The electrical potential energy of a charge q located at a point at potential V is given by

Therefore, if the charge must move between two points at potential V1 and V2, the difference in potential energy of the charge will be

In our problem, the electron (charge e) must travel across a potential difference V. So the energy it will lose traveling from the metal to the detector will be equal to

Therefore, if we want the electron to reach the detector, the minimum energy the electron must have is exactly equal to the energy it loses moving from the metal to the detector:
Answer:
A driver.
Explanation:
Using a driver while at least 350 yds away is better than using a iron, because it will be a waste of the par 4 as it is not as powerful as the driver.
Answer:
7.75 s
Explanation:
Newton's second law:
∑F = ma
35 N = (70 kg) a
a = 0.5 m/s²
Given v₀ = 0 m/s and Δx = 15 m:
Δx = v₀ t + ½ at²
(15 m) = (0 m/s) t + ½ (0.5 m/s²) t²
t = 7.75 s
Answer:
μ = 0.535
Explanation:
On a level floor, normal force = weight.
N = W
Friction force = normal force × coefficient of friction.
F = Nμ
Substitute:
F = Wμ
83 = 155μ
μ = 0.535
Round as needed.
Answer:

Explanation:
Mass of a hockey puck, m = 0.17 kg
Force exerted by the hockey puck, F' = 35 N
The force of friction, f = 2.7 N
We need to find the acceleration of the hockey puck.
Net force, F=F'-f
F=35-2.7
F=32.3 N
Now, using second law of motion,
F = ma
a is the acceleration of the hockey puck

So, the acceleration of the hockey puck is
.