Answer:
1 and 3
Explanation:
because they are going up from 0
Answer:
The options are not shown, so let's derive the relationship.
For an object that is at a height H above the ground, and is not moving, the potential energy will be:
U = m*g*H
where m is the mass of the object, and g is the gravitational acceleration.
Now, the kinetic energy of an object can be written as:
K = (1/2)*m*v^2
where v is the velocity.
Now, when we drop the object, the potential energy begins to transform into kinetic energy, and by the conservation of the energy, by the moment that H is equal to zero (So the potential energy is zero) all the initial potential energy must now be converted into kinetic energy.
Uinitial = Kfinal.
m*g*H = (1/2)*m*v^2
v^2 = 2*g*H
v = √(2*g*H)
So we expressed the final velocity (the velocity at which the object impacts the ground) in terms of the height, H.
Given:
The magnitude of each charge is q1 = q2 = 1 C
The distance between them is r = 1 m
To find the force when distance is doubled.
Explanation:
The new distance is
The force can be calculated by the formula
Here, k is the constant whose value is
On substituting the values, the force will be
Answer:
Work done = (1/2)[(Gmm_e)/(R_e)]
Explanation:
I've attached the explanations below.
