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AlladinOne [14]
3 years ago
13

A team of students conducts a series of experiments to investigate collisions. In the first experiment, the two carts collide wi

th each other on a smooth surface. The carts stick together and continue to move forward. In the second experiment, the two carts collide with each other on a rough surface. The carts stick together and quickly come to rest. In both experiments, the initial speeds of the carts are identical. Is there a difference in the total energy of the two experiments?
Physics
1 answer:
never [62]3 years ago
7 0

Answer: Yes, there is a difference in the total energy of the two experiments because the first two carts that were on a smooth surface had friction which causes two or more objects to stick together.

Explanation:

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In a food chain, which trophic level contains the greatest amount of energy?
evablogger [386]
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6 0
3 years ago
A 30 N rock falls from a 40 m cliff. At what point during its fall are its
vaieri [72.5K]

Answer:

<em>Both energies are equal when the rock has fallen 20 m or equivalently when it is at a height of 20 m.</em>

Explanation:

<u>Potential and Kinetic Energy</u>

The gravitational potential energy is the energy an object has due to its height above the ground. The formula is

U=mgh

Where:

m = mass of the object

g = acceleration of gravity (9.8~m/s^2)

h = height

Note we can also use the object's weight W=mg into the formula:

U=Wh

The kinetic energy is the energy an object has due to its speed:

\displaystyle K=\frac{1}{2}mv^2

Where v is the object's speed.

Initially, the object has no kinetic energy because it's assumed at rest.

The W=30 N rock falls from a height of h=40 m, thus:

U=30*40=1,200 J

Since the sum of the kinetic and potential energies is constant:

U' + K' = 1,200 J

Here, U' and K' are the energies at any point of the motion. Since both must be the same:

U' = K' = 600 J

U'=Wh'=600

Solving for h':

\displaystyle h'=\frac{600}{W}=\frac{600}{30}=20~m

Both energies are equal when the rock has fallen 20 m or equivalently when it is at a height of 20 m.

3 0
3 years ago
A 25N force is applied to a bar that can pivot around its end. The force is r=0.75 m away from the end of an angle at 0= 30. wha
Alecsey [184]

Answer:

<h2>9.375Nm</h2>

Explanation:

The formula for calculating torque τ = Frsin∅ where;

F = applied force (in newton)

r = radius (in metres)

∅ = angle that the force made with the bar.

Given  F= 25N, r = 0.75m and ∅ = 30°

torque on the bar τ  = 25*0.75*sin30°

τ = 25*0.75*0.5

τ = 9.375Nm

The torque on the bar is 9.375Nm

6 0
4 years ago
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