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lana [24]
3 years ago
12

A gas cylinder is connected to a manometer that contains water. The other end of the manometer is open to the atmosphere, which

in Flagstaff is 79 kPa absolute. Draw a sketch of the problem and determine the absolute static pressure in the gas cylinder if the manometer is reading 13 in H2O. Assume the water in the manometer is at 20 °C

Engineering
1 answer:
horrorfan [7]3 years ago
4 0

Answer: the absolute static pressure in the gas cylinder is 82.23596 kPa

Explanation:

Given that;

patm = 79 kPa, h = 13 in of H₂O,

A sketch of the problem is uploaded along this answer.

Now

pA = patm + 13 in of H₂O ( h × density × g )

pA= 79 + (13 × 0.0254 × 9.8 × 1000/1000)

pA = 82.23596 kPa

the absolute static pressure in the gas cylinder is 82.23596 kPa

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Farmers who use conventional tillage use compost more than other farmers.

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8 0
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As you get older your muscles grow. True or False
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Answer:

True

Explanation:

This is a true fact because From the time you are born to around the time you turn 30, your muscles grow larger and stronger.

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Pedro holds a heavy science book over his head for 10 minutes. Petro is doing work during that time. True or False
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True because he is working his arms to lift and hold the weight

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3 years ago
A car accelerates from rest with an acceleration of 5 m/s^2. The acceleration decreases linearly with time to zero in 15 s, afte
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Answer: At time 18.33 seconds it will have moved 500 meters.

Explanation:

Since the acceleration of the car is a linear function of time it can be written as a function of time as

a(t)=5(1-\frac{t}{15})

a=\frac{d^{2}x}{dt^{2}}\\\\\therefore \frac{d^{2}x}{dt^{2}}=5(1-\frac{t}{15})

Integrating both sides we get

\int \frac{d^{2}x}{dt^{2}}dt=\int 5(1-\frac{t}{15})dt\\\\\frac{dx}{dt}=v=5t-\frac{5t^{2}}{30}+c

Now since car starts from rest thus at time t = 0 ; v=0 thus c=0

again integrating with respect to time we get

\int \frac{dx}{dt}dt=\int (5t-\frac{5t^{2}}{30})dt\\\\x(t)=\frac{5t^{2}}{2}-\frac{5t^{3}}{90}+D

Now let us assume that car starts from origin thus D=0

thus in the first 15 seconds it covers a distance of

x(15)=2.5\times 15^{2}-\farc{15^{3}}{18}=375m

Thus the remaining 125 meters will be covered with a constant speed of

v(15)=5\times 15-\frac{15^{2}}{6}=37.5m/s

in time equalling t_{2}=\frac{125}{37.5}=3.33seconds

Thus the total time it requires equals 15+3.33 seconds

t=18.33 seconds

3 0
3 years ago
Bessy's calf weighed 99.19 pounds when calved on March 7th. Her calf gained an average of 1.85 pounds per day for 266 days. What
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591.3

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