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Vaselesa [24]
3 years ago
9

Three identical boxcars are coupled together and are moving at a constant speed of 28.0 m/sm/s on a level, frictionless track. T

hey collide with another identical boxcar that is initially at rest and couple to it, so that the four cars roll on as a unit. Friction is small enough to be neglected.
Required:
a. What is the speed of the four cars?
b. What percentage of the kinetic energy of the boxcars is dissipated in the collision?
c. What happened to this energy?
Physics
1 answer:
frez [133]3 years ago
4 0

Answer:

A) v = 21 m /s

B) 25%

C) ) on collision, this energy in the question appears in the form of the following namely; sound energy, heat energy etc

Explanation:

A) Let m be the mass of any of the cars

Thus:

mass of the three cars = 3m

Formula for kinetic energy = ½mv²

Thus, Kinetic energy of 3 identical and coupled cars = ½ x 3m x 28² = 1176 m

KE = 1176 m

Now mass of 4 coupled cars together = 4m

From conservation of linear momentum, we can find the speed of the four cars. Thus;

m1v1 = m2v2

We are told that the 3 coupled moved together with a speed of 28 m/s

Thus;

4m × v = 3m × 28

v = 3m x 28 / 4m

v = 21 m /s

B) from earlier, we saw the formula for kinetic energy. Thus, kinetic energy with of mass of 4 coupled cars together. Thus;

K = ½ x 4m x 21²

K = 882m

Loss of kinetic energy

ΔK = 1176 m - 882 m

ΔK = 294 m

Therefore, percentage of loss is;

%loss = (294 / 1176 ) x 100

%loss = 25 %

C) on collision, this energy in the question appears in the form of the following namely; sound energy, heat energy etc

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Jim is driving a 2268-kg pickup truck at 19 m/s and releases his foot from the accelerator pedal. The car eventually stops due t
lord [1]

Answer:

The initial kinetic energy of the truck is 409374 J

Explanation:

This problem can be solved in two ways. Let´s solve it first in the easiest way.

The kinetic energy is calculated using this equation:

E = 1/2 · m · v²

Where:

E = kinetic energy

m = mass

v = velocity

Then, the kinetic energy of the truck will be:

E = 1/2 · 2268 kg · (19 m/s)² = 409374 J

And that´s it.

But we can complicate it a bit:

The kinetic energy is the work needed to move an object from rest to a desired velocity. If the object is moving, the work needed to stop it must be of the same magnitude as its kinetic energy (in the opposite direction to the movement).

The equation for work is:

W = F · d

Where:

W= work

F = force

d = distance

We know the magnitude of the force applied to the truck, but we do not know for how much distance that force was applied. The distance can be calculated using the equation for the position of an object moving in a straight line:

x = x0 + v0 · t + 1/2 · a · t²

where

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

But we still do not know the time nor the acceleration.

The acceleration can be obtained from the equation of force:

F = m · a

Where

F = force

m = mass

a = acceleration

Then:

900 N = 2268 kg ·a

a = 900 N /2268 kg = 0.397 m/s²

Now, we can calculate the time needed for the truck to stop. We know that at the final time, the velocity is 0. Then, we can use the equation for velocity to obtain that time:

v = v0 + a · t

Where:

v = velocity at time t

v0 = initial velocity

a = acceleration

t = time

Then:

v = 19 m/s - 0.397 m/s² · t

0 = 19 m/s - 0.397 m/s² · t

-19 m/s / -0.397 m/s² = t (acceleration is negative because it is opposite to the direction of the movement)

t = 47.86 s (The truck stoped at 47.86 s after releasing the foot from the accelerator pedal)

With the time and acceleration, we can calculate the traveled distance.

x = 0 m + 19 m/s · 47.86 s - 1/2 · 0.397 m/s² · (47.86s)²

x = 454.66 m (without rounding the acceleration nor the time, the value will be 454.86 m)

Now, we can calculate the work done to stop the truck which will be of the same magnitude as the kinetic energy:

W = 900 N · 454.66 m = 409194 J

(if you do all the calculations without rounding, you will get the same value as we calculated above using the equation of kinetic energy, 409374 J).

3 0
3 years ago
Question:
valkas [14]

Answer:

the answer is true.

Explanation:

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5 0
3 years ago
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At atmospheric pressure, what is the characteristic boiling point of water, in degrees Celsius?
xxMikexx [17]
Water boils at 100 degrees Celsius at atmospheric pressure.
4 0
4 years ago
Question above! Science 8 please help!
Darina [25.2K]
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3 years ago
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An Olympic runner competing in a long-distance event finishes with a time of 2 hours , 15 minutes , and 23 seconds . The event h
Talja [164]

Answer:

The average speed of the runner is 18.2 km/h.

Explanation:

Hi there!

The average speed is calculated as the traveled distance over time:

speed = traveled distance / elapsed time to cover that distance

We have to find the speed in km/h, so let´s start converting the time into hours:

23 s · (1 min/60 s) = 0.38 min      

15.38 min · (1 h / 60 min) = 0.26 h

Total time: 0.26 h + 2 h = 2.26 h

The distance traveled in km is:

25.5 mi · (1.61 km/ 1 mi) = 41.1 km

Then, the average speed will be:

speed = 41.1 km / 2.26 h = 18.2 km/h

The average speed of the runner is 18.2 km/h.

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