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svlad2 [7]
3 years ago
6

A ball is projected into the air. Where is potential energy the greatest? A) A. B) B. C) C. D) A. and C.

Physics
2 answers:
Ksju [112]3 years ago
5 0

The point with the greatest potential energy is B.

The potential energy of an object is depends on the relative distance between the object and the ground and its mass. The higher the object is from the ground the greater the potential energy posses.

Potential energy (P.E) = mgh

Where m is the mass, g is the gravity and h is the height from the ground to where the object is.

Since the mass and gravity is constant in this case, only the height will determine the point with the greatest P.E and that point is B.

aev [14]3 years ago
4 0

Answer:

B) B

Explanation:

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If your body is dehydrated, the control center in the brain will stimulate the release of a hormone which causes the reabsorptio
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It is an example of negative feedback mechanism

Explanation:

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7 0
3 years ago
A heat engine accepts 200,000 Btu of heat from a source at 1500 R and rejects 100,000 Btu of heat to a sink at 600 R. Calculate
diamong [38]

To solve the problem it is necessary to apply the concepts related to the conservation of energy through the heat transferred and the work done, as well as through the calculation of entropy due to heat and temperatra.

By definition we know that the change in entropy is given by

\Delta S = \frac{Q}{T}

Where,

Q = Heat transfer

T = Temperature

On the other hand we know that by conserving energy the work done in a system is equal to the change in heat transferred, that is

W = Q_{source}-Q_{sink}

According to the data given we have to,

Q_{source} = 200000Btu

T_{source} = 1500R

Q_{sink} = 100000Btu

T_{sink} = 600R

PART A) The total change in entropy, would be given by the changes that exist in the source and sink, that is

\Delta S_{sink} = \frac{Q_{sink}}{T_{sink}}

\Delta S_{sink} = \frac{100000}{600}

\Delta S_{sink} = 166.67Btu/R

On the other hand,

\Delta S_{source} = \frac{Q_{source}}{T_{source}}

\Delta S_{source} = \frac{-200000}{1500}

\Delta S_{source} = -133.33Btu/R

The total change of entropy would be,

S = \Delta S_{source}+\Delta S_{sink}

S = -133.33+166.67

S = 33.34Btu/R

Since S\neq   0 the heat engine is not reversible.

PART B)

Work done by heat engine is given by

W=Q_{source}-Q_{sink}

W = 200000-100000

W = 100000 Btu

Therefore the work in the system is 100000Btu

4 0
3 years ago
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bezimeni [28]

Answer:

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Answer:

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