Answer:
724.3J/Kg.K
Explanation:
CHECK THE COMPLETE QUESTION BELOW
Compute the specific heat capacity at constant volume of nitrogen (N2) gas.and compare with specific heat of liquid water. The molar mass of N2 is 28.0 g/mol.
The specific heat capacity can be computed by using expression below
c= CV/M
Where c= specific heat capacity
M= molar mass
CV= molar hear capacity
Nitrogen is a diatomic element, the Cv can be related to gas constant with 5/2R
Where R= 8.314J/mol.k
Molar mass= 28 ×10^-3Kg/mol
If we substitute to the expression, we have
c= (5R/2)/(M)
=5R/2 × 1/M
=(5×8.314) /(2×28 ×10^-3)
=724.3J/Kg.K
Hence, the specific heat capacity at constant volume of nitrogen (N2) gas is
724.3J/Kg.K
The specific heat of liquid water is about 4182 J/(K kg) which is among substance with high specific heat, therefore specific heat of Nitrogen gas is 724.3J/Kg.K which is low compare to that of liquid water.
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Answer:
a) Q1= Q2= 11.75×10^-6Coulombs
b) Q1 =15×10^-6coulombs
Q2 = 38.75×10^-6coulombs
Explanation:
a) For a series connected capacitors C1 and C2, their equivalent capacitance C is expressed as
1/Ct = 1/C1 + 1/C2
Given C1 = 3.00 μF C2 = 7.75μF
1/Ct = 1/3+1/7.73
1/Ct = 0.333+ 0.129
1/Ct = 0.462
Ct = 1/0.462
Ct = 2.35μF
V = 5.00Volts
To calculate the charge on each each capacitors, we use the formula Q = CtV where Cf is the total equivalent capacitance
Q = 2.35×10^-6× 5
Q = 11.75×10^-6Coulombs
Since same charge flows through a series connected capacitors, therefore Q1= Q2=
11.75×10^-6Coulombs
b) If the capacitors are connected in parallel, their equivalent capacitance will be C = C1+C2
C = 3.00 μF + 7.75 μF
C = 10.75 μF
For 3.00 μF capacitance, the charge on it will be Q1 = C1V
Q1 = 3×10^-6 × 5
Q1 =15×10^-6coulombs
For 7.75 μF capacitance, the charge on it will be Q2 = 7.75×10^-6×5
Q2 = 38.75×10^-6coulombs
Note that for a parallel connected capacitors, same voltage flows through them but different charge, hence the need to use the same value of the voltage for both capacitors.
Answer:
true
Explanation:
I just had a test on this
Average acceleration = (change in speed) / (time for the change)
In this case . . .
Average acceleration = (speed at the end - speed at the beginning)/10sec.
We can't be any more specific without knowing the difference
between the beginning speed and the ending speed.
