Answer:
The heat capacity for the second process is 15 J/K.
Explanation:
Given that,
Work = 100 J
Change temperature = 5 k
For adiabatic process,
The heat energy always same.
We need to calculate the number of moles and specific heat
Using formula of heat
Put the value into the formula
We need to calculate the heat
Using formula of heat
Put the value into the formula
We need to calculate the heat capacity for the second process
Using formula of heat
Put the value into the formula
Hence, The heat capacity for the second process is 15 J/K.
The initial momentum of the system can be expressed as,
The final momentum of the system can be given as,
According to conservation of momentum,
Plug in the known expressions,
Initially, the second mass move towards the first mass therefore the initial speed of second mass will be taken as negative and the recoil velocity of first mass is also taken as negative.
Plug in the known values,
Thus, the final velocity of second mass is 2.99 m/s.
Answer:
R₁ = 50.77 Ω
Explanation:
Since, we know that:
Electric Power = P = VI
but from Ohm's Law:
V = IR
(or) I = V/R
Therefore,
P = V²/R
(OR) R = V²/P
where,
V = Battery Voltage
R = Resistance of combination
FOR SERIES COMBINATION:
R = Rs = (57 V)²/48 W
Rs = 67.69 Ω
but, we know that:
Rs = R₁ + R₂
R₁ + R₂ = 67.69 Ω
R₁ = 67.69 Ω - R₂ __________ eqn (1)
FOR PARALLEL COMBINATION:
R = Rp = (57 V)²/256 W
Rp = 12.69 Ω
but, we know that:
Rp = (R₁R₂)/(R₁ + R₂) = 12.69 Ω
using eqn (1) and value of R₁ + R₂, we get
Rp = 12.69 = R₂(67.69 - R₂)/67.69
859.08 = 67.69 R₂ - R₂²
R₂² - 67.69 R₂ + 859.08 = 0
Solving this quadratic equation we get the answers:
Either, R₂ = 50.76 Ω
Either, R₂ = 16.92 Ω
Since, it is stated in the question that R₁ > R₂. Therefore, we choose the second value. So,
<u>R₂ = 16.92 Ω</u>
using this value in eqn (1), we get:
R₁ = 67.69 Ω - 16.92 Ω
<u>R₁ = 50.77 Ω</u>
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