Explanation:
Impulse is change in momentum:
I = Δp
I = (1400 kg) (27 m/s) - (1400 kg) (0 m/s)
I = 37800 kg m/s
Impulse is also average force times time:
I = F ΔT
37800 kg m/s = F (7 s)
F = 5400 N
Since it was dropped, it should be the speed of gravity which is 9.8 meters/second
Answer B is the correct answer
We know that kinetic energy , where m is the mass of object and v is the velocity of object.
In this case only velocity is the variable, mass remains constant.
So point having higher velocity has higher kinetic energy.
When it leaves the racket, the ball will be having a certain height, but just before it reaches the ground it will not having any height. So maximum velocity of ball is at that time when it reaches just above the ground.
So option B is the correct answer.
Explanation:
Given:
Δy = d
v₀ = 0
a = g
Find: v
v² = v₀² + 2aΔy
v² = (0)² + 2(g)(d)
v = √(2gd)
If we double d:
√(2g (2d))
√(4gd)
Compared to the first velocity:
√(4gd) / √(2gd)
√2
So the new velocity is √2 greater than the first velocity.
√2 v
The speed of light (electromagnetic radiation) is equal to 299 792 458
m / s or 3x10^8 m/s in scientific notation.
So with this information, we could now look for the
distance. Solution:
Take note that μs means microseconds.
Speed of light * microseconds travelled * actual amount of microseconds
(3x10^8 m/s) (45.0 μs) (1x10^-6 s/μs) = 13,500 m.