Answer:
1.99×10^-4coulombs
Explanation:
The charge (Q) across the resistor the directly proportional to the voltage (V) where capacitance of the capacitor(C) is the proportionality constant. Mathematically, Q = CV
If V is the voltage across the resistor, V = IR (according to ohm's law) where I is the current in the resistor and R is the resistance.
We need to calculate the voltage on the resistor first when 0.18A current is passed through it.
V = 0.18 × 185
V = 33.3Volts
The charge Q on the resistor will be;
Q = CV
Were C = 6.00 μF, V = 33.3
Q= 6×10^-6 ×33.3
Q = 0.0001998
Q= 1.99×10^-4Coulombs
Answer:
2 x 10⁻³ volts
Explanation:
B = magnetic of magnetic field parallel to the axis of loop = 1 T
= rate of change of area of the loop = 20 cm²/s = 20 x 10⁻⁴ m²
θ = Angle of the magnetic field with the area vector = 0
E = emf induced in the loop
Induced emf is given as
E = B
E = (1) (20 x 10⁻⁴ )
E = 2 x 10⁻³ volts
E = 2 mV
Answer:
The distance of separation is 
Explanation:
The mass of the each ball is 
The negative charge on each ball is 
Now we are told that the lower ball is restrained from moving this implies that the net force acting on it is zero
Hence the gravitational force acting on the lower ball is equivalent to the electrostatic force i.e

=> 
here k the the coulomb's constant with a value 
So
![0.01 * 9.8 = \frac{ 9*10^9 *[1*10^{-6} * 1*10^{-6}]}{d}](https://tex.z-dn.net/?f=0.01%20%2A%209.8%20%20%3D%20%20%5Cfrac%7B%209%2A10%5E9%20%2A%5B1%2A10%5E%7B-6%7D%20%2A%201%2A10%5E%7B-6%7D%5D%7D%7Bd%7D)

Answer:

Explanation:
First of all, we need to find the pressure exerted on the sphere, which is given by:

where
is the atmospheric pressure
is the water density
is the gravitational acceleration
is the depth
Substituting,

The radius of the sphere is r = d/2= 1.1 m/2= 0.55 m
So the total area of the sphere is

And so, the inward force exerted on it is

Answer:

Explanation:
As we know that water from the fountain will raise to maximum height

now by energy conservation we can say that initial speed of the water just after it moves out will be




Now we can use Bernuolli's theorem to find the initial pressure inside the pipe


