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3 years ago

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Which of these correctly describes whether a girl holding a ball in the same position is doing work on the ball?

1 answer:

aksik [14]3 years ago

4 0

Answer:

The girl is doing work on the ball because the energy in her muscles changed, even though the ball is not displaced.

Explanation:

The complete question is...

Which of these correctly describes whether a girl holding a ball in the same position is doing work on the ball?

-The girl is doing work on the ball because the energy of the ball changed, even though it is not displaced.

-The girl is doing work on the ball because the energy in her muscles changed, even though the ball is not displaced.

-The girl is doing no work on the ball because the ball is not displaced.

-The girl is doing no work on the ball because she is exerting a net force on the ball.

Holding up a ball costs energy, which is used to counter the work that would have otherwise be done on the ball by gravity. Although no physical distance is moved, we should consider the fact that by holding the ball, the girls hand exerts physical force to hold the ball in place. Also, there is a potential gravitational work on the ball due to gravity, but the force exerted by the girls hand does an equivalent of this gravito-potential work in order to counter it and hold the ball in place. All these activities eventually lead to a change in energy in her hand muscle to show that energy is expended.

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The linear impulse delivered by the hit of a boxer is 202 N · s during the 0.244 s of contact. What is the magnitude of the aver

zlopas [31]

Answer: Magnitude of the average force exerted on the glove by the other boxer is 827.86 N (approximately 828 N).

Explanation: Impulse is defined as the force acting on an object for a short period or interval of time.

Mathematically it is given by the relation:

Impulse = Force $\times$ Time

According to the numerical values given in the question, I = 202 Ns and T = 0.244 s

So, Force F = $\frac{Impulse}{Time}$ = $\frac{202}{0.244}$ = 827.86 N

Magnitude of the average force exerted on the glove by the other boxer is 827.86 N (approximately 828 N).

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3 years ago

The precision of a laboratory instrument is ± 0.05 g. The accepted value for your measurement is 7.92 g. Which measurements are

skelet666 [1.2K]

Answer:

7.89 7.91

Explanation:

The ranges of measurement lie between 7.92-0.05 and 7.92+0.05

7.87g and 7.97g

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3 years ago

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zhenek [66]

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A packing crate rests on a horizontal surface. It is acted on by three horizontal forces: 600 N to the left, 200 N to the right,

egoroff_w [7]

Answer:

The resultant force would (still) be zero.

Explanation:

Before the 600-N force is removed, the crate is not moving (relative to the surface.) Its velocity would be zero. Since its velocity isn't changing, its acceleration would also be zero.

In effect, the 600-N force to the left and 200-N force to the right combines and acts like a 400-N force to the left.

By Newton's Second Law, the resultant force on the crate would be zero. As a result, friction (the only other horizontal force on the crate) should balance that 400-N force. In this case, the friction should act in the opposite direction with a size of 400 N.

When the 600-N force is removed, there would only be two horizontal forces on the crate: the 200-N force to the right, and friction. The maximum friction possible must be at least 200 N such that the resultant force would still be zero. In this case, the static friction coefficient isn't known. As a result, it won't be possible to find the exact value of the maximum friction on the crate.

However, recall that before the 600-N force is removed, the friction on the crate is 400 N. The normal force on the crate (which is in the vertical direction) did not change. As a result, one can hence be assured that the maximum friction would be at least 400 N. That's sufficient for balancing the 200-N force to the right. Hence, the resultant force on the crate would still be zero, and the crate won't move.

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3 years ago

Consult Interactive Solution 10.37 to explore a model for solving this problem. A spring is compressed by 0.0647 m and is used t

padilas [110]

Answer:

$\omega=32.14\ rad/s$

Explanation:

Given that,

The compression in the spring, x = 0.0647 m

Speed of the object, v = 2.08 m/s

To find,

Angular frequency of the object.

Solution,

We know that the elation between the amplitude and the angular frequency in SHM is given by :

$v=\omega\times A$

A is the amplitude

In case of spring the compression in the spring is equal to its amplitude

$\omega=\dfrac{v}{A}$

$\omega=\dfrac{2.08\ m/s}{0.0647\ m}$

$\omega=32.14\ rad/s$

So, the angular frequency of the spring is 32.14 rad/s.

4 0

3 years ago