Answer:
<em>Velocity</em><em> </em><em>-</em><em>time</em><em> </em><em>graph</em><em> </em>
Explanation:
hope it helps ✌️✌️
I believe this is what you have to do:
The force between a mass M and a point mass m is represented by

So lets compare it to the original force before it doubles, it would just be the exact formula so lets call that F₁
So F₁ = G(Mm/r^2)
Now the distance has doubled so lets account for this in F₂:
F₂ = G(Mm/(2r)^2)
Now square the 2 that gives you four and we can pull that out in front to give
F₂ =
G(Mm/r^2)
Now we can replace G(Mm/r^2) with F₁ as that is the value of the force before alterations
now we see that:
F₂ =
F₁
So the second force will be 0.25 (1/4) x 1600 or 400 N.
Answer:
It would point up.
Explanation:
Since I am at the earth's geographic north magnetic pole, the place on the earth's surface that compasses point toward, the north pole of the compass would also point towards the earth's geographic north magnetic pole, since all other compasses point toward there.
Since the compass is free to swivel in any direction, the compass would point up, since it is at the earth's geographic north magnetic pole, the place on the earth's surface that compasses point toward.
So, the compass would point up.
Answer:
the question is incomplete, the complete question is
"A circular coil of radius r = 5 cm and resistance R = 0.2 ? is placed in a uniform magnetic field perpendicular to the plane of the coil. The magnitude of the field changes with time according to B = 0.5 e^-t T. What is the magnitude of the current induced in the coil at the time t = 2 s?"
2.6mA
Explanation:
we need to determine the emf induced in the coil and y applying ohm's law we determine the current induced.
using the formula be low,

where B is the magnitude of the field and A is the area of the circular coil.
First, let determine the area using
where r is the radius of 5cm or 0.05m

since we no that the angle is at
we determine the magnitude of the magnetic filed


the Magnitude of the voltage is 0.000532V
Next we determine the current using ohm's law

