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DIA [1.3K]
2 years ago
8

Water is pumped from a lower reservoir to a higher reservoir by a pump that provides 20 kW of shaft power. The free surface of t

he upper reservoir is 45 m higher than that of the lower reservoir. If the flow rate of water is measured to be 0.03 m3/s, determine mechanical power that is converted to thermal energy during this process due to frictional effects.
Physics
1 answer:
kifflom [539]2 years ago
7 0

Answer:

6.5 kW

Explanation:

Input power = 20 kW = 20000 W

h = 45 m

Volume flow per second = 0.03 m^3 /s

mass flow per second = volume flow per second x density of water

                                    = 0.03 x 1000 = 30 kg/s

Output power = m g h / t = 30 x 10 x 45 = 13500 W

Power converted in form of heat = Input power - Output power

                                                       = 20000 - 13500 = 6500 W = 6.5 kW

Thus, mechanical power converted into heat is 6.5 kW.

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Which letter correctly identifies the part of the hydrologic cycle that is most directly affected by impervious building materia
masha68 [24]

Infiltration

Explanation:

The component of the hydrologic cycle affected by impervious building such as concrete and asphalt is infiltration.

  • Water infiltration is a major component of the hydrologic cycle.
  • Concretes and other materials can prevent water from going down into the earth.
  • This affects the ground water system in place.
  • It leads to increase in surface run off and might cause inundation of an area.
  • Infiltration is a very important component of water cycle.
  • It takes water to plant root and recharges groundwater systems.
  • Impervious structures takes this capability away.

learn more:

Biogeochemical cycle brainly.com/question/3509510

#learnwithBrainly

3 0
2 years ago
A 75-g bullet is fired from a rifle having a barrel 0.540 m long. Choose the origin to be at the location where the bullet begin
Mashutka [201]

The given question is incomplete. The complete question is as follows.

A 75-g bullet is fired from a rifle having a barrel 0.540 m long. Choose the origin to be at the location where the bullet begins to move. Then the force (in newtons) exerted by the expanding gas on the bullet is 14,000 + 10,000x − 26,000x^{2}, where x is in meters. Determine the work done by the gas on the bullet as the bullet travels the length of the barrel.

Explanation:

We will calculate the work done as follows.

     W = \int_{0}^{0.54} F dx

         = \int_{0}^{0.54} (14,000 + 10,000x - 26,000x^{2}) dx

         = [14000x + 5000x^{2} - 8666.7x^{3}]^{0.54}_{0}

         = 7560 + 1458 - 1364.69

         = 7653.31 J

or,      = 7.65 kJ       (as 1 kJ = 1000 J)

Thus, we can conclude that the work done by the gas on the bullet as the bullet travels the length of the barrel is 7.65 kJ.

5 0
2 years ago
In an isolated system, the total heat given off by warmer substances equals the total heat energy gained by cooler substances. N
galina1969 [7]

Answer:

The temperature of the cooler substance was close to the room temperature. Therefore, the system experienced less change

Explanation:

Generally, in a closed system containing two bodies at different temperatures, there is a flow of heat energy from the body at a higher temperature to the body at a lower temperature. The effect is more significant when there is a large temperature difference between the bodies. However, if the temperature difference is small or insignificant, the change will be less.

3 0
1 year ago
A bar of silicon is 4 cm long with a circular cross section. If the resistance of the bar is 270 ω at room temperature, what is
vova2212 [387]

solution:

consider the following data\\
length of slicon bar with circular cross section is 4cm or 0.04m\\
at room temperature resistance of the slicon bar is 270\Omega \\
represent the resistance in mathematical from\\
r=p\frac{1}{A}---1\\
where r is resistance and l is the length \\
A is cross sectional area\\
it is clear that resistivity of the silicon meterial is 6.4\times^2 \Omega.m\\
substitute 6.4\times10^2 for p,270\Omega for R and 0.04m for l i equation (1).\\270=(604\times10^2)\frac{0.04}{A}\\
rewrite the equation\\
a=(6.4\times10^2)\frac{(0.04)}{270}\\
=0.9481m^2\\
write the formula for the circular cross sectional area of silicon bar.\\
A=\pi r^2\\
substitute 0.9481 for A in the above equation\\
\pi r^2=0.9481
r^2=\frac{0.9481}{3.14},since \pi =3.14\\
0.30194\\
further simplified\\
r^2=0.30194\\
\sqrt{0.30194}\\
\cong 0.1509m\\
\cong 150.1mm

7 0
2 years ago
A wooden disk of mass m and radius r has a string of negligible mass is wrapped around it. If the disk is allowed to fall and th
Tju [1.3M]

Answer:

a = \frac{2}{3}g

T = \frac{mg}{3}

Explanation:

As the disc is unrolling from the thread then at any moment of the time

We have force equation as

mg - T = ma

also by torque equation we can say

TR = I\alpha

TR = \frac{1}{2}mR^2(\frac{a}{R})

T = \frac{1}{2}ma

Now we have

mg - \frac{1}{2}ma = ma

mg = \frac{3}{2}ma

a = \frac{2}{3}g

Also from above equation the tension force in the string is

T = \frac{1}{2}ma

T = \frac{mg}{3}

7 0
2 years ago
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