Complete Question
The complete question is shown on the first uploaded image
Answer:
The workdone is 
Explanation:
From the question we are told that
The initial Volume is 
The final volume is 
The external pressure is
Generally the change in volume is

Substituting values we have


Generally workdone is mathematically represented as

W is negative because the working is done on the environment by the system which is indicated by volume increase
Substituting values


Now 
Therefore 

Answer:
U₂ = 400 KJ
Explanation:
Given that
Initial energy of the tank ,U₁= 800 KJ
Heat loses by fluid ,Q= - 500 KJ
Work done on the fluid ,W= - 100 KJ
Sign -
1.Heat rejected by system - negative
2.Heat gain by system - Positive
3.Work done by system = Positive
4.Work done on the system-Negative
Lets take final internal energy =U₂
We know that
Q= U₂ - U₁ + W
-500 = U₂ - 800 - 100
U₂ = -500 +900 KJ
U₂ = 400 KJ
Therefore the final internal energy = 400 KJ
Answer:
w = 4,786 rad / s
, f = 0.76176 Hz
Explanation:
For this problem let's use the concept of angular momentum
L = I w
The system is formed by the two discs, during the impact the system remains isolated, we have the forces are internal, this implies that the external torque is zero and the angular momentum is conserved
Initial Before sticking
L₀ = 0 + I₂ w₂
Final after coupling
= (I₁ + I₂) w
The moments of inertia of a disk with an axis of rotation in its center are
I = ½ M R²
How the moment is preserved
L₀ = 
I₂ w₂ = (I₁ + I₂) w
w = w₂ I₂ / (I₁ + I₂)
Let's reduce the units to the SI System
d₁ = 60 cm = 0.60 m
d₂ = 40 cm = 0.40 m
f₂ = 200 min-1 (1 min / 60 s) = 3.33 Hz
Angular velocity and frequency are related.
w₂ = 2 π f₂
w₂ = 2π 3.33
w₂ = 20.94 rad / s
Let's replace
w = w₂ (½ M₂ R₂²) / (½ M₁ R₁² + ½ M₂ R₂²)
w = w₂ M₂ R₂² / (M₁ R₁² + M₂ R₂²)
Let's calculate
w = 20.94 8 0.40² / (12 0.60² + 8 0.40²)
w = 20.94 1.28 / 5.6
w = 4,786 rad / s
Angular velocity and frequency are related.
w = 2π f
f = w / 2π
f = 4.786 / 2π
f = 0.76176 Hz
The equivalent resistance of several devices connected in parallel is given by

where

are the resistances of the various devices. We can see that every time we add a new device in parallel, the term

increases, therefore the equivalent resistance of the circuit

decreases.
But Ohm's law:

tells us that if the equivalent resistance decreases, the total current in the circuit increases. The power dissipated through the circuit (and so, the heat produced) depends on the square of the current:

therefore if there are too many devices connected in parallel, this can be a problem because there could be too much power dissipated (and too much heat) through the circuit.
Radioactive "decay" means particles and stuff shoot OUT of a nucleus.
After that happens, there's less stuff in the nucleus than there was before.
So the new mass number is always less than the original mass number.