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Alecsey [184]
3 years ago
11

Three observers watch a train pull away from a station toward the right of the platform. Observer A is in one of the train’s car

s; Observer B is on the station platform; and Observer C is on another train traveling in the opposite direction along a parallel track. How would the observer in each frame of reference describe the motion of the train? According to Newton’s laws of motion, how would the motion in the station platform’s frame of reference change if the conductor applied the brakes to the train? Does universal gravitation affect the train? Explain why or why not.
Physics
1 answer:
blagie [28]3 years ago
3 0

Describing motion from each frame of reference:

Since observer A is on one of the train's cars, he will find that he is at rest with respect to the train even when it is pulling away because,  he is also moving with respect to the train.

Observer B on the stationary platform will observe that the train is pulling away from the station towards the right of the platform. as is described


Observer C will notice that the train is approaching him in the opposite direction at a speed which is the sum of the speeds at which both the train are traveling


If the conductor applies brakes on the train, since the platform is a stationary frame of reference. The motion will be observed as a simple decelerationg. he will observe that the force due to the brakes will cause the velocity of the train with respect to the train to decrease


Universal law of gravitation effects all objects alike . There will be a constant force of gravity acting on the train that will keep the train on its tracks. The tracks in turn exert a reaction force on the train. There will not be any affect on the train's motion as such( assuming that the train is moving along the tracks and gravitational force is exactly perpendicular)

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A book falls off a shelf that is 10.0 m tall. What is the velocity at which the book hits the ground?
Elena L [17]

Answer:

14 m/s

Explanation:

The motion of the book is a free fall motion, so it is an uniformly accelerated motion with constant acceleration g=9.8 m/s^2 towards the ground. Therefore we can find the final velocity by using the equation:

v^2 = u^2 + 2gd

where

u = 0 is the initial speed

g = 9.8 m/s^2 is the acceleration

d = 10.0 m is the distance covered by the book

Substituting data, we find

v=\sqrt{0^2 + 2(9.8 m/s^2)(10.0 m)}=14 m/s

8 0
4 years ago
Scientist have changed the model of the atom.what experimental evidence led them to change from the previous model .
Thepotemich [5.8K]

Rutherford's model of the atom (ESAAQ) Rutherford carried out some experiments which led to a change in ideas around the atom. His new model described the atom as a tiny, dense, positively charged core called a nucleus surrounded by lighter, negatively charged electrons.

7 0
4 years ago
if you have to apply 40 n of force on a crowbar to lift a 400 n rock what is the actual machanical andvatage of the crow bar
Eduardwww [97]
M.A. = 400/40

M.A. = 10

Hope this helps!
7 0
3 years ago
You have two lightweight metal spheres, each hanging from an insulating nylon thread. One of the spheres has a net negative char
kkurt [141]

Answer:attract each other

Explanation:

When two-sphere, one with a negative charge and another neutral is brought close together but do not touch then they try to attract each other.

This because of the polarization of the neutral sphere as it is placed in the vicinity of a negatively charged sphere. The negatively charged sphere will induce the positive charge in the neutral sphere and they will attract each other according to Columb law.

8 0
3 years ago
A 6.0 g marble is fired vertically upward using a spring gun. The spring must be compressed 9.4 cm if the marble is to just reac
RoseWind [281]

Answer:

a) \Delta U_{g} = 12.945\,J, b) \Delta U_{k} = 12.945\,J, c) k = 2930.059\,\frac{N}{m}

Explanation:

a) The change in the gravitational potential energy of the marble-Earth system is:

\Delta U_{g} = (0.06\,kg)\cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (22\,m)

\Delta U_{g} = 12.945\,J

b) The change in the elastic potential energy of the spring is equal to the change in the gravitational potential energy, then:

\Delta U_{k} = 12.945\,J

c) The spring constant of the gun is:

\Delta U_{k} = \frac{1}{2} \cdot k \cdot x^{2}

k = \frac{2\cdot \Delta U_{k}}{x^{2}}

k = \frac{2\cdot (12.945\,J)}{(0.094\,m)^{2}}

k = 2930.059\,\frac{N}{m}

4 0
3 years ago
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