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Inessa05 [86]
3 years ago
5

A physical change

Physics
1 answer:
Wittaler [7]3 years ago
6 0

Answer: D) occurs when water is evaporated.

Explanation:

A physical change is defined as a change in which there is alteration in shape, size etc. No new substance gets formed in these reactions.

A chemical change is defined as a change in which a change in chemical composition takes place. A new substance is formed in these reactions.

a).  iron rusts :the chemical reaction occurs by combination of iron with oxygen , thus a chemical change.

b) sugar is heated into caramel. the chemical reaction occurs by decomposition , thus a chemical change.

c) glucose is converted into energy within your cells: the chemical reaction occurs by decomposition , thus a chemical change.

d) when water is evaporated : Only the state changes and thus a physical change and cannot be reversed.

e) when propane is burned for heat: the chemical reaction as propane combines with oxygen to form carbon dioxide and water , thus a chemical change.

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Answer:

(a) 20,154.1 J

(b) 95,223.5 J

Explanation:

The expression for the moment of inertia for the uniform solid cylinder is as follows;

I= \frac{1}{2}mr^{2}

Here, I is the moment of inertia, r is the radius and m is the mass of the object.

The expression for the rotational kinetic energy is as follows;

K= \frac{1}{2}I\omega ^{2}

Here, K is the rotational kinetic energy and \omega is the angular velocity.

(a)

Calculate the moment of inertia of the smaller solid cylinder.

I= \frac{1}{2}mr^{2}

Put m= 3.44 kg and r= 0.356 m.

I= \frac{1}{2}(3.44)(0.356)^{2}

I= 0.218 kg m^{2}

Calculate the rotational kinetic energy of the smaller cylinder.

K= \frac{1}{2}I\omega ^{2}

Put I= 0.218 kg m^{2} and \omega = 430 rads^{-1}.

K= \frac{1}{2}(0.218)(430) ^{2}

K= 20,154.1 J

Therefore, the rotational kinetic energy for the smaller cylinder is 20,154.1 J.

(b)

Calculate the moment of inertia of the larger solid cylinder.

I= \frac{1}{2}mr^{2}

Put m= 3.44 kg and r= 0.775 m.

I= \frac{1}{2}(3.44)(0.775)^{2}

I= 1.03 kg m^{2}

Calculate the rotational kinetic energy of the smaller cylinder.

K= \frac{1}{2}I\omega ^{2}

Put I= 1.03 kg m^{2} and \omega = 430 rads^{-1}.

K= \frac{1}{2}(1.03)(430) ^{2}

K= 95,223.5 J

Therefore, the rotational kinetic energy for the larger cylinder is 95,223.5 J.

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