Maritime polar air mass comes from the northern part of the pasific ocean and that air mass goes to Utah.
I couldnt type it out so here's a picture
Answer:
Explanation:
Given that,
Mass of counterweight m= 4kg
Radius of spool cylinder
R = 8cm = 0.08m
Mass of spool
M = 2kg
The system about the axle of the pulley is under the torque applied by the cord. At rest, the tension in the cord is balanced by the counterweight T = mg. If we choose the rotation axle towards a certain ~z, we should have:
Then we have,
τ(net) = R~ × T~
τ(net) = R~•i × mg•j
τ(net) = Rmg• k
τ(net) = 0.08 ×4 × 9.81
τ(net) = 3.139 Nm •k
The magnitude of the net torque is 3.139Nm
b. Taking into account rotation of the pulley and translation of the counterweight, the total angular momentum of the system is:
L~ = R~ × m~v + I~ω
L = mRv + MR v
L = (m + M)Rv
L = (4 + 2) × 0.08
L = 0.48 Kg.m
C. τ =dL/dt
mgR = (M + m)R dv/ dt
mgR = (M + m)R • a
a =mg/(m + M)
a =(4 × 9.81)/(4+2)
a = 6.54 m/s
The ball at the top of the ramp has potential energy but when it rolls down or when it is rolling down it has kinetic. The ball has kinetic energy when it is moving or rolling down the ramp but there is Potential energy when the ball is at the top or is sitting at the bottom still. when the ball is rolling down the ramp it has heat energy because of how it get speed and it starts to heat up.
The solution for this problem is:
Work it backwards: spring U = ½kx² = ½ * 4000N/m * (0.10m) ²
= 20 J
so the pre-impact of KE = 20 J = ½ * m * v² = ½ * 0.40kg * v²
v = 10 m/s → for the bullet/block.
Now conserve the momentum:
50g * u = 400g * 10m/s
u = 80 m/s is the arrow
velocity
