Answer:
final displacement lf = 0.39 m
Explanation:
from change in momentum equation:
![\delta p = m \sqrt(2g * y/x)* [\sqrt li + \sqrt lf]](https://tex.z-dn.net/?f=%5Cdelta%20p%20%3D%20m%20%5Csqrt%282g%20%2A%20y%2Fx%29%2A%20%5B%5Csqrt%20li%20%2B%20%5Csqrt%20lf%5D)
given: m = 0.4kg, y/x = 19/85, li = 1.9 m,
\delta p = 1.27 kg*m/s.
putting all value to get the final displacement value
![1.27 = 0.4\sqrt(2*9.81 *(19/85))* [\sqrt 1.9 + \sqrt lf]](https://tex.z-dn.net/?f=1.27%20%3D%200.4%5Csqrt%282%2A9.81%20%2A%2819%2F85%29%29%2A%20%5B%5Csqrt%201.9%20%2B%20%5Csqrt%20lf%5D)
final displacement lf = 0.39 m
Answer:
a=F/m
a=12N/3kg (here newton can be written as kgm/s^2 so kg will be cancelled)
a=4m/s^2
Explanation:
Answer:
The acceleration of the car, a = -3.75 m/s²
Explanation:
Given data,
The initial velocity of the airplane, u = 75 m/s
The final velocity of the plane, v = 0 m/s
The time period of motion, t = 20 s
Using the I equations of motion
v = u + at
a = (v - u) / t
= (0 - 75) / 20
= -3.75 m/s²
The negative sign indicates that the plane is decelerating
Hence, the acceleration of the car, a = -3.75 m/s²
Answer:
A
Explanation:
because thats what I put and got it right
