v1f = -0.16 ms
Explanation:
Use the conservation law of linear momentum:
m1v1i + m2v2i = m1v1f + m2v2f
where
v1i = v2i = 0
m1 = 160 kg
m2 = 0.50 kg
v2f = 50m/s
v1f = ?
So we have
0 = (160 kg)v1f + (0.5 kg)(50 m/s)
v1f = -(25 kg-m/s)/(160 kg)
= -0.16 m/s
Note: the negative sign means that its direction is opposite that of the arrow.
Answer:
7.45
Explanation:
At 3.0 m PE = m*g*3.0, KE = m*g*(4.0 - 3.0)
At 2.0 m PE = m*g*2.0, KE = m*g*(4.0 - 2.0)
At 3.0 m PE = m*g*1.0, KE = m*g*(4.0 - 1.0)
At 0 m PE = 0, KE = m*g*h = total energy = 7.45 J
Answer:
120°
Explanation:
Draw a free body diagram. There are three forces acting on the traffic light. Two tension forces acting along the cables, and weight.
The tension forces have an angle θ between them. That means each tension force forms an angle of θ/2 with respect to the vertical. So the y component of each tension force is:
Ty = T cos (θ/2)
Sum of the forces in the y direction:
∑F = ma
Ty + Ty − W = 0
2 Ty = W
Substituting:
2 T cos (θ/2) = W
If W = T, then:
2 W cos (θ/2) = W
2 cos (θ/2) = 1
cos (θ/2) = 1/2
θ/2 = 60°
θ = 120°