Here in all such collision type question we can use momentum conservation as we can see that there is no external force on this system
as we know that
now from above equation we have
so the speed of combined system is 2 m/s
Multiply by (1000 meters / 1 km).
Then multiply by (1 hour / 3600 seconds).
Both of those fractions are equal to ' 1 ', because the top
and bottom numbers are equal, so the multiplications
won't change the VALUE of the 72 km/hr. They'll only
change the units.
(72 km/hour) · (1000 meters / 1 km) · (1 hour / 3600 seconds)
= (72 · 1000 / 3600) (km·meter·hour / hour·km·second)
= 20 meter/second
Answer:
The force applied is 32 N
Explanation:
F = ma
F = 8 × 4
F = 32 N
Using the kinematic equation d = V_0 * t + 1/2 * a * t^2, where d is height you can rewrite this to be d = 1/2*g*t^2 or 4.9t^2
g = a because this is a free fall
d = 1/2 * 9.81m/s^2 * 2.5^2
d = 30.65625m
d = 30.7m
