Answer:
331.75 V
Explanation:
Given:
Number of turns of the coil, N = 40 turns
Area, A = 0.06 m²
Magnetic Field, B = 0.4 T
Frequency, f = 55 Hz
Maximum induce emf, E₀ = NABω
but ω = 2πf
Maximum induce emf, E₀ = NAB(2πf₀)
Maximum induce emf, E₀ = 2πNABf₀
Where;
N is number of turns of the coil
A is area
B is magnetic field
ω is the angular velocity
f is the frequency
E₀ = 2 × π × 40 × 0.06 × 0.4 × 55
E₀ = 342.81 V
The maximum induced emf is 331.75 V
Answer:
F = 2.49 x 10⁻⁹ N
Explanation:
The electrostatic force between two charged bodies is given by Colomb's Law:
where,
F = Electrostatic Force = ?
k = colomb's constant = 9 x 10⁹ N.m²/C²
q₁ = charge on proton = 1.6 x 10⁻¹⁹ C
q₂ = second charge = 1.4 C
r = distace between charges = 0.9 m
Therefore,
<u>F = 2.49 x 10⁻⁹ N</u>
Answer:
It would be 2600
Explanation:
M/S stands for meters per second. If it moved 1 meter for 2600 seconds, than it would be 2600. You just multiply 2600 by 1! I hope this helps :D
Answer:
Energy May be measured in joule
