Answer:21.45 m/s
Explanation:
Given
Mass of sport car=920 kg
Mass of SUV=2300 kg
distance to which both car skid is 2.4 m
coefficient of friction ()=0.8
Let u be the initial velocity of both car at the starting of skidding
and they finally come to zero velocity
s=2.4 m
u=6.13 m/s
so before colliding sport car must be travelling at a speed of
(conserving momentum)
v=21.45 m/s
Answer:
The force has been reduced by 8018 N
Explanation:
The impulse exerted on the car during the crash is equal to the product of the force exerted and the duration of the collision, and it is also equal to the change in momentum of the car. So we can write:
where:
F is the force exerted on the car
is the duration of the collision
m = 1400 kg is the mass of the car
is the change in velocity of the car
We can re-write the equation as
In the 1st collision, the time is 1.5 seconds, so the force is
In the 2nd collision, the time is increased to 2.2 seconds, so the force is
Therefore, the force has been reduced by:
k =
k = (6.626×10-¹⁹/590 × 10-⁹ )^{2} /2 × 1.673 × 10-²⁷
k = (1.12 × 10-³⁰)^2/3.346×10-²⁷
k = 1.25 × 10-⁶⁰ /3.346×10-²⁷
k = 0
ldk why, my answer is coming this :(