According to the Hooke’s law formula, the force is proportional to the displacement of the spring. <em>(C) </em>
Answer:
f = 5 cm
Explanation:
using the thin lens equation, given as follows:

where,
f = focal length = ?
do = the distance of object from lens = 20 cm
di = the distance of image from lens = 6.6667 cm
Therefore,

<u>f = 5 cm</u>
Answer:
Guysi hate math answer this guy plsss ssss
I think the answer is Gamma Rays
Answer:
The acceleration of the car will be 
Explanation:
We have given that distance from stop sign s = 200 m
Time t = 0.2 sec
We have to find the constant acceleration
Now from second equation of motion 


So the acceleration of the car will be 