Answer:
The magnetic field through the wire must be changing
Explanation:
According to Faraday's law, the induced emf, ε in a metallic conductor is directly proportional to the rate of change of magnetic flux,Φ through it. This is stated mathematically as ε = dΦ/dt.
Now for the wire, the magnetic flux through it is given by Φ = ABcosθ where A = cross-sectional area of wire, B = magnetic field and θ = angle between A and B.
So, dΦ/dt = dABcosθ/dt
Since A and B are constant,
dΦ/dt = ABdcosθ/dt = -(dθ/dt)ABsinθ
Since dθ/dt implies a change in the angle between A and B, since A is constant, it implies that B must be rotating.
So, <u>for an electric current (or voltage) to be produced in the wire, the magnetic field must be rotating or changing</u>.
The Moon is 3.8 108 m from Earth and has a mass of 7.34 1022 kg. 5.97 1024 kg is the mass of the Earth.
<h3>What kind of gravitational pull does the moon have on the planet?</h3>
On the surface of the Moon, the acceleration caused by gravity around 1.625 m/s2 which is 16.6% greater than on the surface of the Earth 0.166.
<h3>What does the Earth's center's gravitational pull feel like?</h3>
Gravity is zero if you are in the centre of the earth since everything around you is pulling "up" (up is the only direction).
<h3>Where is the Earth's and the moon's gravitational centre?</h3>
It is around 1700 kilometres below Earth's surface.
To know more about gravitational force visit:-
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<span>The force of static friction F equals the coefficient of friction u times the normal force N the object exerts on the surface: F = uN. N is the centripetal force of the wall on the people; N = ma_N, where m is the mass of the people and a_N is the centripetal acceleration.
The people will not slip down if F is greater than the force of gravitation: F = uma_N > mg, or u > g/a_N.
a_N is the velocity v of the people squared divided by the radius of the room r: a_N = v^2/r.
The circumference of the room is 2 pi r = 28.3 m. So v = 28.3 * 0.8 m/sec = 22.6 m/sec.
So a_N = 114 m/sec^2.
g = 9.81 m/sec^2, so u must be at least 9.81/114 = 0.086.</span>
Complete Question:
A 10 kg block is pulled across a horizontal surface by a rope that is oriented at 60° relative to the horizontal surface.
The tension in the rope is constant and equal to 40 N as the block is pulled. What is the instantaneous power (in W) supplied by the tension in the rope if the block when the block is 5 m away from its starting point? The coefficient of kinetic friction between the block and the floor is 0.2 and you may assume that the block starting at rest.
Answer:
Power = 54.07 W
Explanation:
Mass of the block = 10 kg
Angle made with the horizontal, θ = 60°
Distance covered, d = 5 m
Tension in the rope, T = 40 N
Coefficient of kinetic friction, 
Let the Normal reaction = N
The weight of the block acting downwards = mg
The vertical resolution of the 40 N force, 
Power, 
