Answer:
Do not see a picture or graph but suspect it would show the golf ball falling faster and striking the ground slightly before the soccer ball.
Probably D: Soccer ball was affected by air resistance more than the golf ball.
Explanation:
Even though heavier, friction loss of the greater surface area soccer ball will counter pull of gravity more than the compact golf ball.
In a vacuum, (no friction) both objects fall at the same rate regardless of mass.
The impulse is given by the magnitude of the force times the time interval:

We can see that the force F of the 4 skateboard is always the same (33 N), this means that the skateboard with greatest impulse is the one with the largest time interval, so skateboard C).
Answer:
The charge is 
Explanation:
Given that,
Distance = 2.5 mm
Electric field = 800 NC
Length 
We need to calculate the linear charge density
Using formula of linear charge density


Put the value into the formula


We need to calculate the charge
Using formula of charge

Put the value into the formula


Hence, The charge is 
<span>
Of course. Wind is air in motion, and the gases in air are composed of
all the usual familiar stuff ... atoms, molecules, mass, etc. That's how
the wind moves things ... it has momentum and kinetic energy, which
get transferred to the things that move in the wind.</span>
Answer:
D) 21
Explanation:
When gas absorbs light , electron at lower level jumps to higher level .
and the difference of energy of orbital is equal to energy of radiation absorbed.
Here energy absorbed is equivalent to wavelength of 91.63 nm
In terms of its energy in eV , its energy content is eual to
1243.5 / 91.63 = 13.57 eV. This represents the difference the energy of orbit .
Electron is lying in lowest or first level ie n = 1.
Energy of first level
= - 13.6 / 1² = - 13.6 eV.
Energy of n th level = - 13.6 / n². Let in this level electron has been excited
Difference of energy
= 13.6 - 13.6 / n² = 13.57 ( energy of absorbed radiation)
13.6 / n² = 13.6 - 13.57 = .03
n² = 13.6 / .03 = 453
n = 21 ( approx )