It would be D) the rope is pulled to the right. This is because their is a greater force in that direction.
Gravitational force = G · (mass₁) · (mass₂) / (distance)
(distance²) = G · (mass₁) · (mass₂) / (Gravitational force)
G = 6.67 x 10⁻¹¹ n-m² / kg² (the "gravitational constant")
Distance² = (6.67 x 10⁻¹¹ n-m² / kg²) (28,500 kg) (2.2 x 10⁸ kg) / (39 N)
Distance² = (6.67 · 28,500 · 2.2 x 10⁻³ N-m²) / (39N)
Distance² = (418.209 N-m²) / (39N)
Distance² = 10.72 m²
<em>Distance = 3.275 meters</em>
An absurd scenario, but that's by golly what the math says with the numbers provided. I guess it's a teeny tiny planet orbiting 3.275 meters outside a teeny tiny black hole.
Answer:
We know that the gravitational force between two objects of mass M1 and M2 that are at a distance R, is given by:
F = G*(M1*M2)/R^2
Where G is a constant.
If you reduce one of the masses, then the gravitational force between the objects will change.
So if we take un account the Earth and the Sun, when you reduce the mass of Earth, the force between Earth and the Sun will decrease, and this will change the orbit of the Earth around the Sun.
(The orbit also depends on the gravitational force between the Earth and the other planets in the system, and all those forces also change, which also has an impact in the orbit change)
Answer:
Explanation:
Given that,
We have two capacitors connected in series
C1=2.0-μF
C2=4.0-μF
Then the equivalent of their series connection
1/Ceq = ½ + ¼
1/Ceq= (2+1)/4
1/Ceq=¾
Taking the reciprocal
Ceq= 4/3 μF
The capacitors are connected to a battery of 1kv
V=1000Volts
We know that,
Q=CV
Where Q is charge
C is capacitance and
V is voltage
Then, Q=4/3 ×1000
Q=4000/3 -μC
Since the capacitors are in series, then the charge pass through them, so each charge on the capacitors are 4000/3 μF
After the capacitor has been charge, the capacitor are disconnect and reconnected in parallel to each other,
For parallel connection, they have the same voltage but different charges.
When connected in parallel, there is a charge redistribution,
And the total charge will be 2•4000/3=8000/3 -μF
Then, Q1 +Q2= 8000/3 μF
Now the charge on each capacitor will be, let them have a common voltage V
Q=CV
Then, Q1=C1V
Q1= 2×V=2V
Q2= 4×V=4V
Then, Q1+Q2=8000/3
4V+2V=8000/3
6V=8000/3
V=8000/(3×6)
V=4000/9
V=444.44Volts
Now, Q1=2V
Q1=2×4000/9
Q1=8000/9 μF
Also, Q2=4V
Q2=4×4000/9
Q2=16000/9 μF
It is possible to demonstrate that the maximum distance occurs when the angle at which the projectile is fired is

.
In fact, the laws of motions on both x- and y- directions are


From the second equation, we get the time t at which the projectile hits the ground, by requiring

, and we get:

And inserting this value into Sx(t), we find

And this value is maximum when

, so this is the angle at which the projectile reaches its maximum distance.
So now we can take again the law of motion on the x-axis

And by using

, we find the value of the initial velocity v0: