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

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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.

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A toy rocket, launched from the ground, rises vertically with an acceleration of 28 m/s 2 for 9.7 s until its motor stops. Disre

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And the velocity achieved in part 1:

$v_1=(0m/s)+(28m/s^2)(9.7s)=271.6m/s$

We do the same for part 2, but now we must consider that the initial height is the one achieved in part 1 ( $y_{02}=1317.26m$ ) and its initial velocity is the one achieved in part 1 ( $v_{02}=271.6m/s$ ), now in free fall, which means with a downwards acceleration $a_{2}=-9,8m/s^2$ . For the data we have it's faster to use the formula $v_f^2=v_0^2+2ad$ , where d will be the displacement, or difference between maximum height and starting height of part 2, and the final velocity at maximum height we know must be 0m/s, so we have: