Answer:
<h3>62.5N</h3>
Explanation:
The pressure at one end of the piston is equal to the pressure on the second piston.
Pressure = Force/Area
F1/A1 = F2/A2
Given
F1 = 250N
A1 = 2.0m²
A2 = 0.5m²
F2 = ?
Substituting the given values in the formula;
250/2 = F2/0.5
cross multiply
250*0.5 = 2F2
125 = 2F2
F2 = 125/2
F2 = 62.5N
Hence the force needed to lift this piston if the area of the second piston is 0.5 m^2 is 62.5N
Answer:
d. Not enough information is given to answer this question.
Explanation:
From first law of thermodynamics
Q= W + ΔU
Q=Heat ,W= Work , ΔU=Change in internal energy
If work done by the gas :
It means that W and Q both are positive
Q- W = ΔU
Ii Q > W ,then temperature of the gas will increase.
If Q< W ,Then temperature of the gas will decreases.
If work done on the gas:
Q positive but W will be negative
Q- W = ΔU
Q= W or Q>W or Q< W ,then temperature of the gas will increase.
There are three cases because they did not give any information about the work.That is why option d is correct.
Answer:
The acceleration of the car, a = -3.75 m/s²
Explanation:
Given data,
The initial velocity of the airplane, u = 75 m/s
The final velocity of the plane, v = 0 m/s
The time period of motion, t = 20 s
Using the I equations of motion
v = u + at
a = (v - u) / t
= (0 - 75) / 20
= -3.75 m/s²
The negative sign indicates that the plane is decelerating
Hence, the acceleration of the car, a = -3.75 m/s²
Answer: It is a transverse wave ^^
Explanation:
Answer:
Explanation:
Initial velocity of mailbag u = 2 m/s
acceleration downwards a = g = 9.8 m/s²
time t = 3 s
a ) final velocity v = ?
v = u + at
= 2 + 9.8 x 3
= 31.4 m /s
b )
s = ut + 1/2 g t²
s is relative displacement of mailbag
u = relative initial velocity of mailbag = 0
relative acceleration = g = 9.8 m /s²
time t = 3 s
s = 0 + 1/2 x 9.8 x 3²
= 44.1 m
relative displacement of mailbag = 44.1 m .
