Answer:
An apple in free fall accelerates toward the Earth with a free fall acceleration, g. The force of the apple on the Earth also causes the Earth to accelerate toward the falling apple. By Newton's Third Law, the force of the Earth on the apple is exactly equal and opposite to the force of the apple on the Earth. By Newton,s Second law, the force of the Earth on the apple is equal to the mass of the apple times g , the accelerations due to gravity. And, the force of the the apple on the Earth is equal to the mass of the Earth times the acceleration of the Earth toward the apple. In conclusion, the magnitude of the forces are equal, or
F ( apple on the Earth) = F( the Earth on the apple) or
M( mass of the earth) x a( the acceleration of the earth toward the apple) = m(mass of the apple) x g( the acceleration of the apple toward the Earth) or
a = (m/M) g
Explanation:
<span>2002 seconds, or 33 minutes, 22 seconds.
First, let's calculate how many joules it will take to lift 78 kg against gravity for 1100 meters. So:
78 kg * 9.8 m/s^2 * 1100 m = 840840 kg*m^2/s^2
Now a watt is defined as kg*m^2/s^3, so a division of the required joules should give us a convenient value of seconds. So:
840840 kg*m^2/s^2 / 420 kg*m^2/s^3 = 2002 seconds.
And 2002 seconds is the same as 33 minutes, 22 seconds.</span>
The moon orbits quite fast: it moves about 0.5 degrees per hour in the sky. In 24 hours it moves 13 degrees. The moon's observed motion eastward results from its physical motion of the moon along its orbit around the Earth. so I think its A .
Answer:
a) 0 < r < R: E = 0, R < r < 2R: E = KQ/r^2, r > 2R: E = 2KQ/r^2
b) See the picture
Explanation:
We can use Gauss's law to find the electric field in all the regions:
EA = qen/e0 where qen is the enclosed charge
Remember that the electric field everywhere outside a sphere is:
E(r) = q/(4*pi*eo*r^2) = Kq/r^2
a)
- For 0 < r < R: There is not enclosed charge because all of it remains on the outer layer of the conducting sphere, therefore E = 0 EA = 0/e0 = 0 E = 0
- For R < r < 2R: Here the enclosed charge is equal Q E = Q/(4*pi*eo*r^2) = KQ/r^2
- For r > 2R: Here the enclosed charge is equal 2Q E = Q/(4*pi*eo*r^2) + Q/(4*pi*eo*r^2) = 2Q/(4*pi*eo*r^2) = 2KQ/r^2
b) At the beginning there is no electric field this is why you see a line in zero, In R the electric field is maximum and then it starts to decrease exponentially with the distance and finally in 2R the field increase a little due to the second sphere to then continue decreasing exponentially with the distance
