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MArishka [77]
2 years ago
15

Need help ASAP, 1 MC

Physics
1 answer:
stepladder [879]2 years ago
3 0

Answer:

The first one is the only one that is true all the time

Explanation:

The second one may be true if friction is high enough.

The other three are false all the time

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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

8 0
3 years ago
Velocity of a car that travled 120 miles in 2 hours
Sophie [7]

60 Miles per hour 60 times 2 is 120

4 0
3 years ago
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A car travels due east 14 miles in 18 minutes. Then, the car turns around and returns to its starting point, taking an additiona
N76 [4]

Answer:

(a) The average speed is 0.85 milles/minute

(b) The average velocity is zero

Explanation:

In order to answer part (a) and (b) you have to apply the formulas for average speed and average velocity which are:

<em>-Average speed formula:</em>

v=\frac{d}{t}

where d is the total distance traveled and t is the tota time

Replacing the given values:

v=\frac{14+14}{18+15}\\v=\frac{28}{33} \\v=0.85 milles/minute

Notice that you have to replace the total distance, which is 14 milles for the go plus 14 milles for the return. The same for the total time.

<em>-Average velocity formula:</em>

V = Δx/Δt

Where V is the velocity vector, Δx is the displacement and Δt is the change in time

V= \frac{X2-X1}{t2-t1}

Where X2 is the final position and X1 is the initial position

In this case X1= 0 i and X2=0 i (i is the unit vector in the x direction). So, the displacement is zero.

Therefore, the average velocity is:

V= 0 i [milles/minute]

4 0
3 years ago
If you had a rock with a volume of 237 ml and a density of 4.52 g/ml how much mass would it have?
larisa [96]

Explanation:

Mass = Volume * Density

= 237ml * (4.52g/ml)

= 1071.2g.

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During combustion reactions, explain why the energy of the reactants must exceed the total energy of the products​
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Answer:

In these reactions the products are higher in energy than the reactants. ... This barrier is due to the fact that to make CO2 and H2O we have to break 4 carbon-hydrogen bonds and some ...

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