Explanation:
An perfect mass less spring, attached at one end and with a free mass attached at the other end, will have a distinct frequency of oscillation depending on its constant spring and mass. On the other hand, a spring with mass along its length will not have a characteristic frequency of oscillation.
Alternatively, based on its spring constant and mass per length, it will now have a wave Speed. It would be possible to use all wavelengths and frequencies, as long as the component fλ= S, where S is the spring wave size. If that sounds like longitudinal waves, like solid sound waves.
It would be option C. It rotates, or spins, on its axis, but it revolves around the sun.
The answer for that is True.
Answer:
Explanation:
(Assuming the cell in the circuit has 0 internal resistance)
Ohm's Law is given as:
Voltage is Current multiplied by Resistance.
We can rearrange this formula to give us:
Now we can plug in our values
Gravitational force between 2 objects . . .
F = G · m₁ · m₂ / D²
-- You said that F = 3.5 x 10²² Newtons.
-- G = the gravitational constant = 6.67 x 10⁻¹¹ N m² / kg²
-- You want to find D .
F = G · m₁ · m₂ / D²
Multiply each side by D² . . . D² · F = G · m₁ · m₂
Divide each side by F . . . D² = G · m₁ · m₂ / F
So finally . . . D = √(G · m₁ · m₂ / F )
D = √(6.67 x 10⁻¹¹ N·m²/kg² · Earth mass · Sun mass / 3.5 x 10²² N)
<em>D = 4.37 x 10⁻¹⁷ · √(Earth mass · Sun mass) </em> meters
