C) A current is induced in the coiled wire, which lights the light bulb
The moving magnetic field creates electricity which lights the light bulb
Hope it helps!
If u mean pressure, pressure = Force/Area
Correct me if I am wrong :D
Answer:
250 m/s
Explanation:
The mass of the bullet, m₁ = 100 g = 0.1 kg
The mass of the gun, m₂ = 5 kg
The backward velocity of the gun, v₂ = -5 m/s
Given that the momentum is conserved, we have;
The total initial momentum = The total final momentum
The gun and the bullet are at rest, therefore, we have;
The initial momentum = 0
The total final momentum = m₁·v₁ + m₂·v₂
Where;
v₁ = The forward velocity of the bullet
Therefore, we get;
m₁·v₁ + m₂·v₂ = 0
0.1 kg × v₁ + 5 kg × (-5 m/s) = 0
0.1 kg × v₁ = 5 kg × 5 m/s
v₁ = (5 kg × 5 m/s)/(0.1 kg) = 250 m/s
The forward velocity of the bullet, v₁ = 250 m/s
Answer:
A) 1.4167 × 10^(-11) F
B) r_a = 0.031 m
C) E = 3.181 × 10⁴ N/C
Explanation:
We are given;
Charge;Q = 3.40 nC = 3.4 × 10^(-9) C
Potential difference;V = 240 V
Inner radius of outer sphere;r_b = 4.1 cm = 0.041 m
A) The formula for capacitance is given by;
C = Q/V
C = (3.4 × 10^(-9))/240
C = 1.4167 × 10^(-11) F
B) To find the radius of the inner sphere,we will make use of the formula for capacitance of spherical coordinates.
C = (4πε_o)/(1/r_a - 1/r_b)
Rearranging, we have;
(1/r_a - 1/r_b) = (4πε_o)/C
ε_o is a constant with a value of 8.85 × 10^(−12) C²/N.m
Plugging in the relevant values, we have;
(1/r_a - 1/0.041) = (4π × 8.85 × 10^(−12) )/(1.4167 × 10^(-11))
(1/r_a) - 24.3902 = 7.8501
1/r_a = 7.8501 + 24.3902
1/r_a = 32.2403
r_a = 1/32.2403
r_a = 0.031 m
C) Formula for Electric field just outside the surface of the inner sphere is given by;
E = kQ/r_a²
Where k is a constant value of 8.99 × 10^(9) Nm²/C²
Thus;
E = (8.99 × 10^(9) × 3.4 × 10^(-9))/0.031²
E = 3.181 × 10⁴ N/C
C. sound waves and water waves
