Solution :-
Given :
Distance 1 = 30 km
Distance 2 = 70 km
We know that speed = distance/time
and, Average speed = total distance/total time taken
When the train acquired a speed of 30 km/hr, the time taken = 30/30 = 1 hour
Average speed = 9distance 1 + distance 2)/(time 1 + time 2)
AS time 2 or t2 is time taken for the second part of the journey of 70 km
⇒ 40 = 100/(1 + t2)
⇒ 40 + 40t2 = 100
⇒ 40t2 = 100 - 40
⇒ 40t2 = 60
⇒ t2 = 60/40
⇒ t2 = 1.5
So, t2 or time taken to travel the second part of the journey is 1.5 hours.
Speed of the second part of the journey = distance 2/time 2
⇒ 70/1.5
⇒ 46.666 km/hr or 46.7 km/hr.
Hence the answer is = 46.666 km/hr or 46.7 km/hr.
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Efficiency = (energy that does the job) / (total energy used)
= (45 J) / (120J)
I think you can handle the division.
Weight = mass * gravity = 60 kg * 3.75 m/s² = 225 N
<span>Option D.</span>
The displacement of a moving object is the straight-line distance between the place it starts from and the place where it stops.
The displacement of anything moving along a circular track depends on how far around it goes before it stops. The greatest displacement it can possibly have is the diameter of the track ... 100m on this particular one ... because that's as far apart as two places on a circle can ever be.
The most interesting case is when the object goes around the circle exactly once. Then it stops at the same place it started from, the distance between the starting point and ending point is zero, and after all that motion, the displacement is zero.
