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Viefleur [7K]
3 years ago
15

Consider a locomotive and the rest of a freight train to be a single object. Suppose the locomotive is pulling the train up a hi

ll. Describe the action and reaction forces that cause the locomotive to move up the hill, such as the reaction force and gravity.
Physics
1 answer:
murzikaleks [220]3 years ago
6 0

Answer:

From lowest to highest acceleration:

3rd train

2nd train

1st train

Explanation:

The acceleration of an object can be found by using Newton's second law:

where

a is the acceleration

F is the net force on the object

m is the mass of the object

We notice that for equal values of the forces F, the acceleration a is inversely proportional to the mass, m. Therefore, greater mass means lower acceleration, and viceversa.

So, the train with lowest acceleration is the one with largest mass, i.e. the 3rd train consisting of 50 equally loaded freight cars. Then, the 2nd train has larger acceleration, since it consists of 50 empty freight cars (so its mass is smaller). Finally, the 1st train (a single empty car) is the one with largest acceleration, since it is the train with smallest mass.

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Define the difference between the rigid body problem and a single particle problem.
agasfer [191]
. In single particle problem whole mass is concentrated at a single point so it has a single displacement, single velocity and single acceleration. while, in rigid body mass is distributed
3 0
4 years ago
Please solve this question ​
lesantik [10]

Answer:

88200 Pa

it is because

height =9m

density=1000kg/m(cube)

gravity = 9.8m/s(square)

now,

P=d×g×h

= 1000×9.8×9

=88200pa

8 0
3 years ago
1. A woman driving in a car at 21 m/s pushes down on the gas pedal. 6 seconds later, she is moving at 36 m/s. What was her avera
Furkat [3]

a=Vf-Vi/t=36-21/6=15/6=2.5m/s^2

5 0
3 years ago
A van is traveling with an initial velocity of 12 m/s. The driver takes a time of 45 seconds to speed up to a velocity of 20 m/s
Rufina [12.5K]
  • Initial velocity=u=12m/s
  • Final velocity=v=20m/s
  • Time=t=45s

\\ \rm\hookrightarrow Acceleration=\dfrac{v-u}{t}

\\ \rm\hookrightarrow Acceleration=\dfrac{20-12}{45}

\\ \rm\hookrightarrow Acceleration=\dfrac{8}{45}

\\ \rm\hookrightarrow Acceleration=0.1m/s^2

Now

  • Distance=s

\\ \rm\hookrightarrow v^2-u^2=2as

\\ \rm\hookrightarrow (20)^2-12^2=2(0.1)s

\\ \rm\hookrightarrow 400-144=0.2s

\\ \rm\hookrightarrow 256=0.2s

\\ \rm\hookrightarrow s=\dfrac{256}{0.2}

\\ \rm\hookrightarrow s=1280m

4 0
3 years ago
A ball of moist clay falls 17.3 m to the ground. It is in contact with the ground for 24.0 ms before stopping. (a) What is the a
gizmo_the_mogwai [7]

Answer:

Acceleration,  767.08\ m/s^2

Explanation:

Given that,

Height from a ball falls the ground, h = 17.3 m

It is in contact with the ground for 24.0 ms before stopping.

We need to find the average acceleration the ball during the time it is in contact with the ground.

Firstly, find the velocity when it reached the ground. So,

v^2=u^2+2ah

u = initial velocity=0 m/s

a = acceleration=g

v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 17.3} \\\\v=18.41\ m/s

It is in negative direction, u = -18.41 m/s

Let a is average acceleration of the ball. Consider, v = and u = -18.41 m/s.

a=\dfrac{v-u}{t}\\\\a=\dfrac{0-18.41}{24\times 10^{-3}}\\\\a=767.08\ m/s^2

So, the average acceleration of the ball during the time it is in contact is 767.08\ m/s^2.

4 0
3 years ago
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