By ideal gas theory, cylinder b has the higher temperature.
We need to know about the ideal gas theory to solve this problem. The ideal gas can be represented by
P . V = n . R . T
where P is the pressure, V is volume, n is the number of molecules, R is the ideal gas constant and T is temperature.
From the question above, we know that
Pa = Pb = P
na = 3nb
Find the temperature of the cylinder a
P . V = n . R . Ta
Ta = P . V /( na . R )
Substitute na
Ta = P . V /( (3nb) . R )
Ta = (1/3) x (P . V /( (nb . R ))
Find the temperature of the cylinder b
P . V = n . R . Tb
Tb = P . V /( nb . R )
The cylinder a temperature is 3 times smaller than the temperature in cylinder b.
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Complementary angles are those whose sum is 90 degrees, right?, then let's find A, A=72 degrees, because its mid point is 36 deg. Then, B = 18 degrees and its mid point is 9 degrees! :)
Answer:
The 50-W bulb glows more brightly than the 100-W bulb
Explanation:
The bulb has a rating of 100 W under 110 V . So it will glow with full brightness when it is fed 110 V . When bulbs are in parallel combination , each bulb receives 110 V . So they glow with full brightness .
When they are in series combination , 110 supply voltage gets distributed between the , thus , reducing the voltage appearing on each of them less than 110 V . So their brightness is reduced.
resistance = V
Bulb having high wattage rating has low resistance resulting in higher current . In the second case , both have same current as they are in series combination . So more heat will be generated in bulb having more resistance . Since 50 W bulb has higher resistance , it will glow brighter than 100 W bulb.
To isolate v₁ from the given equation, subtract aΔt from both members of the equation and simplify:
Therefore, the formula for v₁ is:
