Answer:
v = √(2gh)
Explanation:
Given:
Δy = h
v₀ = 0
a = g
Find: v
v² = v₀² + 2aΔy
v² = (0)² + 2gh
v = √(2gh)
You can also prove this with conservation of energy.
PE = KE
mgh = ½mv²
v = √(2gh)
Complete Question
The complete question shown on the first uploaded image
Answer:
a)
The force on Q due to dipole is Attractive
b)
The charge Q exerts attractive force on the dipole
c)
Yes from the above parts, force depends on the sign of charge
d)
![F = kQq[\frac{d^{2}+2rd}{r^{2}(d+r)^{2}} ]](https://tex.z-dn.net/?f=F%20%3D%20kQq%5B%5Cfrac%7Bd%5E%7B2%7D%2B2rd%7D%7Br%5E%7B2%7D%28d%2Br%29%5E%7B2%7D%7D%20%5D)
e)
The magnitude o force decrease by a factor of 8.0 times
Explanation:
The explanation is shown on the second uploaded image
Answer: V=IR
Explanation: for a series circuit connected to a battery supply, the total emf across the circuit is given as
E = I(R + r) and by expanding, we have that E =IR + It
Where r is the internal resistance of the battery
I is the total current flowing in the circuit
R total load resistance in the circuit.
E is the total emf of the circuit.
The total emf is the sum of 2 separate voltages.
"IR" which is the terminal voltage and "Ir" which is the loss voltage.
The teenila voltage is the voltage flowing in the circuit based on the equivalent resistance of the circuit while the loss voltage is the wasted voltage based on the internal resistance of the battery source.
