Using the mirror formula.
1/v + 1/u = 1/f
1/9.5 + 1/3.2 = 1/f
1/f = 3.2 + 9.5 / 9.5 * 3.2
1/f = 4.82 cm
Radius = 2f
Radius = 2 x 4.82
Radius = 9.64 cm
B) 7.87 m/s
The gravitational pull is the rate of change of velocity which is the acceleration. Formula for acceleration is;

Given:
• Initial velocity = 0m/s; I dropped the ball, and didn't throw it, so it was at rest firstly
• Time taken = 2.40s
• Acceleration = 3.28m/s^2
We're require to find the final velocity, at which the ball hit the ground with. Ignoring air resistance, keep in mind that the velocity of an object increases as it comes closer to the ground.


The relevant equation we can use in this problem is:
h = v0 t + 0.5 g t^2
where h is height, v0 is initial velocity, t is time, g is
gravity
Since it was stated that the rock was drop, so it was free
fall and v0 = 0, therefore:
h = 0 + 0.5 * 9.81 m/s^2 * (4.9 s)^2
<span>h = 117.77 m</span>
Answer: The magnitude of the current in the second wire 2.67A
Explanation:
Here is the complete question:
Two straight parallel wires are separated by 7.0 cm. There is a 2.0-A current flowing in the first wire. If the magnetic field strength is found to be zero between the two wires at a distance of 3.0 cm from the first wire, what is the magnitude of the current in the second wire?
Explanation: Please see the attachments below