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A boy throws a ball of mass 0.22 kg straight upward with an initial speed of 29 m/s. When the ball returns to the boy, its speed

maksim [4K]

Answer:

The work is -67.76 J

Explanation:

The law of conservation of energy is considered one of one of the fundamental laws of physics and states that the total energy of an isolated system remains constant. except when it is transformed into other types of energy.

This is summed up in the principle that energy can neither be created nor destroyed in the universe, only transformed into other forms of energy.

In this case you must calculate the loss of kinetic energy. This loss is actually the work done against the resistive force in the air. Friction is the only force other than gravity that acts on the ball.

So, the loss of kinetic energy is $\frac{1}{2} *m*(vf^{2} -vi^{2} )$

You know:

mass=m=0.22 kg
Initial velocity of the ball: $vi= 29 \frac{m}{s}$

Final velocity of the ball: $vf= 15 \frac{m}{s}$

Replacing:

$\frac{1}{2} *0.22 kg*(15^{2} -29^{2} )$ = -67.76 J

Friction work is always negative because friction is always against displacement.

5 0

3 years ago

I need help on 18,19,20 and 21

Dmitriy789 [7]

When a relationship between two different things is shown in a fraction it is a ratio.

hope this helps :)

8 0

3 years ago

Determine the minimum work per unit of heat transfer from the source reservoir that is required to drive a heat pump with therma

Sloan [31]

Answer:

The minimum work per unit heat transfer will be 0.15.

Explanation:

We know the for a heat pump the coefficient of performance ( $C_{HP}$ ) is given by

$C_{HP} = \dfrac{Q_{H}}{W_{in}}$

where, $Q_{H}$ is the magnitude of heat transfer between cyclic device and high-temperature medium at temperature $T_{H}$ and $W_{in}$ is the required input and is given by $W_{in} = Q_{H} - Q_{L}$ , $Q_{L}$ being magnitude of heat transfer between cyclic device and low-temperature $T_{L}$ . Therefore, from above equation we can write,