The equilibrium temperature of aluminium and water is 33.2°C
We know that specific heat of aluminium is 0.9 J/gm-K, and that of water is 1 J/gm-K
Now we can calculate the equilibrium temperature
(mc∆T)_aluminium=(mc∆T)_water
15.7*0.9*(53.2-T)=32.5*1*(T-24.5)
T=33.2°C
Answer:
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Answer:
a) v1 = 5.52m/s
b) v2 = -1.52m/s
c) v3 = 4.62m/s
d) vt = 3.85m/s
Explanation:
The velocity of the football wide receiver is his displacement per unit time.
Velocity v = (displacement d)/time t
v = d/t .....1
For each of the cases, equation 1 would be used to calculate the velocity.
a) v1 = d1/t1
d1= 16m
t1 = 2.9s
v1 = 16m/2.9s
v1 = 5.52m/s
b) v2 = d2/t2
d2 = -2.5m
t2 = 1.65s
v2 = -2.5/1.65
v2 = -1.52m/s
c) v3 = d3/t3
d3 = 24m
t3 = 5.2s
v3 = 24/5.2
v3 = 4.62m/s
d) vt = dt/tt
dt = 16m - 2.5m + 24m = 37.5m
tt = 2.9 + 1.65 + 5.2 = 9.75s
vt = 37.5/9.75
vt = 3.85m/s
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Answer:
The maximum amount of work is 
Explanation:
From the question we are told that
The temperature of the environment is 
The volume of container A is 
Initially the number of moles is 
The volume of container B is 
At equilibrium of the gas the maximum work that can be done on the turbine is mathematically represented as
![W = P_A V_A ln[ \frac{V_B}{V_A} ]](https://tex.z-dn.net/?f=W%20%3D%20%20P_A%20V_A%20%20ln%5B%20%5Cfrac%7BV_B%7D%7BV_A%7D%20%5D)
Now from the Ideal gas law
So substituting for 
in the equation above
![W = nRT ln [\frac{V_B}{V_A} ]](https://tex.z-dn.net/?f=W%20%3D%20%20nRT%20ln%20%5B%5Cfrac%7BV_B%7D%7BV_A%7D%20%5D)
Where R is the gas constant with a values of 
Substituting values we have that
![W = 1.2 * (8.314) * (280) * ln [\frac{3.5}{2} ]](https://tex.z-dn.net/?f=W%20%3D%201.2%20%2A%20%288.314%29%20%2A%20%28280%29%20%2A%20ln%20%5B%5Cfrac%7B3.5%7D%7B2%7D%20%5D)
