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Allushta [10]
3 years ago
9

2. Determine the units of the quantity described by each of the following

Physics
1 answer:
notka56 [123]3 years ago
3 0

Explanation:

a.) kg(m/s)(1/s) = force (N)

b.) (kg/s)(m/s) = force (N)

c.) (kg/s)(m/s) = force (N)

d.) (kg/s)(m/s) = force (N)

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Explanation:

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A 75-kg piano is hoisted on a crane and delivered throughout the window of a 6th story apartment (20 meters above ground). What
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Answer:

P = 14700 J

Explanation:

Given that,

Mass of a piano, m = 75 kg

It is delivered throughout the window of a 6th story apartment which is 20 m above the ground.

We need to find the potential energy of the piano. It is given by :

P = mgh

Putting all the values,

P = 75 kg × 9.8 m/s² × 20 m

P = 14700 J

So, the potential energy of the piano is 14700 J.

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3 years ago
a falling skydiver slows from a speed of 52 m/s to 8 m/s in 0.8 sec as the parachute opens. what is the diver's acceleration and
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How does the cell membrane work to protect the cell?
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Consider a Carnot heat-engine cycle executed in a closed system using 0.025 kg of steam as the working fluid. It is known that t
ArbitrLikvidat [17]

Answer:

The temperature of the steam during the heat rejection process is 42.5°C

Explanation:

Given the data in the question;

the maximum temperature T_H in the cycle is twice the minimum absolute temperature T_L in the cycle

T_H  = 0.5T_L

now, we find the efficiency of the Carnot cycle engine

η_{th = 1 - T_L/T_H

η_{th = 1 - T_L/0.5T_L

η_{th = 0.5

the efficiency of the Carnot heat engine can be expressed as;

η_{th = 1 - W_{net/Q_H

where W_{net is net work done, Q_H is is the heat supplied

we substitute

0.5 = 60 / Q_H

Q_H = 60 / 0.5

Q_H = 120 kJ

Now, we apply the first law of thermodynamics to the system

W_{net = Q_H - Q_L

60 = 120 -  Q_L

Q_L = 60 kJ

now, the amount of heat rejection per kg of steam is;

q_L = Q_L/m

we substitute

q_L = 60/0.025

q_L = 2400 kJ/kg

which means for 1 kilogram of conversion of saturated vapor to saturated liquid , it takes 2400 kJ/kg of heat ( enthalpy of vaporization)

q_L = h_{fg = 2400 kJ/kg

now, at  h_{fg  = 2400 kJ/kg from saturated water tables;

T_L = 40 + ( 45 - 40 ) ( \frac{2400-2406.0}{2394.0-2406.0}\\} )

T_L = 40 + (5) × (0.5)

T_L = 40 + 2.5

T_L = 42.5°C  

Therefore, The temperature of the steam during the heat rejection process is 42.5°C  

4 0
3 years ago
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