Block test MCQ
1. Average value very close to actual value , hence accurate. The deviation of each value far from the average value , hence not precise.
2. prefixes M is power +6, u is -6 and G is +9. Finally arrange them 5, 5x10*2 , 5x10*3, 5x10*5.
4. more precise would mean also more decimal places.
5. search lowest velocity means smallest gradient at that time in displacement time graph.
6. Equate same height for the 2 cases.
7. At highest point the vertical velocity is zero but horizontal is not zero. As long as object is in the air the acceleration is not zero.
8. Solve vertically using s = ut + 0.5 at*2.
9. Resolving the tension into x and y comp will give tan@=a/g where @ can be found.
10. Resolving Fcos35 = 25N.
11. The common acceleration is 15/11=1.36ms*-2. Contact force can be found by considering X .
12. Use u1-u2 =v2 -v1. and COM . Finally both v1 and v2 are positive , hence towards Y.
13. Drag force increases with velocity , hence finally acceleartion = 0, hence resultant force is decreasing.
14. Resolving vertically
15.Area of large triangle (40mm) minus Area of smaller triangle (30mm) . Area of triangle = 0.5kx*2.
16. Upwards = downward forces. Take moments about L , the unknown x = 4units.(atQ)
17. Work done against friction = friction x distance moved.
18. Velocity is not constant, hence displacement is not constant, hence gpe cannot be increasing uniformly as gpe = mg h .
19. Velocity increase uniformly from zero as acc is constant, hence power is the same shape too . Power is force x velocity where force is constant.
20. mgh/t is only 70%. The electrical power is 100%.
