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Dmitry_Shevchenko [17]
4 years ago
5

A(n) 30 kg boy rides a roller coaster. The acceleration of gravity is 9.8 m/s 2 . With what force does he press against the seat

when the car moving at 7.3 m/s goes over a crest whose radius of curvature is 15 m? Answer in units of N.
Physics
1 answer:
forsale [732]4 years ago
5 0

Answer:

Force, F = 187.42 N

Explanation:

It is given that,

Mass of boy, m = 30 kg

Acceleration due to gravity, a=9.8\ m/s^2

Radius of curvature of the roller coaster, r = 15 m

Speed of the car, v = 7.3 m/s

The force acting on the boy are force of gravity and the centripetal force. The net force acting on him is as follows :

F=mg-\dfrac{mv^2}{r}

F=m(g-\dfrac{v^2}{r})

F=30\times (9.8-\dfrac{(7.3)^2}{15})

F = 187.42 N

So, he press against the seat with a force is 187.42 N. Hence, this is the required solution.

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The Latent heat of fusion is also known as the Enthalpy of Fusion. It is a kind pf change in enthalpy.

<h3>Brief explanation of Latent Heat of Fusion? </h3>

The latent heat of fusion, is the change in enthalpy that takes place when energy, typically heat, is provided to a specific quantity of a substance to causes it to change from a solid state to a liquid state at a constant pressure.

For instance, 1 kilogram of ice melts at 0 °C at a variety of pressures, absorbing 333.55 kJ of energy without causing a change in temperature.

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<h3>Calculation</h3>

The formula of latent heat of fusion is L= q/m

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So, the latent heat of fusion is 4 j/kg

To know more about latent Heat of Fusion visit:

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