The magnitude of e.m.f induced in the loop when t = 2 s is 31 Volts.
<h3>emf induced in the loop</h3>
The magnitude of e.m.f induced in the loop is calculated as follows;
emf = dФ/dt
Ф = 6t² + 7t
dФ/dt = 12t + 7
at t = 2 seconds
emf = dФ/dt = 12(2) + 7 = 31 V
Thus, the magnitude of e.m.f induced in the loop when t = 2 s is 31 Volts.
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Answer:
Explanation: P = 300 W and t = 2 min = 120 s
Energy Q = Pt = 300 W · 120 s = 36 000 J.
Thus, plate can not produce 45 000 J heat.
Add 100 ML of sulfuric acid to 400 ML of base as an 80H B.
An increase in voltage causes the flow of electric current to INCREASE. An increase in resistance causes the flow of electric current to DECREASE.
100% right
The magnitude of work done by the gas is 279 J and the sign is negative so W = -279 J as work is done by the system.
<u>Explanation:</u>
According to first law of thermodynamics, the change in internal energy of the system is equal to the sum of the heat energy added or released from the system with the work done on or by the system. If the heat energy is added to the system to perform a certain work, then the heat energy is taken as positive, while it will be negative when the heat energy is released from the system.
Similarly, in this case, the heat energy of 597 J is added to the system. So the heat energy will be positive, while the gas expansion occurs means work is done by the system.
ΔU = Q+W
Since ΔU is the change in internal energy which is given as 318 J and the heat energy added to the system is Q = 597 J.
Then the work done by the gas = ΔU - Q = 318 J - 597 J = - 279 J.
As the work is done by the system, so it will be denoted in negative sign and the magnitude of work done by the gas is 279 J.
