Answer:
t = 5.89 s
Explanation:
To calculate the time, we need the radius of the pulley and the radius of the sphere which was not given in the question.
Let us assume that the radius of the pulley () = 0.4 m
Let the radius of the sphere (r) = 0.5 m
w = angular speed = 150 rev/min = (150 × 2π / 60) rad/s = 15.708 rad/s
Tension (T) = 20 N
mass (m) = 3 kg each
Substituting values:
Answer:
v = 1.28 m/s
Explanation:
Given that,
Maximum compression of the spring,
Spring constant, k = 800 N/m
Mass of the block, m = 0.2 kg
To find,
The velocity of the block when it first reaches a height of 0.1 m above the ground on the ramp.
Solution,
When the block is bounced back up the ramp, the total energy of the system remains conserved. Let v is the velocity of the block such that,
Initial energy = Final energy
Substituting all the values in above equation,
v = 1.28 m/s
Therefore the velocity of block when it first reaches a height of 0.1 m above the ground on the ramp is 1.28 m/s.
Answer:
(a) W = 1329.5 J = 1.33 KJ
(b) ΔU = 24.27 KJ
Explanation:
(a)
Work done by the gas can be found by the following formula:
where,
W = Work = ?
P = constant pressure = (0.991 atm)() = 100413 Pa
ΔV = Change in Volume = 18.7 L - 5.46 L = (13.24 L)() = 0.01324 m³
Therefore,
W = (100413 Pa)(0.01324 m³)
<u>W = 1329.5 J = 1.33 KJ</u>
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(b)
Using the first law of thermodynamics:
ΔU = ΔQ - W (negative W for the work done by the system)
where,
ΔU = change in internal energy of the gas = ?
ΔQ = heat added to the system = 25.6 KJ
Therefore,
ΔU = 25.6 KJ - 1.33 KJ
<u>ΔU = 24.27 KJ</u>