Explanation:
hope this would be the right answer ..
Answer:
Explanation:
Given
Pressure, Temperature, Volume of gases is

Let P & T be the final Pressure and Temperature
as it is rigid adiabatic container therefore Q=0 as heat loss by one gas is equal to heat gain by another gas


where Q=heat loss or gain (- heat loss,+heat gain)
W=work done by gas
change in internal Energy of gas
Thus from 1 & 2 we can say that




where 



and 
Answer:
Answer: <u>Height</u><u> </u><u>is</u><u> </u><u>0</u><u>.</u><u>2</u><u>0</u><u>4</u><u> </u><u>m</u>
Explanation:
At the highest point, it is called the maximum height.
• From third newton's equation of motion:

• At maximum height, v is zero
• u is initial speed
• g is -9.8 m/s²
• s is the height

Answer:
The force on q₁ due to q₂ is (0.00973i + 0.02798j) N
Explanation:
F₂₁ = 
Where;
F₂₁ is the vector force on q₁ due to q₂
K is the coulomb's constant = 8.99 X 10⁹ Nm²/C²
r₂₁ is the unit vector
|r₂₁| is the magnitude of the unit vector
|q₁| is the absolute charge on point charge one
|q₂| is the absolute charge on point charge two
r₂₁ = [(9-5)i +(7.4-(-4))j] = (4i + 11.5j)
|r₂₁| = 
(|r₂₁|)² = 148.25

= 0.050938(0.19107i + 0.54933j) N
= (0.00973i + 0.02798j) N
Therefore, the force on q₁ due to q₂ is (0.00973i + 0.02798j) N
A particle that is smaller than an atom or a cluster of particles.