Answer: Option D: 5.5×10²Joules
Explanation:
Work done is the product of applied force and displacement of the object in the direction of force.
W = F.s = F s cosθ
It is given that the force applied is, F = 55 N
The displacement in the direction of force, s = 10 m
The angle between force and displacement, θ = 0°
Thus, work done on the object:
W = 55 N × 10 m × cos 0° = 550 J = 5.5 × 10² J
Hence, the correct option is D.
Efficiency = useful energy out / total energy in x 100
= 100/400 x 100
=0.25 x 100
= 25%
25%
First method
initial distance = 16m
final distance= 43 m
total distance covered= final -initial
=43m -16m
=27m
Second method
Si= 16m
Sf =43 m
t= 12 s
first we will find V
V = (Sf-Si)/ t
V =( 43- 16)/ 12
V = 27/12 ⇒ V= 9/4
V= distance / time
distance= V×time
distance = (9/4) ×12
distance =27
C) alternately increase and decrease
Answer:
1) 1 2) 1 3)4 4) 1 5) 1 6)1 7)3 8)3 9)2 10)3
Explanation:
