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8090
1 month ago
13

Shows a closed tank holding air and oil to which is connected a U-tube mercury manometer and a pressure gage. Determine the read

ing of the pressure gage, in lbf/in^2. (gage). The densities of the oil and mercury are 55 and 845, respectively, each in lb/ft^3. Let g=32.2ft/s^2.
Engineering
1 answer:
Daniel [329]1 month ago
3 0
To determine the pressure gauge reading, we note that the closed tank filled with oil and air has a pressure of P₁, where the oil's pressure at a specific height in the U-tube of mercury is p₁gh₁, and the mercury pressure acting on the air in the gauge is p₂gh₂. Thus, we aim to find P₁-P₂, assuming equal heights: h₁=h₂=h.
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A 10-ft-long simply supported laminated wood beam consists of eight 1.5-in. by 6-in. planks glued together to form a section 6 i
pantera1 [306]
Point B is where Q_B = 101.25 \ in^3 shows the highest Q value at section a–a. The missing diagram that was supposed to accompany this question can be found in the attached file. Based on the given information, we must identify the point with the greatest Q value at section a–a. To achieve this, we're working with the attached image. The image reveals that there are 8 blocks stacked vertically on the right-hand side, totaling 12 inches; thus, each block measures 1.5 inches in height. Additionally, the blocks are divided into sections marked as A, B, C, and D. For point A, we use the formula Q representing the moment of Area A. For point B, we set Q as the moment of Area B, and similarly for points C and D. However, point D has no area, resulting in a moment of 0.
3 0
1 month ago
Retype the below code. Fix the indentation as necessary to make the program work. if 'New York' in temperatures: if temperatures
Daniel [329]

Answer: Below is the correctly formatted code:

if 'New York' in temperatures:

if temperatures['New York'] > 90:

   print('The city is melting!')

else:

   print('The temperature in New York is', temperatures['New York'])

else:

   print('The temperature in New York is unknown.')

Explanation:

This is the entire program:

temperatures = {

   'Seattle': 56.5,

   'New York': 105,

   'Kansas City': 81.9,

   'Los Angeles': 76.5

}

if 'New York' in temperatures:

if temperatures['New York'] > 90:

   print('The city is melting!')

else:

   print('The temperature in New York is', temperatures['New York'])

else:

   print('The temperature in New York is unknown.')

The output will be:

The city is melting!

This program mainly comprises nested if statements. The first check determines if 'New York' exists within the temperature data. When confirmed, it moves to another if statement to see if New York's temperature exceeds 90. If true, it outputs the message, "The city is melting!". If false, it presents the actual temperature of New York. If neither condition is satisfied, it defaults to showing that the temperature is unknown. Given that New York's temperature is 105, the output reflects that the city is melting!

Proper indentation is crucial in Python, as it distinguishes code blocks without needing brackets, enhancing readability. Observe how the nested if statements are structured within the outer if statement. The else statements align appropriately beneath their respective if declarations, and print statements follow the necessary indentation.

4 0
3 months ago
A test machine that kicks soccer balls has a 5-lb simulated foot attached to the end of a 6-ft long pendulum arm of negligible m
choli [298]

Answer:

a) v₂ = 26.6 ft/s

b) v₂ = 31.9 ft/s

Explanation:

a) Applying the principle of energy conservation for the soccer arm from the point of impact to its lowest point:

mgh=\frac{1}{2} mv^{2} \\v=\sqrt{2gh} =\sqrt{2*32.2*6} =19.66ft/s

The ball's velocity in the tangential direction is given by:

v₂*sinθ = v₁*sin30

With the values:

if v₁ = 0

θ = 0º

The restitution coefficient is:

e=\frac{v_{2}*cosO-v_{2}cos30 }{v*cos30-(-v_{1}*cos30 )} \\0.8=\frac{v_{2}cos0-v_{2}*cos30 }{19.66*cos30-(0*cos30)} \\v_{2} =\frac{v_{2}-13.6 }{cos30}

where O=θ

The aggregate momentum equation is:

mv-mv_{1} =mv_{2} +mv_{2} cos(O+30)\\, where O = θ

mv-mv_{1} =m(\frac{v_{2}-13.6 }{cos30} )+mv_{2} cos(O+30)

When we incorporate gravitational acceleration:

mgv-mgv_{1} =mg(\frac{v_{2}-13.6 }{cos30} )+mgv_{2} cos(O+30)\\5*19.66-1*0=5*(\frac{v_{2}-13.6 }{cos30} )+1*v_{2} cos(O+30)

Isolating v₂ results in:

v₂ = 26.6 ft/s

b) To find the ball's velocity, we utilize:

v₂ * sinθ = v₁ * sin30

if v₁ = 10 ft/s

then v₂ * sinθ = 10 * sin30

thus sinθ = 5/v₂

The restitution coefficient remains:

e=\frac{v_{2}*cosO-v_{2}cos30 }{v*cos30-(-v_{2}*cos30) } \\0.8=\frac{v_{2}cosO-v_{2}cos30 }{19.66*cos30-(-10*cos30)} \\v_{2} =\frac{v_{2}cos30-20.55}{cos30}

where O = θ

mgv-mgv_{1} =mgv_{2} +mgv_{2} cos(O+30)\\5*19.66-1*0=5v_{2} +v_{2} cos(O+30)\\98.3=5v_{2}+v_{2} cosOcos30-v_{2} sinOsin30

Solving for v₂ yields:

v₂ = 31.9 ft/s

6 0
2 months ago
The water level in a large tank is maintained at height H above the surrounding level terrain. A rounded nozzle placed in the si
pantera1 [306]

Answer:

Explanation: Kindly refer to the attached files for a detailed solution process.

Take note that the diagram file is the first one and is positioned accordingly.

8 0
1 month ago
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