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2 years ago

A piston having a diameter of 5.48 inches and a length of 9.50 in slides downward with a

Engineering

1 answer:

inna [77]2 years ago

4 0

Answer:

$V = 4.585 \times 10^{-3} ft/s$

Explanation:

surface Area is given as $= \pi dL$

here , d is diameter of piston

L is distance travelled by the piston

$a = \pi 5.48\times 9.5$

velocity of piston

$W = \mu A \frac{V}{t}$

where,

w is weight of piston, $\mu$ dynamic viscosity of fluid, t is thickness of fluid

$0.5 = 0.016 \times \frac{163.55}{12^2} \frac{V}{\frac{0.002}{12}}$

$0.5 = 109.033 V$

SOLVING FOR V

$V = 4.585 \times 10^{-3} ft/s$

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Which of the following statements about glycogen metabolism is FALSE?

Ratling [72]

Answer:

Glycogen is the primary energy source for muscle and liver cells.

Explanation:

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Glycogen is not as reduced as fatty acids are and consequently not as energy rich. Why do animals store any energy as glycogen? Why not convert all excess fuel into fatty acids? Glycogen is an important fuel reserve for several reasons. The controlled breakdown of glycogen and release of glucose increase the amount of glucose that is available between meals. Hence, glycogen serves as a buffer to maintain blood-glucose levels. Glycogen's role in maintaining blood-glucose levels is especially important because glucose is virtually the only fuel used by the brain, except during prolonged starvation. Moreover, the glucose from glycogen is readily mobilized and is therefore a good source of energy for sudden, strenuous activity. Unlike fatty acids, the released glucose can provide energy in the absence of oxygen and can thus supply energy for anaerobic activity.

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3 years ago

The most important rating for batteries is the what

kifflom [539]

Answer:

I'm completely sure that the answer is: The most important rating for batteries is the ampere-hour rating. Ampere-hour is the battery discharge rating. It's used as a measure of charge in your device. It indicates how long your device will work without charging.

Explanation:

Hope this helped!

7 0

2 years ago

If a signal is transmitted at a power of 250 mWatts (mW) and the noise in the channel is 10 uWatts (uW), if the signal BW is 20M

Bess [88]

Answer:

C = 292 Mbps

Explanation:

Given:

- Signal Transmitted Power P = 250mW

- The noise in channel N = 10 uW

- The signal bandwidth W = 20 MHz

Find:

what is the maximum capacity of the channel?

Solution:

-The capacity of the channel is given by Shannon's Formula:

C = W*log_2 ( 1 + P/N)

- Plug the values in:

C = (20*10^6)*log_2 ( 1 + 250*10^-3/10)

C = (20*10^6)*log_2 (25001)

C = (20*10^6)*14.6096

C = 292 Mbps

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2 years ago

As Becky was driving "Old Betsy," the family station wagon, the engine finally quit, being worn out after 171,000 miles. It can

andriy [413]

Answer:

1) 4.361 x 10 raised to power 8 revolutions

2) 1.744 x 10 raised to power 9 firings

3) 2.18 x 10 raised to power 8 intake strokes

Explanation:

The step by step explanation is as shown in the attachment

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3 years ago

A Pelton wheel is supplied with water from a lake at an elevation H above the turbine. The penstock that supplies the water to t

gayaneshka [121]

Answer:

Following are the proving to this question:

Explanation:

$\frac{D_1}{D} = \frac{1}{(2f(\frac{l}{D}))^{\frac{1}{4}}}$

using the energy equation for entry and exit value :

$\to \frac{p_o}{y} +\frac{V^{2}_{o}}{2g}+Z_0 = \frac{p_1}{y} +\frac{V^{2}_{1}}{2g}+Z_1+ f \frac{l}{D}\frac{V^{2}}{2g}$

where

$\to p_0=p_1=0\\\\\to Z_0=Z_1=H\\\\\to v_0=0\\\\AV =A_1V_1 \\\\\to V=(\frac{D_1}{D})^2 V_1\\\\\to V^2=(\frac{D_1}{D})^4 V^{2}_{1}$

$= (\frac{1}{(2f (\frac{l}{D} ))^{\frac{1}{4}}})^4\ V^{2}_{1}\\\\$

$= \frac{1}{(2f (\frac{l}{D}) )} \ V^{2}_{1}\\$

$\to \frac{p_o}{y} +\frac{V^{2}_{o}}{2g}+Z_0 =\frac{p_1}{y} +\frac{V^{2}_{1}}{2g}+Z_1+ f \frac{l}{D}\frac{V^{2}}{2g} \\\\$

$\to 0+0+Z_0 = 0 +\frac{V^{2}_{1} }{2g} +Z_1+ f \frac{l}{D} \frac{\frac{1}{(2f(\frac{l}{D}))}\ V^{2}_{1}}{2g} \\\\\to Z_0 -Z_1 = +\frac{V^{2}_{1}}{2g} \ (1+f\frac{l}{D}\frac{1}{(2f(\frac{l}{D}) )} ) \\\\\to H= \frac{V^{2}_{1}}{2g} (\frac{3}{2}) \\\\\to \frac{V^{2}_{1}}{2g} = H(\frac{3}{2})$

L.H.S = R.H.S

7 0

2 years ago