As per the question the initial speed of the car [ u] is 42 m/s.
The car applied its brake and comes to rest after 5.5 second.
The final velocity [v] of the car will be zero.
From the equation of kinematics we know that
[ here a stands for acceleration]
![0=42 +5.5a](https://tex.z-dn.net/?f=0%3D42%20%2B5.5a)
![a =\frac{-42}{5.5} m/s^2](https://tex.z-dn.net/?f=a%20%3D%5Cfrac%7B-42%7D%7B5.5%7D%20m%2Fs%5E2)
![a= -7.64 m/s^2](https://tex.z-dn.net/?f=a%3D%20-7.64%20m%2Fs%5E2)
Here a is taken negative as it the car is decelerating uniformly.
We are asked to calculate the stopping distance .
From equation of kinematics we know that
[here S is the distance]
![= 42*5.5 +\frac{1}{2} [-7.64] [5.5]^2 m](https://tex.z-dn.net/?f=%3D%2042%2A5.5%20%2B%5Cfrac%7B1%7D%7B2%7D%20%5B-7.64%5D%20%5B5.5%5D%5E2%20m)
[ans]
Answer:
I believe the answer for #1 is D and the answer for #2 is B
Explanation:
I hope this is correct and helps
Answer:
3675 J
Explanation:
Gravitational Potential Energy =
× mass × g × height
( g is the gravitation field strength )
Mass = 50 kg
G = 9.8 N/kg ( this is always the same )
Height = 15 m
Gravitational Potential Energy =
× 50 ×9.8 × 15
= 3675 J
M/s^2 is the correct answer
The combined amount of kinetic and potential energy of its molecules