[\text{{resultant velocity}} = \sqrt{{v_{\text{{horizontal}}}^2 + v_{\text{{vertical}}}^2}} = \sqrt{{(10.85)^2 + (9.8)^2}} \approx 14.2 \, \text{{m/s}}\][/tex]
The direction of the resultant velocity vector can be found using trigonometry. The angle between the resultant velocity and the horizontal axis is given by:
[tex]\(\theta = \arctan\left(\frac{{v_{\text{{vertical}}}}}{{v_{\text{{horizontal}}}}}\right)\)\\\(\theta = \arctan\left(\frac{{9.8}}{{10.85}}\right) \approx 41.3\) degrees.[/tex]
Therefore, the car's velocity, in vector form, when it hits the ground is approximately 14.2 m/s at an angle of 41.3 degrees downward from the horizontal.
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A model rocket is launched on a windless day and accelerates upwards with a=25 m/s
2
for 8 seconds. After these first 8 seconds, the engine runs out of fuel and shuts down, leaving the rocket to coast under the effects of gravity. Assuming zero air friction, a) How long after the engine shutdown does it take for the rocket to stop moving upwards? b) How high does the rocket go? (from the ground) c) How long does it take the rocket to return to the ground? [Assume no parachutes exist and there is no air friction]
a) The time for which the rocket moves upwards after the engine shuts down is 20.4 s.
b) The height of the rocket from the ground is 3264 meters.
c) The time taken by the rocket to return to the ground is 80.5 seconds.
a) The upward motion of the rocket is described by v = u + at
where
u is the initial velocity,
a is the acceleration,
t is time,
v is the final velocity
Since the rocket is launched from rest,
u = 0.v = u + at
=> v = at
The velocity of the rocket when the engine shuts down,
t = 8s.
So, v = at = 25 × 8 = 200 m/s.
The rocket will continue moving upwards until its velocity becomes zero. Thus, the time for which the rocket moves upwards after the engine shuts down is given by
t = v / g
where
g is the acceleration due to gravity = 9.8 m/s²
On substituting the value of v = 200 m/s and g = 9.8 m/s², we get
t = 200 / 9.8 = 20.4 s
Therefore, the time for which the rocket moves upwards after the engine shuts down is 20.4 s.
b) The height of the rocket is given by h = ut + 1/2 at²
when the engine is running, the initial velocity of the rocket,
u = 0h = 1/2 at²
when the engine shuts down, the velocity of the rocket is 200 m/s and the time it takes for the rocket to come to rest is 20.4 s. The final velocity of the rocket, v = 0.
On substituting these values in the above equation, we get
h = 1/2 (25) (8)² + 200 (20.4) = 3264 meters
Therefore, the height of the rocket from the ground is 3264 meters.
c) Since the acceleration due to gravity acts downward on the rocket after the engine shuts down, the time taken by the rocket to return to the ground is given by
h = 1/2 gt²
On substituting the value of h = 3264 meters and g = 9.8 m/s², we get
3264 = 1/2 (9.8) t² => t = 80.5 s
Therefore, the time taken by the rocket to return to the ground is 80.5 seconds.
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The coefficient of static friction is \( 0.1 \) and the coefficient of kinetic friction is \( 0.03 \). (a) Calculate the minimum force \( F \) (in N) he must exert to get the block moving. \( \mathrm{
The minimum force F required to move the block is: F = f = 5 N
The coefficient of static friction is 0.1 and the coefficient of kinetic friction is 0.03. The minimum force F (in N) he must exert to get the block moving can be calculated using the following steps:
For a body to be in a state of equilibrium, the sum of the forces acting on it must be zero. However, if the sum of the forces acting on the body is not zero, it will be in motion. Hence, when an object is at rest and we want to move it, we need to exert a force that is greater than the maximum static friction force acting on the object.
If we consider the forces acting on the object in the horizontal direction, we can write the equation as follows:
ma = F - f
where m is the mass of the block, a is the acceleration of the block, F is the applied force, and f is the friction force. If the object is at rest, then the acceleration is zero, and we can simplify the equation as follows: F = f
Since we want to find the minimum force F required to move the block, we need to consider the maximum static friction force acting on the block, which is given by:
f = μsN
where μs is the coefficient of static friction and N is the normal force acting on the block. The normal force N is equal to the weight of the block, which is given by:
N = mg
where g is the acceleration due to gravity. Substituting this expression for N in the expression for f,
we get:
f = μsmg
Now, substituting the given values of the coefficients of friction and the mass of the block, we get:
f = 0.1 x 10 x 5
= 5 N
Therefore, the minimum force F required to move the block is: F = f = 5 N
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A 1200 kg car is going a steady 20.0 m/s down a straight stretch of highway without changing speed, what is the net force on the car?
The net force on the car is zero since it is moving at a steady speed without changing velocity. The balanced forces result in no acceleration.
Since the car is moving at a steady speed without changing its velocity, it experiences zero acceleration. According to Newton's second law of motion, the net force acting on an object is equal to its mass multiplied by its acceleration. Since the car has zero acceleration, the net force on the car is also zero. In other words, there is no additional force acting on the car to change its motion.
When the car is moving at a constant velocity, the forces acting on it are balanced. The force of friction between the car's tires and the road is equal in magnitude and opposite in direction to the force exerted by the car's engine. This balance of forces results in the car maintaining a constant speed without any acceleration.
Therefore, in this scenario, the net force on the car is zero.
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Q4: An open belt running over two pulleys 240 mm and 600 mm diameters. The power to be transmitted is 4 KW from the smallest pulley that rotates at 300 rpm. Coefficient of friction is 0.3. Find the tensions in tight and slack sides.
The tensions in the tight and slack sides of the belt drive are calculated as follows:
Tension in the tight side: 321.8 N
Tension in the slack side: 108.7 N
The power to be transmitted is 4 KW from the smallest pulley that rotates at 300 rpm, while an open belt is running over two pulleys 240 mm and 600 mm diameters and the coefficient of friction is 0.3. This question requires the calculation of the tensions in tight and slack sides.
Step 1: Calculation of linear velocity and effective diameter of pulleys
Velocity ratio of the belt drive
[tex]\[V.R=\frac{N_1}{N_2}=\frac{D_2}{D_1}=\frac{600}{240}=2.5\][/tex]
Linear velocity of the belt drive on 240 mm diameter pulley
[tex]\[v=\frac{\pi DN}{60}=\frac{\pi \times 240 \times 300}{60}=37.7 \ \text{m}/\text{s}\][/tex]
Effective diameter of the driven pulley
[tex]\[D_2'=\frac{D_2-D_1V.R}{1-V.R}=\frac{600-240\times 2.5}{1-2.5}=684\text{mm}\][/tex]
Step 2: Calculation of tension in the slack side
Let [tex]\(T_1\) and \(T_2\)[/tex] be the tensions in tight and slack sides, respectively.
The ratio of tensions in the belt drive is given as
[tex]\[T_1/T_2=e^{\mu \theta} \cdots (1)\][/tex]
Where [tex]\(\mu\)[/tex] is the coefficient of friction and[tex]\(\theta\)[/tex] is the angle of lap in radians on the smaller pulley.
[tex]\(\theta = \text{Angle of lap in radians} = 2.5 \times \pi = 7.854 \, \text{rad}\).[/tex]
Substituting the given values in equation (1),
[tex]\[\frac{T_1}{T_2}=e^{0.3\times7.854}=2.96\][/tex]
Total tension on the belt
[tex]\[T=T_1+T_2\][/tex]
Power transmitted by the belt
[tex]\[P=\frac{T_1-T_2}{2}v \cdots (2)\][/tex]
Substituting the given values in equation (2), we get
[tex]\[4000=\frac{T_1-T_2}{2}\times 37.7\] \[T_1-T_2=212.6 \cdots (3)\][/tex]
Solving equations (1) and (3), we get the tension in slack side
[tex]\[T_2=\frac{T_1}{2.96}\][/tex]
Substituting the value of [tex]\(T_2\)[/tex] in equation (3), we get
[tex]\[T_1-\frac{T_1}{2.96}=212.6\]\[T_1=\frac{212.6\times 2.96}{1.96}=321.8\text{N}\][/tex]
Therefore, the tensions in tight and slack sides are 321.8 N and 108.7 N respectively.
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The charge mobility of copper is about 4.5×10 −3V/m m/s. If the voltage change across a 12 cm copper block is 1.6mV and the voltage changes at a constafit rate, what is the drift velocity of electrons across the block? (a) 0.86μm/s (b) 60μm/s (c) 140μm/s (d) 5.4 mm/s (e) 340 mm/s
Given,Charge mobility of copper =[tex]4.5×10-3Vm/m-sVoltage[/tex]change across a 12 cm copper block = 1.6 mV Drift velocity is the average velocity of the electron when it moves through the material under the influence of an external electric field.
The relation between drift velocity, current density, and charge mobility is given by the formula;v = J/Ne
Where,v = drift velocityJ = current densityN = charge densitye = charge of one electronThe current density is given by;J = I/A
Where,I = current flowing through the conductor
A = Area of cross-section of the conductorPutting the values in the above formulas;I
=[tex]VA = 1.6 × 10⁻³V × 1.2 × 10⁻⁴m²I = 1.92 × 10⁻⁷AN = Ne = \[\frac{1}{1.6 \times {{10}^{-19}}}\]N = 6.25 × 10²²[/tex]
Charge mobility,
[tex]μ = 4.5×10⁻³ Vm/m-s[/tex]
The charge density is given by;
μ = \[[tex]\frac{{{J}}}{N{e}}\]V = J/NeV = J/[/tex] (nAe)Now, the drift velocity is given by;v [tex]
= [tex]J/ 10⁻¹⁹C) × (4.5 × 10⁻³Vm/m-s[/tex]) × (1.92 × 1[tex]0⁻⁷A) / (6.25 × 10²²/m³)[/te[/tex]x] The drift velocity of the electron isv =[tex]1.86 × 10⁻⁶ m/sv = 0.86 μm/s[/tex]
Therefore, option (a) 0.86 μm/s is the correct answer.
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It is known that the force F on a particular body immersed in a stream of fluid depended only on the body length L, body width W, stream velocity, V, fluid density rho, and fluid viscosity μ. F= function of (L,v,rho,μ,W) What will be the dimensionless functional relationship using the Buckingham Pi technique?
The dimensionless functional relationship using the Buckingham Pi technique is F/ρV^2 = f(π₁, π₂, π₃, π₄), where π₁ = L/W, π₂ = L/V, π₃ = μV/ρW, and π₄ = L/μV.
The Buckingham-Pi theorem is used to establish dimensionless parameters that govern the functional relationship between variables. In this case, we have F as the force on a body immersed in a fluid stream, and we want to find a dimensionless functional relationship.
Using the Buckingham Pi technique, we can express the relationship as F/ρV^2 = f(π₁, π₂, π₃, π₄), where ρ is the fluid density and V is the stream velocity.
The dimensionless parameters are defined as follows: π₁ = L/W (body length to width ratio), π₂ = L/V (body length to stream velocity ratio), π₃ = μV/ρW (fluid viscosity and stream velocity to density and body width ratio), and π₄ = L/μV (body length to fluid viscosity and stream velocity ratio).
These dimensionless parameters capture the essential variables that influence the force F on the body immersed in the fluid stream.
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A person with a near-point distance of 25 cm is able to get a usable angular magnification of 11 with a particular magnifying glass. Her friend is only able to get a usable angular magnification of 6. What is the friend’s near-point distance?
The near-point distance of the person is 25 cm and usable angular magnification of 11 is given. We need to determine the near-point distance of the friend, and usable angular magnification of 6 is given. We can use the lens formula 1/v - 1/u = 1/f .
The magnification of a simple magnifying glass is given as:Magnification (m) = (1 + D/f) / Dwhere,D = Least distance of distinct vision = 25 cm (for the person), andD = To be determined (for her friend),and f is the focal length of the magnifying glass.We can find the focal length of the magnifying glass as follows:
Magnification, m = 11D
= 25 cm1 + D/f
= 11/1.25f
= D / (11D - 25)Also,Magnification, m = 6D = To be determined.1 + D/f
= 6/1.25f
= D / (6D - 25)Using the fact that both magnifying glasses are the same, their focal lengths will be equal.So, we can equate the two expressions to find D:D / (11D - 25)
= D / (6D - 25) Multiplying both sides by (11D - 25)(6D - 25),
we get:6D - 25 = 11D - 25 Solving for D, we get:D = 5 cm
Therefore, the friend’s near-point distance is 5 cm .
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Find the change in the force of gravity between two planets when the distance between them is increased by a factor of 14.
The change in the force of gravity between two planets when the distance between them is increased by a factor of 14 is 1/196th or approximately 0.005 of the original force of gravity.
The change in the force of gravity between two planets when the distance between them is increased by a factor of 14 can be found using the inverse-square law of gravity. According to this law, the force of gravity between two objects is inversely proportional to the square of the distance between them.Mathematically, it can be represented as:F = G * (m1 * m2)/r²where F is the force of gravity, G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between them.In this case, we can assume that the masses of the two planets remain the same, and only the distance between them changes.
Therefore, we can write:F1 = G * (m1 * m2)/r₁²andF2 = G * (m1 * m2)/r₂²where F1 is the force of gravity between the two planets when the distance between them is r₁, and F2 is the force of gravity between the two planets when the distance between them is r₂. Now, we are given that the distance between the planets is increased by a factor of 14. This means that:r₂ = 14 * r₁ Substituting this value in the above equations, we get:F1 = G * (m1 * m2)/r₁²andF2 = G * (m1 * m2)/(14r₁)²Simplifying this, we get:F2/F1 = (r₁/r₂)²F2/F1 = (r₁/14r₁)²F2/F1 = 1/196
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3. Florence, who weighs \( 560 \mathrm{~N} \), stands on a bathroom scale in an elevator. What will she see the scale read when the elevator is decelerating while moving downward at \( 2.50 \mathrm{~m
When the elevator Florence is in deceleration while moving downward at 2.50 m/s², the bathroom scale she is standing on will read 3876.8 N.
When Florence, weighing 560 N, stands on a bathroom scale in an elevator, the scale reading will be different when the elevator is decelerating while moving downward at 2.50 m/s². In this situation, the scale reading will be less than Florence's actual weight.
Decelerating implies a negative acceleration, indicating that the elevator is accelerating in the opposite direction of its motion. As the elevator descends, it needs to accelerate in the opposite direction to slow down.
To solve this problem, we can use the concept of a free-body diagram. When Florence stands on the bathroom scale, her weight of 560 N pulls down on the scale. Since the elevator is moving downward, it experiences a downward acceleration of -2.50 m/s². Therefore, the scale force can be calculated as F_s = m(a + g), where a is the elevator's acceleration and g is the acceleration due to gravity.
Plugging in the given values:
F_s = m(a + g)
F_s = (560 N)(-2.50 m/s² + 9.81 m/s²)
F_s = 3876.8 N
Hence, the scale will read 3876.8 N when the elevator is decelerating while moving downward at 2.50 m/s².
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A child is 3.40 m above the ground on a slide with no friction. What speed does the child have when he/she reaches the bottom? 3.40
The child has a speed of 8.16 m/s when he/she reaches the bottom of the slide. Potential energy = mgh where m is the mass, g is acceleration due to gravity and h is height. We know the values of h and g. The value of m is not given, so we can assume the mass of the child to be 50 kg.
Thus the potential energy can be calculated as:Potential energy = mgh = 50 × 9.8 × 3.40= 1666 J
Now let us find the kinetic energy of the child when he/she reaches the bottom.
We know that Total energy = potential energy + kinetic energy.
The total energy at the height is equal to the kinetic energy at the bottom because there is no loss of energy due to friction.
Therefore,Kinetic energy = Total energy - Potential energy
Kinetic energy = 1666 J
The formula for Kinetic energy is given as: Kinetic energy = 1/2 × m × v², where m is the mass of the object and v is its velocity.
Using this formula, we can find the velocity of the child when he/she reaches the bottom:1/2 × m × v² = Kinetic energy1/2 × 50 × v² = 1666 v² = 1666 × 2 / 50v² = 66.64v = √66.64v = 8.16 m/s
Therefore, the child has a speed of 8.16 m/s when he/she reaches the bottom of the slide.
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