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Gre4nikov [31]
3 years ago
8

When a 1.0-kilogram cart moving with a speed of

Physics
2 answers:
sp2606 [1]3 years ago
5 0
Hello there.

<span>When a 1.0-kilogram cart moving with a speed of
0.50 meter per second on a horizontal surface
collides with a second 1.0-kilogram cart initially
at rest, the carts lock together. What is the speed
of the combined carts after the collision?

</span><span>(3) 0.25 m/s</span>
neonofarm [45]3 years ago
4 0
The type of collision in this system is called inelastic collision where after collision the two objects stick together. We can solve for the final velocity after collision by the expression:

v = m1v1 + m2V2 / (m1+m2)

v = 1(0.50) + (1)(0) / (1+1) = 0.25 m/s 

Therefore, option 3 is the correct answer.
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Answer:

1.F = 256 N

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Read 2 more answers
Consider the same roller coaster. It starts at a height of 40.0 m but once released, it can only reach a height of 25.0 m above
poizon [28]

Answer:

The magnitude of the frictional force between the car and the track is 367.763 N.

Explanation:

The roller coster has an initial gravitational potential energy, which is partially dissipated by friction and final gravitational potential energy is less. According to the Principle of Energy Conservation and Work-Energy Theorem, the motion of roller coster is represented by the following expression:

U_{g,1} = U_{g,2} + W_{dis}

Where:

U_{g,1}, U_{g,2} - Initial and final gravitational potential energy, measured in joules.

W_{dis} - Dissipated work due to friction, measured in joules.

Gravitational potential energy is described by the following formula:

U = m \cdot g \cdot y

Where:

m - Mass, measured in kilograms.

g - Gravitational constant, measured in meters per square second.

y - Height with respect to reference point, measured in meters.

In addition, dissipated work due to friction is:

W_{dis} = f \cdot \Delta s

Where:

f - Friction force, measured in newtons.

\Delta s - Travelled distance, measured in meters.

Now, the energy equation is expanded and frictional force is cleared:

m \cdot g \cdot (y_{1} - y_{2}) = f\cdot \Delta s

f = \frac{m \cdot g \cdot (y_{1}-y_{2})}{\Delta s}

If m = 1000\,kg, g = 9.807\,\frac{m}{s^{2}}, y_{1} = 40\,m, y_{2} = 25\,m and \Delta s = 400\,m, then:

f = \frac{(1000\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (40\,m-25\,m)}{400\,m}

f = 367.763\,N

The magnitude of the frictional force between the car and the track is 367.763 N.

7 0
3 years ago
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