Answer:
Explanation:. What is an excess of electric charge on an
object?
Answer:
T - M g - Ff = M a describes the acceleration of the object
T = M g + Ff if the object moves at constant speed (a = zero)
Ff = μ M g = .3 * 40 * 9.8 = 118 N the force of friction
Ff will be in a direction opposite to the motion
M g = 40 kg * 9.8 m/s = 392 N
a) T = M g + Ff for an object moving upwards
T = 392 + 118 = 510 N
b) T = M g - Ff for an object moving downwards
T = 392 - 118 = 274 N
1.7 Btu
1 watt = 3.41214 Btu/h
1watt * 1h = 3.41214 Btu/h * h
1 = 3.41214 Btu/ (watt*h)/
0.5 watt * h = 0.5 watt*h * 3.41214 Btu/(watt*h) = 1.706 Btu
Complete question is;
The energy flow to the earth from sunlight is about 1.4kW/m²
(a) Find the maximum values of the electric and magnetic fields for a sinusoidal wave of this intensity.
(b) The distance from the earth to the sun is about 1.5 × 10^(11) m. Find the total power radiated by the sun.
Answer:
A) E_max ≈ 1026 V/m
B_max = 3.46 × 10^(-6) T
B) P = 3.95 × 10^(26) W
Explanation:
We are given;
Intensity; I = 1.4kW/m² = 1400 W/m²
Formula for maximum value of electric field in relation to intensity is given as;
E_max = √(2I/(ε_o•c))
Where;
ε_o is electric constant = 8.85 × 10^(-12) C²/N.m²
c is speed of light = 3 × 10^(8) m/s
Thus;
E_max = √(2 × 1400)/(8.85 × 10^(-12) × 3 × 10^(8)))
E_max ≈ 1026 V/m
Formula for maximum magnetic field is;
B_max = E_max/c
B_max = 1026/(3 × 10^(8))
B_max = 3.46 × 10^(-6) T
Formula for the total power is;
P = IA
Where;
A is area = 4πr²
We are given;
Radius; r = 1.5 × 10^(11) m
A = 4π × (1.5 × 10^(11))² = 2.82 × 10^(23) m²
P = 1400 × 2.82 × 10^(23)
P = 3.95 × 10^(26) W
Answer:
It will produce a stronger magnetic field.
Explanation:
Ampere’s law states the magnetic field of a solenoid is directly proportional to the current supplied.
