Explanation:
It is given that,
Let q₁ and q₂ are two small positively charged spheres such that,
.............(1)
Force of repulsion between the spheres, F = 1 N
Distance between spheres, d = 2 m
We need to find the charge on the sphere with the smaller charge. The force is given by :
............(2)
On solving the system of equation (1) and (2) using graph we get,
So, the charge on the smaller sphere is 11.6 micro coulombs. Hence, this is the required solution.
Answer:
7.328m/s
Explanation:
Given parameters:
height of table = 0.68m
final velocity of the ball = 6m/s
Unknown:
Initial velocity of ball = ?
Solution:
To solve this problem, we are going to employ the appropriate motion equation.
We must understand that this fall occurs in the presence of gravity;
V = U + 2gH
Where;
V is the final velocity
U is the initial velocity
g is the acceleration due to gravity
H is the height of the pool table
Since U is the unknown, let us make it the subject of the expression;
U = V - 2gH
U = 6 - (2 x 9.8 x 0.68) = 7.328m/s(deceleration)
The scale is a rest scale reads the support for is not enough net force.
Answer:
C
Explanation:
The period of a pendulum is found by the equation: T = 2pi*sqrt(L/g). Let the original length be L and the original period be T. The length increased by a factor of 4, so it’s new length is 4L. We get that the new period is 2pi*sqrt(4L/g) = 2pi*2sqrt(L/g) = 4pi*sqrt(L/g). We can see that the period increased by a factor of 2 because the original period, T, equals 2pi*sqrt(L/g) and the new period is 4pi*sqrt(L/g) = 2(2pi*sqrt(L/g)) = 2T. Therefore, the new period is 2(1.4) = 2.8
I hope this helps! :)
