Information that is given:
a = -5.4m/s^2
v0 = 25 m/s
---------------------
S = ?
Calculate the S(distance car traveled) with the formula for velocity of decelerated motion:
v^2 = v0^2 - 2aS
The velocity at the end of the motion equals zero (0) because the car stops, so v=0.
0 = v0^2 - 2aS
v0^2 = 2aS
S = v0^2/2a
S = (25 m/s)^2/(2×5.4 m/s^2)
S = (25 m/s)^2/(10.8 m/s^2)
S = (625 m^2/s^2)/(10.8 m/s^2)
S = 57.87 m
Answer:
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and I will be the sky.
you will be the ocean
and I will be the shore.
you will be the trees
and I will be the wind.
whatever we are, you and I will always collide.
There you go! Let me know if it helped.
:)
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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Tysm!
:)
Answer:
Explanation:
Torque is defined as the cross product between the position vector ( the lever arm vector connecting the origin to the point of force application) and the force vector.
Due to the definition of cross product, the magnitude of the torque is given by:
Where 
is the angle between the force and lever arm vectors. So, the length of the lever arm (r) is minimun when 
is equal to one, solving for r:
