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saul85 [17]
3 years ago
5

Which of the following items has the most inertia while at rest?

Physics
2 answers:
dusya [7]3 years ago
8 0
Dog

Because the more mass an object has, the greater it’s inertia.

Mark me Brianliest please:)
AlexFokin [52]3 years ago
7 0

Out of the given options, dog is the correct one as the object with greater mass will have the most inertia while at rest with no external force acting on it.

Answer: Option D

<u>Explanation:</u>

According to the first law of Newton about the force, all objects show resistant in the change in their state and this is what’s called the law of inertia. Now the fact interprets why do heavier objects show more inertia rather than that lighter ones.

This is because inertia has a sole dependent quantity which is mass. The heavier the thing is, the greater its inertia will be. This is why it’s tougher to move heavier objects as compared to the lighter ones.

Try it with a light feather and a dog sleeping on the street. It’s much easier to get the feather displaced than to move the dog from its position. This is because of the mass they have. So, here in this case, it’s a dog that will have the most inertia while at rest with no external force acting on it.

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A single Oreo cookie provides 53 kcal of energy. An athlete does an exercise that involves repeatedly lifting (without accelerat
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Approximately 325 (rounded down,) assuming that g = 9.81\; {\rm N \cdot kg^{-1}}.

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Ensure that all quantities involved are in standard units:

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\begin{aligned} & 53\; {\rm kCal} \times \frac{1\; {\rm kJ}}{4.184\; {\rm kCal}} \times \frac{1000\; {\rm J}}{1\; {\rm kJ}} \approx 2.551 \times 10^{5}\; {\rm J} \end{aligned}.

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Energy required to lift the weight by \Delta h = 0.2\; {\rm m} without acceleration:

\begin{aligned} W &= m\, g\, \Delta h \\ &= 100\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \times 0.2\; {\rm m} \\ &= 196\; {\rm N \cdot m} \\ &= 196\; {\rm J} \end{aligned}.

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\begin{aligned} \text{Energy Input} &= \frac{\text{Useful Work Output}}{\text{Efficiency}} \\ &= \frac{196\; {\rm J}}{0.25} \\ &=784\; {\rm J}\end{aligned}.

Divide 2.551 \times 10^{5}\; {\rm J} by 784\; {\rm J} to find the number of times this weight could be lifted up within that energy budget:

\begin{aligned} \frac{2.551 \times 10^{5}\; {\rm J}}{784\; {\rm J}} &\approx 325 \end{aligned}.

Increasing the efficiency (the denominator) would reduce the amount of energy input required to achieve the same amount of useful work. Thus, the same energy budget would allow this weight to be lifted up for more times.

4 0
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