5.1 m
Explanation:
Let's set the ground as our reference point. Let's also call the dropped ball to be ball #1 and its height above the ground at any time t is given by
(1)
where 10 represents its initial height or displacement of 10 m above the ground. At the same time, the displacement of the second ball with respect to the ground
is given by
(2)
At the instant the two balls collide, they will have the same displacement, therefore

or

Solving for t, we get

We can use either Eqn(1) or Eqn(2) to hind the height where they collide. Let's use Eqn(1):


The electric field strength is inversely related to the square of the distance.so the strength of the electric field is given by

Here,
is constant depend upon medium and its value is
and q is charge and r is the distance.
Given
and we know the charge of proton,
.
Therefore,

Answer:
ΔS=2*m*Cp*ln((T1+T2)/(2*(T1*T2)^1/2))
Explanation:
The concepts and formulas that I will use to solve this exercise are the integration and the change in the entropy of the universe. To calculate the final temperature of the water the expression for the equilibrium temperature will be used. Similarly, to find the change in entropy from cold to hot water, the equation of the change of entropy will be used. In the attached image is detailed the step by step of the resolution.
As ball is projected up in air at an angle of 45 degree without any air resistance
Let the initial speed will be v
now we will have
In x direction

in y direction

now displacement in x direction

displacement in y direction

now from above two equations we have


so above equation is a quadratic equation and hence it will be a parabolic curve
so correct answer will be
<em>C. parabolic curve.</em>