Answer:
F=BILsin90 when perpendicular sin90 =1 30T x50x30 so you can get 45000N
The ball was moving for 0.8seconds
This is because 4/5 x 1 = 0.8
Hope this helps and good luckkkk :)
Explanation:
(a) Draw a free body diagram of the cylinder at the top of the loop. At the minimum speed, the normal force is 0, so the only force is weight pulling down.
Sum of forces in the centripetal direction:
∑F = ma
mg = mv²/RL
v = √(g RL)
(b) Energy is conserved.
EE = KE + RE + PE
½ kd² = ½ mv² + ½ Iω² + mgh
kd² = mv² + Iω² + 2mgh
kd² = mv² + (m RC²) ω² + 2mg (2 RL)
kd² = mv² + m RC²ω² + 4mg RL
kd² = mv² + mv² + 4mg RL
kd² = 2mv² + 4mg RL
kd² = 2m (v² + 2g RL)
d² = 2m (v² + 2g RL) / k
d = √[2m (v² + 2g RL) / k]
Answer:
B) - 5.0 m
Explanation:
B is located on a positive location, 15m from the starting point A. Hence, since E is located a positive distance 10m from A, the difference becomes 10 - 15 = - 5.0 m
