Question

plzzzzzzzzzzzzzzzz quickly

Answer 1

1. (b) length
2. (a) 3
3. (c) M1L2T-1
4. (a)
5. (a)

Let the total distance be d

Then for 1st part,

distance covered = d/2

velocity = v1 (given)

therefore time = distance/speed = d/2 divided by v1 = d/2v1

similarly, for second half, time = d/2v2

SO, AVERAGE VELOCITY = TOTAL DISTANCE/TOTAL TIME = d/(d/2v1 + d/2v2) = 2v1v2/v1+v2(on solving)

so 48 = 2 x 40 x v / (40+v)
     1920 + 48v = 80v
      1920 = 32v
      v = 60 km/hr
6. (c)
  x component of 4N force is 4Cos60 = 4 x 1/2 = 2 N

7. (c)
 ∆K/K x 100 = [ (p2)^2 -(p1)^2 ] / (p1)^2 x100
                     p2 = p1 + 0.2p1 = 1.2 p1
                   
so ∆K/K x 100 = (1.2p1)^2 - (p1)^2 / (p1)^2 x 100
                        = 44% change in kinetic energy

8. (d) 
 if goes then momentum p1 = mv 
 and come again then momentum = p2 = -mv

then change in momentum = Δp = p2 - p1 = |-2mv| = 2mv

9. (d)
F(r) = MgSinθ - μN
 Ma = MgSinθ - μN
 Ma = MgSinθ - μMgCosθ
   a = gSinθ - μgCosθ

10. (c)


i hope it will help you
 regards