B IS YOUR ANSWER.
Soft sound has small amplitude and louder sound has large amplitude. Since, the second wave has large amplitude, it will have the loudest sound.
Manganese has 2 (two) electron that would free floating and able to form a metallic bond.
The electronic configuration of manganese is (Ar) 3d5 4s2. The two electron in 4s orbital are the valence electron which can freely move from one place to another.
Sure !
Start with Newton's second law of motion:
Net Force = (mass) x (acceleration) .
This formula is so useful, and so easy, that you really
should memorize it.
Now, watch:
The mass of the box is 5.25 kilograms, and the box is
accelerating at the rate of 2.5 m/s² .
What's the net force on the box ?
Net Force = (mass) x (acceleration)
= (5.25 kilograms) x (2.5 m/s²)
Net force = 13.125 newtons .
But hold up, hee haw, whoa ! Wait a second !
Bella is pushing with a force of 15.75 newtons, but the box
is accelerating as if the force on it is only 13.125 newtons.
What happened to the rest of Bella's force ? ?
==> Friction is pushing the box in the opposite direction,
and cancelling some of Bella's force.
How much ?
(Bella's 15.75 newtons) minus (13.125 that the box feels)
= 2.625 newtons backwards, applied by friction.
It depends on the indirect taxes and the share market values
To solve this problem we will use the related concepts in Newtonian laws that describe the force of gravitational attraction. We will use the given value and then we will obtain the proportion of the new force depending on the Radius. From there we will observe how much the force of attraction increases in the new distance.
Planet gravitational force



Distance between planet and star

Gravitational force is

Applying the new distance,


Replacing with the previous force,

Replacing our values


Therefore the magnitude of the force on the star due to the planet is 