Answer:
a) The final pressure is 1.68 atm.
b) The work done by the gas is 305.3 J.
Explanation:
a) The final pressure of an isothermal expansion is given by:

Where:
: is the initial pressure = 5.79 atm
: is the final pressure =?
: is the initial volume = 420 cm³
: is the final volume = 1450 cm³
n: is the number of moles of the gas
R: is the gas constant
Hence, the final pressure is 1.68 atm.
b) The work done by the isothermal expansion is:

Therefore, the work done by the gas is 305.3 J.
I hope it helps you!
In spring mass system we know that angular frequency is given as

f = 8.38 Hz


now we know that speed of SHM at its extreme position is given by

here we know that
A = 17.5 cm


so maximum speed is 9.21 m/s
Answer:
A=1
B=-2
Explanation:
Part A and B of the question wasn't given, however, I attached the relevant parts to solve this question as follows.
From part B as attached, it shows that the right option is C which is
2A+3B=-4
Substituting B with 3A-5 then we form the second equation as shown
2A+3(3A-5)=-4
By simplifying the above equation, we obtain
2A+9A-15=-4
Re-arranging, then
11A=-4+15
Finally
11A=11
A=1
To obtain B, we already know that 3A-5 so substituting the value of A into the above then we obtain
B=3(1)-5=-2
Therefore, required values are 1 and -2