Explanation:
Assuming the wall is frictionless, there are four forces acting on the ladder.
Weight pulling down at the center of the ladder (mg).
Reaction force pushing to the left at the wall (Rw).
Reaction force pushing up at the foot of the ladder (Rf).
Friction force pushing to the right at the foot of the ladder (Ff).
(a) Calculate the reaction force at the wall.
Take the sum of the moments about the foot of the ladder.
∑τ = Iα
Rw (3.0 sin 60°) − mg (1.5 cos 60°) = 0
Rw (3.0 sin 60°) = mg (1.5 cos 60°)
Rw = mg / (2 tan 60°)
Rw = (10 kg) (9.8 m/s²) / (2√3)
Rw = 28 N
(b) State the friction at the foot of the ladder.
Take the sum of the forces in the x direction.
∑F = ma
Ff − Rw = 0
Ff = Rw
Ff = 28 N
(c) State the reaction at the foot of the ladder.
Take the sum of the forces in the y direction.
∑F = ma
Rf − mg = 0
Rf = mg
Rf = 98 N
This problem is about the rate of the current. It's important to know that refers to the quotient between the electric charge and the time, that's the current rate.

Where Q = 2.0×10^−4 C and t = 2.0×10^−6 s. Let's use these values to find I.

<em>As you can observe above, the division of the powers was solved by just subtracting their exponents.</em>
<em />
<h2>Therefore, the rate of the current flow is 1.0×10^2 A.</h2>
Answer:
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Explanation:
hoped that helped!!
#16
If we put a resistor in circuit it will slow the speed of current
Let's check ohms law

- So if resistance is more current is less
#17
Again use ohms law


- Voltage must be increased