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Answer:
(A) 0.004 m^3
(B) 48.11 K
Explanation:
Work, W = 2100 J
Pressure, P = 1 x 10^5 Pa
V2 = 0.025 m^3
(A) let the initial volume of the gas is V1.
W = P (V2 - V1)
2100 = 1 x 10^5 (0.025 - V1)
0021 = 0.025 - V1
V1 = 0.025 - 0.021
V1 = 0.004 m^3
Thus, the initial volume of the gas is 0.004 m^3.
(B) Let the temperature of the gas is T
Use ideal gas equation
P1 x V1 = n x R T
1 x 10^5 x 0.004 = 1 x 8.314 x T
T = 48.11 K
Thus, the initial temperature of the gas is 48.11 K.
Answer:
2100 kg*m/s
Explanation:
The proper term for power in this case is to use the physics term "impuls" = p.
1. Find the "power" developed by the gun to fire just 1 bullet.
2. multiply by 60 to find the power produced by the gun per minute.
p = m * v
where
m = mass of the bullet = 50 g = 0.050 kg
v = velocity of the bullet = 700 m/s
p = 0.050 * 700
p = 35 kg*m/s
So the impuls for 1 bullet is 35 kg*m/s
The gun can fire 60 bullets per minute so it's fireing power must there fore be 60 * 35 = 2100 kg*m/s
Answer:
Workdone =0
Explanation:
The body is inclined at θ=30°, s=10m, m=5kg
Workdone by the object = FsCosθ
but the surface is a friction-less surface, this means coefficient of friction =0
F=μR
F=0*R
F=0
W=FScosθ
W=0*SCosθ
W=0