Quantum and state physics
This is a problem of conservation of momentum
Momentum before throwing the rock: m*V = 96.0 kg * 0.480 m/s = 46.08 N*s
A) man throws the rock forward
=>
rock:
m1 = 0.310 kg
V1 = 14.5 m/s, in the same direction of the sled with the man
sled and man:
m2 = 96 kg - 0.310 kg = 95.69 kg
v2 = ?
Conservation of momentum:
momentum before throw = momentum after throw
46.08N*s = 0.310kg*14.5m/s + 95.69kg*v2
=> v2 = [46.08 N*s - 0.310*14.5N*s ] / 95.69 kg = 0.434 m/s
B) man throws the rock backward
this changes the sign of the velocity, v2 = -14.5 m/s
46.08N*s = - 0.310kg*14.5m/s + 95.69kg*v2
v2 = [46.08 N*s + 0.310*14.5 N*s] / 95.69 k = 0.529 m/s
Tsunami and under water volcano
Answer:
(a) 17.37 rad/s^2
(b) 12479
Explanation:
t = 95 s, r = 6 cm = 0.06 m, v = 99 m/s, w0 = 0
w = v / r = 99 / 0.06 = 1650 rad/s
(a) Use first equation of motion for rotational motion
w = w0 + α t
1650 = 0 + α x 95
α = 17.37 rad/s^2
(b) Let θ be the angular displacement
Use third equation of motion for rotational motion
w^2 = w0^2 + 2 α θ
1650^2 = 0 + 2 x 17.37 x θ
θ = 78367.87 rad
number of revolutions, n = θ / 2 π
n = 78367.87 / ( 2 x 3.14)
n = 12478.9 ≈ 12479
