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Wewaii [24]
3 years ago
15

At a waterpark, sleds with riders are sent along a slippery, horizontal surface by the release of a large, compressed spring. Th

e spring with a force constant 37.0 N/cm and negligible mass rests on the frictionless horizontal surface. One end is in contact with a stationary wall. A sled and rider with total mass 66.0 kg are pushed against the other end, compressing the spring 0.370 m. The sled is then released with zero initial velocity.
What is the sled's speed when the spring is still compressed 0.180 m?

Physics
2 answers:
jekas [21]3 years ago
8 0

Answer:

2.2 m/s

Explanation:

<u>solution:</u>

To calculate change in stored energy at desired extension

ΔU = 1/2*k*(δx)^2

     = 1/2*3700*(0.37^2-0.180^2)

     = 201 N.m

use work energy theorem

ΔU = ΔK = 1/2*m*v^2 = 201

              = 2.2 m/s

<u>note:</u>

calculation maybe wrong but method is correct.

sweet [91]3 years ago
5 0

Explanation:

Below is an attachment containing the solution.

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A 70 ft rope hangs from a helicopter above this room. The rope has a mass per unit length of 2 lb/ft. In order to be rescued fro
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Answer:

The work done to get you safely away from the test is  2.47 X 10⁴ J.

Explanation:

Given;

length of the rope, L = 70 ft

mass per unit length of the rope, μ = 2 lb/ft

your mass, W = 120 lbs

mass of the 70 ft rope  = 2 lb/ft x 70 ft

                                         = 140 lbs.

Total mass to be pulled to the helicopter, M = 120 lbs  + 140 lbs  

                                                                       = 260 lbs

The work done is calculated from work-energy theorem as follows;

W = Mgh

where;

g is acceleration due gravity = 32.17 ft/s²

h is height the total mass is raised = length of the rope = 70 ft

W = 260 Lb x 32.17 ft/s²  x 70 ft

W = 585494 lb.ft²/s²

1 lb.ft²/s² = 0.0421 J

W = 585494 lb.ft²/s²  = 2.47 X 10⁴ J.

Therefore, the work done to get you safely away from the test is  2.47 X 10⁴ J.

4 0
2 years ago
An electron is released from rest at the negative plate of a parallel plate capacitor. The charge per unit area on each plate is
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Answer:

v=1.08\times 10^7\ m/s

Explanation:

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The charge per unit area of each plate, \dfrac{Q}{A}=1.69\times 10^{-7}\ C/m^2

Separation between the plates, d=1.75\times 10^{-2}\ m

An electron is released from rest, u = 0

Using equation of kinematics,

v^2-u^2=2ad..........(1)

Acceleration of the electron in electric field, a=\dfrac{qE}{m}............(2)

Electric field, E=\dfrac{\sigma}{\epsilon_o}............(3)

From equation (1), (2) and (3) :

v=\sqrt{\dfrac{2q\sigma d}{m\epsilon_o}}

v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\times 1.69\times 10^{-7}\times 1.75\times 10^{-2}}{9.1\times 10^{-31}\times 8.85\times 10^{-12}}}

v = 10840393.1799 m/s

or

v=1.08\times 10^7\ m/s

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3 0
3 years ago
In 2000, NASA placed a satellite in orbit around an asteroid. Consider a spherical asteroid with a mass of 1.40×1016 kg and a ra
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v is the speed

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h is the altitude of the satellite

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M is the mass of the asteroid

Solving the equation for v, we find

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v=\sqrt{\frac{(6.67\cdot 10^{-11})(1.40\cdot 10^{16}kg)}{8200 m+6000 m}}=8.11 m/s

B) 11.47 m/s

The escape speed of an object from the surface of a planet/asteroid is given by

v=\sqrt{\frac{2GM}{R+h}}

where:

G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}

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3 years ago
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