Answer:
V = 3.54 m/s
Explanation:
Using the conservation of energy:

so:

where w is te weigh of kelly, h the distance that kelly decends, m is the mass of kelly and V the velocity in the lowest position.
So, the mass of kelly is:
m = 425N/9.8 = 43.36 Kg
and h is:
h = 1m-0.36m =0.64m
then, replacing values, we get:

Solving for v:
V = 3.54 m/s
Answer:
hmax = 1/2 · v²/g
Explanation:
Hi there!
Due to the conservation of energy and since there is no dissipative force (like friction) all the kinetic energy (KE) of the ball has to be converted into gravitational potential energy (PE) when the ball comes to stop.
KE = PE
Where KE is the initial kinetic energy and PE is the final potential energy.
The kinetic energy of the ball is calculated as follows:
KE = 1/2 · m · v²
Where:
m = mass of the ball
v = velocity.
The potential energy is calculated as follows:
PE = m · g · h
Where:
m = mass of the ball.
g = acceleration due to gravity (known value: 9.81 m/s²).
h = height.
At the maximum height, the potential energy is equal to the initial kinetic energy because the energy is conserved, i.e, all the kinetic energy was converted into potential energy (there was no energy dissipation as heat because there was no friction). Then:
PE = KE
m · g · hmax = 1/2 · m · v²
Solving for hmax:
hmax = 1/2 · v² / g
Answer:
Option (A) , (b) and (d) are correct option
Explanation:
According to Coulomb's law electric force between two charges is given by

From the relation we can say that force is directly proportional to magnitude of charges and inversely proportional to distance between them '
So if we increase the distance then force will decrease
Increase if any of the charge get larger
If force is attractive then both the charge will be of different sign and is force is repulsive then both the charges of same sign
From above conclusion we can say that (a), (b) and (d) are correct option
Answer : The correct option is, (C) 17 m/s
Explanation :
Formula used :

where,
K.E = kinetic energy = 6.8 J
m = mass of object = 46 g = 0.046 kg (1 kg = 1000 g)
v = velocity
Now put all the given values in the above formula, we get:




Therefore, the ball's velocity be as it leaves the cannon is, 17 m/s