Answer:
D, Im not very sure about this but I would say D
Explanation:
I tried
Answer:
a) 2.5 m/s²
b) 6.12 m/s
Explanation:
Tension of rope = T = 356N
Weight of material = W = 478 N
Distance from the ground = s = 7.5 m
Acceleration due to gravity = g = 9.81 m/s²
Mass of material = m = 478/9.81 = 48.72
Final velocity before the bundle hits the ground = v
Initial velocity = u = 0
Acceleration experienced by the material when being lowered = a
a) W-T = ma
⇒478-356 = 48.72×a

⇒a = 2.5 m/s²
∴ Acceleration achieved by the material is 2.5 m/s²
b) v²-u² = 2as
⇒v²-0 = 2×2.5×7.5
⇒v² = 37.5
⇒v = 6.12 m/s
∴ Velocity of the material before hitting the ground is 6.12 m/s
Answer:
The work done by friction was 
Explanation:
Given that,
Mass of car = 1000 kg
Initial speed of car =108 km/h =30 m/s
When the car is stop by brakes.
Then, final speed of car will be zero.
We need to calculate the work done by friction
Using formula of work done



Put the value of m and v



Hence, The work done by friction was 
The answer is 18000 J
I hope this helps!^^ , if you need the work to be shown please tell me, I hope you have a great day!^^
Given :
Initial speed of car A is 15 m/s and initial speed of car B is zero.
Final speed of car A is zero and final speed of car B is 10 m/s.
To Find :
What fraction of the initial kinetic energy is lost in the collision.
Solution :
Initial kinetic energy is :

Final kinetic energy is :

Now, fraction of initial kinetic energy loss is :

Therefore, fraction of initial kinetic energy loss in the collision is 1.25 .