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balu736 [363]
3 years ago
15

An object is placed 60 cm in front of a concave mirror whose focal length is 20 cm. What is

Physics
1 answer:
Salsk061 [2.6K]3 years ago
6 0
Şsşsşssşggg 44466. 555
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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
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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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