The top box is cell and I’ll look at the others
Given :
Height from which ball is dropped , h = 40 m .
Acceleration due to gravity , g= 10 m/s² .
Initial velocity , u = 0 m/s .
To Find :
Velocity when ball covered 20 m and velocity when it hit the ground .
Solution :
Now , height when ball covered 20 m distance is , 40 - 20 = 20 m .
By equation of motion :
Now , distance covered when body reaches ground is , 40 m .
Putting value h = 40 m in above equation , we get :
Hence , this is the required solution .
Usually, you have to take two years of physical sciences with lab. It doesn't have to be physics, it can be chemistry or biology.
Answer:
very hard others will answer it
Explanation:
hard
Answer:
v = 2.94 m/s
Explanation:
When the spring is compressed, its potential energy is equal to (1/2)kx^2, where k is the spring constant and x is the distance compressed. At this point there is no kinetic energy due to there being no movement, meaning the net energy in the system is (1/2)kx^2.
Once the spring leaves the system, it will be moving at a constant velocity v, if friction is ignored. At this time, its kinetic energy will be (1/2)mv^2. It won't have any spring potential energy, making the net energy (1/2)mv^2.
Because of the conservation of energy, these two values can be set equal to each other, since energy will not be gained or lost while the spring is decompressing. That means
(1/2)kx^2 = (1/2)mv^2
kx^2 = mv^2
v^2 = (kx^2)/m
v = sqrt((kx^2)/m)
v = x * sqrt(k/m)
v = 0.122 * sqrt(125/0.215) <--- units converted to m and kg
v = 2.94 m/s