Answer:
Final mass=0.89kg
Final pressure=5.6bar
Explanation:
To find mass,m=v/v1
But v1=vf + x(vg-vf)
Vf= 0.001093m^3/kg
Vg= 0.3748m^3/kg
V1= 0.001093+0.5(0.3748-0.001093)
V1= 0.225m^3/kg
M= 0.20/0.225 =0.89kg
Final pressure will be:
V/V1= P/P1
Cross multiply
VP1=V1P
P1= 0.225×5/0.2
P1=:5.6 bar
The atomic bomb explosion, depends on the energy released, W, the elapsed time, t, and the air density which altogether lead to the vaporisation of the contents within the mushroom cloud created.
<h3>What is an atomic bomb?</h3>
An atomic bomb is type of nuclear weapon that releases a vast amount of energy upon explosion in the form of fission reaction.
During an explosion of atomic bomb, a cloud of mushroom fire is formed which vaporises anything found within it. The extent of destruction depends on:
Therefore, the atomic bomb explosion, depends on the energy released, W, the elapsed time, t, and the air density which altogether lead to the vaporisation of the contents within the mushroom cloud created.
Learn more about nuclear energy here:
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Answer:
figure out what the letter mean , the figure how it's worth
Answer:
Well, temperature is simply an average measure of the kinetic energy for particles of matter. Another way of putting it would be that temperature simply describes the average vibration of particles. Because the motion of all particles is random, they don’t all move at the same speed and in the same direction.
Answer:
-27 mJ
Explanation:
The work done in moving the three charges an infinite distance from each other is the electric potential energy of the system. Let q₁ = q₂ = 6 μC = 6 × 10⁻⁶ C, q₃ = -3.5 μC = -3.5 × 10⁻⁶ C. Let r₁ be the distance between q₁ and q₃. r₁ = 2 m. Let r₂ be the distance between q₂ and q₃. r₂ = 2 m. Let r₃ be the distance between q₁ and q₂. r₃ = √(2 - 0)² + (0 - 2)² = √4 = 2 m
The potential energy is thus U
U = kq₁q₂/r₃ + kq₁q₃/r₁ + kq₂q₃/r₂
Since q₁ = q₂ = q and r₁ = r₂ = r₃ = r
U = kq(q + 2q₃)/r =
U = 9 × 10⁹ Nm²/C² × 6 × 10⁻⁶ C(6 × 10⁻⁶ C + 2 × -3.5 × 10⁻⁶ C)/2 m
U = 9 × 10⁹ Nm²/C² × 6 × 10⁻⁶ C(6 × 10⁻⁶ C - 7 × 10⁻⁶ C)/2 m
U = 9 × 10⁹ Nm²/C² × 6 × 10⁻⁶ C(- 1 × 10⁻⁶ C)/2 m
U = -54 × 10⁻³ Nm²/2 m
U = -27 × 10⁻³ Nm
U = -27 × 10⁻³ J
U = -27 mJ
So a work of -27 mJ must be done to move the charges an infinite distance away from each other