Answer:
3.7 m/s
Explanation:
M = 444 kg
U = 5 m/s
m = 344 kg
u = - 5 m/s
Let the velocity of train is V and the car s v after the collision.
As the collision is elastic
By use of conservation of momentum
MU + mu = MV + mv
444 x 5 - 344 x 5 = 444 V + 344 v
500 = 444 V + 344 v
125 = 111 V + 86 v .... (1)
By using the formula of coefficient of restitution ( e = 1 for elastic collision)
-5 - 5 = V - v
V - v = - 10
v = V + 10
Substitute the value of v in equation (1)
125 = 111 V + 86 (V + 10)
125 = 197 V + 860
197 V = - 735
V = - 3.7 m/s
Thus, the speed of first car after collision is 3.7 m/s. negative sign shows that the direction is reverse as before the collision.
Answer:
Maximum height reached by the ball is 32 meters.
Explanation:
It is given that,
If a baseball is project upwards from the ground level with an initial velocity of 32 feet per second, then it's height is a function of time. The equation is given as :
...........(1)
t is the time taken
s is the height attained as a function of time.
Maximum height achieved can be calculated as :
-16 t + 32 = 0
t = 2 seconds
Put the value of t in equation (1) as :
s = 32 meters
So, the maximum height reached by the ball is 32 meters. Hence, this is the required solution.
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Answer:
The direction of electric field and equipotential line at the same point are always PERPENDICULAR TO THE ELECTRIC FIELD.
Explanation:
Equipotential surface is a three dimensional part of equipotential lines.
Equipotential lines are a type of contour lines that is use to trace lines that have the same altitude on the map and the altitude is the electric potential.
Equipotential lines are always perpendicular to electric potential because the lines creates three dimension equipotential surface.
