Answer:
R = 715.4 N
L = 166.6 N
Explanation:
ASSUME the painter is standing right of center
Let L be the left rope tension
Let R be the right rope tension
Sum moments about the left end to zero. Assume CCW moment is positive
R[5] - 20(9.8)[5/2] - 70(9.8)[5/2 + 2] = 0
R = 715.4 N
Sum moments about the right end to zero
20(9.8)[5/2] + 70(9.8)[5/2 - 2] - L[5] = 0
L = 166.6 N
We can verify by summing vertical forces
116.6 + 715.4 - (70 + 20)(9.8) ?=? 0
0 = 0 checks
If the assumption about which side of center the paint stood is incorrect, the only difference would be the values of L and R would be swapped.
Answer:
U = – 0.12J
Explanation:
Given N = 10 turns, I = 5A, r = 5×10-²m
B^ = 0.05 T iˆ+ 0.3 T kˆ
Magnitude of the magnetic field vector B = √(0.05²+0.3²) = 0.304T
Area = πr² = π(5×10-²)² = 7.85×10-³m²
Magnetic moment μ = NIA
μ = 10×5×7.85×10-³ = 0.3925Am²
U = -μ•B = –0.3925×0.304 = –0.12J
The sign is negative because the magnetic moment is aligned with the magnetic field.
Answer:
so angular velocity is 7.13128 sec−1
Explanation:
velocity v = 2.2 m/s
displacement s = 220 mm = 0.220 m
distance d = 510 mm = 0.510 m
to find out
angular velocity
solution
we know that
angular velocity will be velocity ( v) / (displacement² + distance²) .....1
now put all these value in equation 1 and we get angular velocity i.e.
angular velocity = velocity ( v) / (displacement² + distance²)
angular velocity = 2.2 / (0.22² + 0.51²)
angular velocity = 2.2 / 0.3085
angular velocity = 7.13128
so angular velocity is 7.13128 sec−1
Answer:
Explanation:
We must use conservation of linear momentum before and after the collision, 
Before the collision we have:
where these are the masses are initial velocities of both players.
After the collision we have:
since they clong together, acting as one body.
This means we have:
Or:
Which for our values is:
Torque = r x F
|F| = mg = 60 * 10 N = 600 N ( assuming g ~ 10m/s^2)
distance of fulcrum = torque / Force = 90/600 m = .15 m.
