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guajiro [1.7K]
1 year ago
10

A large tank is filled with water to a depth of 38 m. An opening is located 14 m

Physics
1 answer:
Nataly_w [17]1 year ago
5 0

When a large tank is filled with water to a depth of 38 m. An opening is located 14 m above the bottom of the tank then the water would  emerge from the opening at a velocity of 21.70 m/s

<h3>What is mechanical energy?</h3>

Mechanical energy is the combination of all the energy in motion represented by total kinetic energy and the total potential energy stored energy in the system which is represented by total potential energy.

As total mechanical energy ias s the sum of all the kinetic as well potential energy stored in the system.

ME = KE + PE

The total kinetic energy

KE = 1/2*m*v²

Similarly, the total potential energy

PE = m*g*h

As given in the problem When a large tank is filled with water to a depth of 38 m. An opening is located 14 m above the bottom of the tank

height of the water above the opening

h= 38-14

h= 24 m

The potential energy of water above the opening is getting converted into the kinetic energy

m*g*h =  1/2*m*v²

v = √(2*g*h)

By substituting the respective values of the height and gravity

v = √(2*9.81*24)

v = 21.70 m/s

Thus, when water is poured into a sizable tank until it reaches a 38 m depth. If a hole were to be made at a height of 14 m above the tank's bottom, water would rush out of it at a speed of 21.70 m/s.

Learn more about mechanical energy from here

brainly.com/question/12319302

#SPJ1

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You wish to cool a 1.83 kg block of tin initially at 88.0°C to a temperature of 57.0°C by placing it in a container of kerosene
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Answer:

0.273 liters are needed to accomplish this task without boiling.

Explanation:

The minimum boiling point of kerosene is 150\,^{\circ}C. According to this question, we need to determine the minimum volume of liquid such that heat received is entirely sensible, that is, with no phase change.

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\rho_{k}\cdot V_{k}\cdot c_{k}\cdot (T-T_{k,o}) = m_{t}\cdot c_{t}\cdot (T_{t,o}-T) (1)

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V_{k} - Volume of kerosene, measured in cubic meters.

c_{k}, c_{t} - Specific heats of the kerosene and tin, measured in joule per kilogram-Celsius.

T_{k,o}, T_{t,o} - Initial temperatures of kerosene and tin, measured in degrees Celsius.

T - Final temperatures of the kerosene-tin system, measured in degrees Celsius.

Please notice that the block of tin is cooled at the expense of the temperature of the kerosene until thermal equilibrium is reached.

From (1), we clear the volume of kerosene:

V_{k} = \frac{m_{t}\cdot c_{t}\cdot (T_{t,o}-T)}{\rho_{k}\cdot c_{k}\cdot (T-T_{k,o})}

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V_{k} = \frac{(1.83\,kg)\cdot \left(218\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (88\,^{\circ}C-57\,^{\circ}C)}{\left(820\,\frac{kg}{m^{3}} \right)\cdot \left(2010\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (57\,^{\circ}C-24\,^{\circ}C)}

V_{k} = 2.273\times 10^{-4}\,m^{3}

V_{k} = 0.273\,L

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