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3 years ago

How much work does gravity do on the compressor during this displacement?

Physics

2 answers:

gayaneshka [121]3 years ago

6 0

Explanation and Examples

let the mass of the compressor be

mass (m):

height in x axis is (h1)

height in y axis be (h2):

Height difference: h2-h1

displacement x force:

mass x gravity x height

(m)*9.8*(height difference) = ___ J

Since gravity is forcing down, it would be negative!

Put the values that you require and get the answer.

evablogger [386]3 years ago

4 0

Answer:

W = mg × (r₂ - r₁)

Explanation:

Work done is the product of force acting and the displacement. In this case, we have to find the work done by the gravity on the compressor during displacement.

Let r₁ and r₂ are the initial and the final position of the body. So, displacement will be, r₂ - r₁.

Work done, W = F × d

Since, F = mg

W = mg × (r₂ - r₁)

Hence, the work done by the gravity on the compressor is " mg × (r₂ - r₁) Joules".

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Water flows through a cast steel pipe (k = 50 W m.K, ε = 0.8) with an outer diameter of 104mm and 2 mm wall thickness. Calculate

masha68 [24]

Answer:

The heat loss per unit length is $\frac{Q}{L} = 2981 W/m$

Explanation:

From the question we are told that

The outer diameter of the pipe is $d = 104mm = \frac{104}{1000} = 0.104 m$

The thickness is $D = 2mm = \frac{2}{1000} = 0.002m$

The temperature of water is $T = 90^oC = 90 + 273 = 363K$

The outside air temperature is $T_a = -10^oC = -10 +273 = 263K$

The water side heat transfer coefficient is $z_1 = 300 W/ m^2 \cdot K$

The heat transfer coefficient is $z_2 = 20 W/m^2 \cdot K$

The heat lost per unit length is mathematically represented as

$\frac{Q}{L} = \frac{2 \pi (T - Ta)}{ \frac{ln [\frac{d}{D} ]}{z_1} + \frac{ln [\frac{d}{D} ]}{z_2}}$

Substituting values

$\frac{Q}{L} = \frac{2 * 3.142 (363 - 263)}{ \frac{ln [\frac{0.104}{0.002} ]}{300} + \frac{ln [\frac{0.104}{0.002} ]}{20}}$

$\frac{Q}{L} = \frac{628}{0.2107}$

$\frac{Q}{L} = 2981 W/m$

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3 years ago

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Andrew [12]

A.) reference group

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Dennis_Churaev [7]

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A parallel-plate capacitor is connected to a battery. After it becomes charged, the capacitor is disconnected from the battery a

Inessa [10]

Answer:

The potential difference between the plates increases

Explanation:

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$q = CV$ (1)

where

q = charge

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$C = \frac{\epsilon_{o}A}{D}$ (2)

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3 years ago