Must be isolated from the temperature of the gas
Answer:
The time of flight of the ball is 1.06 seconds.
Explanation:
Given 

Also, 

Let us say the velocity in the x-direction is
and in the y-direction is
. And acceleration in the x-direction is
and in the y-direction is
.
Also,
is distance covered in x and y direction respectively. And
is the time taken by the ball to hit the backboard.
We can write
. Where
is velocity of ball.
Now,


Also,
.
Plugging this value in


So, the time of flight of the ball is 1.06 seconds.
Answer:
the mass of the air in the classroom = 2322 kg
Explanation:
given:
A classroom is about 3 meters high, 20 meters wide and 30 meters long.
If the density of air is 1.29 kg/m3
find:
what is the mass of the air in the classroom?
density = mass / volume
where mass (m) = 1.29 kg/m³
volume = 3m x 20m x 30m = 1800 m³
plugin values into the formula
1.29 kg/m³ = <u> mass </u>
1800 m³
mass = 1.29 kg/m³ ( 1800 m³ )
mass = 2322 kg
therefore,
the mass of the air in the classroom = 2322 kg
FVJDJFN.s<ldF KN,M c":F,BJ TNHIJRT IHJYODIFG
Answer:

Explanation:
<u>LC Circuit</u>
It's a special circuit made of three basic elements: The AC source, a capacitor, and an inductor. The charge, current, and voltage are oscillating when there is an interaction between the electric and magnetic fields of the elements. The following variables will be used for the formulas:
= charge of the capacitor in any time 
= initial charge of the capacitor
=angular frequency of the circuit
= current through the circuit in any time 
The charge in an LC circuit is given by

The current is the derivative of the charge

We are given

It means that
![q(t_1) = q_0 \, cos (\omega t_1 )=q_1\ .......[eq 1]](https://tex.z-dn.net/?f=q%28t_1%29%20%3D%20q_0%20%5C%2C%20cos%20%28%5Comega%20t_1%20%29%3Dq_1%5C%20.......%5Beq%201%5D)
![i(t_1) = - \omega q_0 \, sin(\omega t_1)=i_1.........[eq 2]](https://tex.z-dn.net/?f=i%28t_1%29%20%3D%20-%20%5Comega%20q_0%20%5C%2C%20sin%28%5Comega%20t_1%29%3Di_1.........%5Beq%202%5D)
From eq 1:

From eq 2:

Squaring and adding the last two equations, and knowing that


Operating

Solving for 

Now we know the value of
, we repeat the procedure of eq 1 and eq 2, but now at the second time
, and solve for 

Solving for 

Now we replace the given values. We'll assume that the placeholder is a pi for the frequency, i.e.




Finally

