Answer:
A title
Explanation:
Because this is middle school.
Answer:
a) p = m1 v1 + m2 v2
, b) dp / dt = m1 a1 + m2 a2
, c) It is equivalent to force
dp / dt = 0
Explanation:
In this problem we have two blocks and the system is formed by the two bodies.
Part A. Initially they ask us to find the moment of the whole system
p = m1 v1 + m2 v2
Part B.
Find the derivative
dp / dt = m1 dv1dt + m2 dv2 / dt
dp / dt = m1 a1 + m2 a2
Part C.
Let's analyze the dimensions
m a = [kg] [m / s2] = [N]
It is equivalent to force
Part d
Acceleration is due to a net force applied
Part e
The acceleration of block 1 is due to the force exerted by block 2 during the moment change
Part f
Force of block 1 on block 2
True f12 = m1a1 f21 = m2a2
Part g
By the law of action and reaction are equal magnitude F12 = f21
Part H
dp / dt = 0
Isolated system F12 = F21 and the masses are constant. The total moment is only redistributed
Answer:
0.83 ω
Explanation:
mass of flywheel, m = M
initial angular velocity of the flywheel, ω = ωo
mass of another flywheel, m' = M/5
radius of both the flywheels = R
let the final angular velocity of the system is ω'
Moment of inertia of the first flywheel , I = 0.5 MR²
Moment of inertia of the second flywheel, I' = 0.5 x M/5 x R² = 0.1 MR²
use the conservation of angular momentum as no external torque is applied on the system.
I x ω = ( I + I') x ω'
0.5 x MR² x ωo = (0.5 MR² + 0.1 MR²) x ω'
0.5 x MR² x ωo = 0.6 MR² x ω'
ω' = 0.83 ω
Thus, the final angular velocity of the system of flywheels is 0.83 ω.
First, we have to calculate the normal forces on different surfaces.The normal force on the 4.00 kg, N1 = (4)(9.8) = 39.2 N. The normal force on the 10.0 kg, N2 = (14)(9.8) = 137.2 N. Looking at the 10.0 kg block, the static forces that counteract the pulling force equals the sum of the friction from the two surfaces. Fc = N1 * 0.80 + N2 * 0.80 = 141.12 N. Since the counter force is less than the pulling force, the blocks start to move and hence, kinetic frictions are considered.
Therefore, f1 = uk * N1 = (0.60)(39.2) = 23.52 N.
If I’m understanding the question then 11.76cm^2 in two sig figs is 12cm^2. I don’t see a part c to this question though so if I’m not giving you the answer you need please advise and I will answer what you need! Thank you!!