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morpeh [17]
2 years ago
11

Derive P=4T/R from fluid mechanics​

Physics
1 answer:
Nutka1998 [239]2 years ago
5 0

To Find: Diameter of soap bubble =? Solution: By Laplace's law of spherical membrane for a bubble. (Pi– Po) = 4T / r.

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A lens of focal length 15.0 cm is held 10.0 cm from a page (the object ). Find the magnification .
nevsk [136]

Answer:

Magnification, m = 3

Explanation:

It is given that,

Focal length of the lens, f = 15 cm

Object distance, u = -10 cm

Lens formula :

\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}

v is image distance

\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{15}+\dfrac{1}{(-10)}\\\\v=-30\ cm

Magnification,

m=\dfrac{v}{u}\\\\m=\dfrac{-30}{10}\\\\m=3

So, the magnification of the lens is 3.

3 0
3 years ago
What is the thermal energy of an object?
mr Goodwill [35]
The thermal energy of an object is the energy contained in the motion and vibration of its molecules. Thermal energy is measured through temperature. The energy contained in the small motions of the object's molecules can be broken up into a combination of microscopic kinetic energy and potential energy.
5 0
3 years ago
Which, at the same temperature, will cool slower on a hot day
kykrilka [37]
The answer would be b.
6 0
3 years ago
A Celcius and a Fahrenheit thermometer are dipped in boiling water. The water temperature is lowered until the Fahrenheit thermo
Sholpan [36]

Answer:

(d) 22.22 °C

Explanation:

Because in 1 Celcius is 1.8 fahrenheit apart from 2 celcius, so

when °C is 0 f is 32, when °C is 1 = 33.8 , so it rises with quantities of 1.8 fahrenheit

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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