Answer:
A current will not be induced.
Explanation:
The wire is not part of a closed circuit.
The electrical charges have no place to flow to.
Answer:
(a) the deceleration of the player is -80.36 m/s²
(b) the time the collision last is 0.093 s
Explanation:
Given;
Initial velocity of the football player, u = 7.50 m/s
Final velocity of the football player, v = 0
distance traveled = compression of the pad, s = 0.35 m
Part (a) the deceleration of the player
v² = u² + 2as
0 = 7.5² + (2 x 0.35)a
0 = 56.25 + 0.7a
- 56.25 = 0.7a
a = -56.25 / 0.7
a = -80.36 m/s²
Part (b) the time the collision last
v = u + at
t = (v - u)/a
t = (0 - 7.5)/ -80.36
t = - 7.5 / -80.36
t = 0.093 s
Answer:
Both charges must have the same charge, Qt/2.
Explanation:
Let the two charges have charge Q1 and Q2, respectively.
Use Coulombs's Law to find an expression for the force between the two charges.
, where
Ke is Coulomb's contant and
r is the distance between the charges.
We know from the question that
Q1 + Q2 = Qt
So,
Q2 = Qt - Q1

Simplify to obtain,

In order to find the value of Q1 for which F is the maximum, we will use the optimization technique of calculus.
Differentiate F with respect to Q1,

Equate the differential to 0, to obtain the value of Q1 for which F is the maximum.

It follows that
.