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nexus9112 [7]
3 years ago
12

Weather balloons are filled with only a small amount of helium because the ____________________ of the balloon will increase as

the air pressure decreases at higher altitudes.
Physics
1 answer:
Shkiper50 [21]3 years ago
7 0
Weather balloons are filled with only a small amount of helium because the __Volume__. of the balloon will increase as the air pressure decreases at higher altitudes.
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Assuming that Bernoulli's equation applies, compute the volume of water ΔV that flows across the exit of the pipe in 1.00 s . In
OLEGan [10]

Answer:

discharge rate (Q) = 0.2005 m^{3} / s

Explanation:

if you read the question you would see that some requirements are missing, by using search engines, you can get the complete question as stated below:

Water flows steadily from an open tank as shown in the figure. (Figure 1) The elevation of point 1 is 10.0m , and the elevation of points 2 and 3 is 2.00 m . The cross-sectional area at point 2 is 4.80x10-2m ; at point 3, where the water is discharged, it is 1.60x10-2m. The cross-sectional area of the tank is very large compared with the cross-sectional area of the pipe. Part A Assuming that Bernoulli's equation applies, compute the volume of water DeltaV that flows across the exit of the pipe in 1.00 s . In other words, find the discharge rate \Delta V/Delta t. Express your answer numerically in cubic meters per second.

solution:

time = 1 s

elevation of point 1 (z1) = 10 m

elevation of point 2 (z2) = 2 m

elevation of point 3 (z3) = 2 m

cross section area of point 2 = 4.8 x 10^{2} m

cross section area of point 3 = 1.6 x 10^{2} m

g

acceleration due to gravity (g) = 9.8 m/s^{2}

find the discharge rate at point 3 which is the exit pipe.

discharge rate (Q) = A3 x V3

where A3 is the cross sectional area at point 3 and V3 is the velocity of the fluid and can be gotten by applying Bernoulli's equation below

\frac{P1}{ρg} +  \frac{V1^{2} }{2g} + Z1 =  \frac{P3}{ρg} + \frac{V3^{2} }{2g} + Z3

pressure at point 1 (P1) is the same as pressure at point 3 (P3), and at point 1, the velocity (V1) = 0. therefore the equation now becomes

\frac{P1}{ρg} + Z1 =  \frac{P1}{ρg} + \frac{V3^{2} }{2g} + Z3

Z1 = \frac{V3^{2} }{2g} + Z3

V3 = \sqrt{2g(Z1-Z3)}

V3 = \sqrt{2 x 9.8 x (10 - 3)}

V3 = 12.53 m/s

discharge rate (Q) = A3 x V3 = 1.6 x 10^{-2} x 12.53

discharge rate (Q) = 0.2005 m^{3} / s

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