Answer:
3.86×10⁶ Newton/coulombs
Explaination:
Applying,
E = F/q....................... Equation 1
Where E = Electric Field, F = Force, q = charge.
From the question,
Given: F = 5.4×10⁻¹ N, q = -1.4×10⁻⁷ coulombs
Substitute these values into equation 1
E = 5.4×10⁻¹/ -1.4×10⁻⁷
E = -3.86×10⁶ Newtons/coulombs
Hence the magnitude of the electric field created by the
negative test charge is 3.86×10⁶ Newton/coulombs
Answer:
Explanation:
Given that,
The current flowing in the circuit, I = 3 A
The power of the battery, P = 25 W
We need to find the resistance of the battery. We know that the power of the battery is given by the formula as follows :
Put all the values to find R.
So, the resistance is equal to .
Answer:
the watt is the unit of power or radiant flax
Answer:
T²= 4π²R³/GM
Explanation:
First we know that
Fg= Fc
Because centripetal force must equal gravitational force
So
GMm/R² = Mv²/R
But velocity is 2πR/T
So by substitution we have
GMm/R²= M (2πR/T)/T
We have
T²= 4π²R³/GM as period
Answer: a) 0.315 (V/L)
Explanation:
From Conservation of angular momentum, we know that
L1 = L2 ,
Therefore MV L/2 = ( Irod + Ib) x W
M/4 x V x L/2 = (M (L/2)^2 + 1/3xMxL^2) x W
M/8 X VL = (ML^2/16 + ML^2 /3 )
After elimination we have,
V/8 = 19/48 x L x W
W = 48/8 x V/19L = 6/19 x V/L
Therefore W = (0.136)X V/L