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likoan [24]
4 years ago
12

The pressure at the bottom of a glass filled with water (r 5 1 000 kg/m3 ) is P. The water is poured out and the glass is filled

with ethyl alcohol (r 5 806 kg/m3 ). The pressure at the bottom of the glass is now (a) smaller than P (b) equal to P (c) larger than P (d) indeterminate.
Physics
1 answer:
bearhunter [10]4 years ago
3 0

Answer:

Option (a) is the correct answer

Explanation:

Option (a) smaller than P is the correct answer for the given question. This is because the pressure (P) at any depth 'd' from the top surface of the water with density (ρ)  is given as:

P = ρ × g × d

where,

g is the acceleration due to the gravity

thus from the above equation it can be concluded that the pressure at any depth 'd' is directly proportional to the density of the water.

thus,

in the given case the density of the water is lower than that in the first case.

Hence, the option (a) is the correct answer

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As the temperature rises the solubility of all gases in water will _________
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3 0
3 years ago
Read 2 more answers
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
[7 pts] 1. Suppose that a car starts from rest, its engine providing an acceleration of 4 ft/s2 while air resistance provides 0.
kumpel [21]

Answer:

a) Initial Value Problem

dv/dt = 4 - 0.1v

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X = 14,390 ft

Explanations:

The complete explanations of each of the sections contained in the question are in the files attached to this solution.

4 0
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