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Answer:
1. The magnitude of the force from the spring on the object is zero on <em>Equilibrium.</em>
2. The magnitude of the force from the spring on the object is a maximum on <em>The top and bottom.</em>
3. The magnitude of the net force on the object is zero on <em>The Bottom.</em>
4. The magnitude of the force on the object is a maximum on <em>the Top.</em>
Explanation:
<em>1. Because the change in position delta X is zero.</em>
<em>2. Because of delta X.</em>
<em>3. Beacuse, the force of gravity and the force of the spring oppose each other to keep the block at rest, away from the equilibrium position.</em>
<em>4. Because, the force of the spring from compressiom and the force of gravity both act on the mass.</em>
Answer:
a) v, v
b) 2mv^2
c) Elastic collion
Explanation:
(a) The velocity of the second particle after the collision is (v2x,v2y)=(v,−v). From momentum conservation in x-direction
Here x, y represent direction.They are not variable. 1 and 2 represent before and after.
2vm=v1xm+v2xm, we find v1x=v.
From momentum conservation in y-direction
0 =v1ym+v2ym, we findv1y=v.
(b) By energy conservation principle
Before: K=1/2m(2v)^2=2mv^2.
After: K=1/2m(v^2(1x)+v^2(1y))+12m(v22x+v22y)=2mv^2
(c) The collision is elastic
Let the vector position of the object in the (x-y) plane be
The applied force is
By definition, the applied torque is
Answer:
Answer: I think it’s 20cm.
