Answer:
Explanation:
Let the charge on bead A be q nC and the charge on bead B be 28nC - qnC
Force F between them
4.8\times10^{-4} = 
=120 x 10⁻⁸ = 9 x q(28 - q ) x 10⁻⁹
133.33 = 28q - q²
q²- 28q +133.33 = 0
It is a quadratic equation , which has two solution
q_A = 21.91 x 10⁻⁹C or q_B = 6.09 x 10⁻⁹ C
·The acceleration of gravity is proportional to
1 / (the square of the distance from the center) .
When we're on the surface, we're 1 radius from the center of the Earth,
and the acceleration of gravity is 9.8 m/s² .
The boy's weight = (mass) · (gravity) = (50kg) · (9.8 m/s²)
= 490 newtons .
At the distance of 5 radii from the center (4 radii altitude from the surface),
the acceleration of gravity is
(9.8 m/s²) · (1/5)² = 0.39 m/s² .
The boy's weight is (mass) · (gravity) = (50kg) · (0.39 m/s²)
= 19.6 newtons .
Just as we expected, his weight at that distance is
(19.6 / 490) = 0.04 = 1/25 = 1/5² of his weight on the surface.
Answer:
0.5 kg
Explanation:
» <u>Concepts</u>
Newton's second law, the Law of Acceleration, states that F = ma, where F = Force in Newtons, m = mass in kg, and a = acceleration in m/s^2.
» <u>Application</u>
We are asked to find the mass of the ball using the equation F = ma. We're also given the force and acceleration, so the equation looks like 5 = 10(m).
» <u>Solution</u>
Step 1: Divide both sides by 10.
Thus, the mass of the ball is 0.5 kg.
I assume L=120 yards as the length of the football field.
1) The average speed is given by the total distance covered by the player divided by the time taken.
The total distance covered to go from one goal line to the other and then back to the 40-yards line is
And the time taken is t=22.4 s, so the average speed of the player is
2) The find the average velocity, we should also consider the direction (and the sign) of the velocity.
In the the first part of the motion, the player goes from one goal line to the other one, so he covers 120 y. However, in the second part of the motion he goes back by 80 y. Therefore, the net displacement of the player is
and so, the average velocity is
The magnet is attracted to the ring since the north pole of the current loop is above the ring and the south pole is below the ring.
