Complete Question
A 100-W (watt) light bulb has resistance R=143Ω (ohms) when attached to household current, where voltage varies as V=V0sin(2πft), where V0=110 V, f=60 Hz. The power supplied to the bulb is P=V2R J/s (joules per second) and the total energy expended over a time period [0,T] (in seconds) is 
Compute U if the bulb remains on for 5h
Answer:
The value is 
Explanation:
From the question we are told that
The power rating of the bulb is 
The resistance is 
The voltage is ![V = V_o sin [2 \pi ft]](https://tex.z-dn.net/?f=V%20%20%3D%20%20V_o%20%20sin%20%5B2%20%5Cpi%20ft%5D)
The energy expanded is
The voltage 
The frequency is 
The time considered is 
Generally power is mathematically represented as
=> ![P = \frac{( 110 sin [2 \pi * 60t])^2}{ 144}](https://tex.z-dn.net/?f=P%20%3D%20%20%5Cfrac%7B%28%20110%20%20sin%20%5B2%20%5Cpi%20%2A%2060t%5D%29%5E2%7D%7B%20144%7D)
=> ![P = \frac{ 110^2 [ sin [120 \pi t])^2}{ 144}](https://tex.z-dn.net/?f=P%20%3D%20%20%5Cfrac%7B%20110%5E2%20%5B%20sin%20%5B120%20%5Cpi%20t%5D%29%5E2%7D%7B%20144%7D)
So
![U = \int\limits^T_0 { \frac{ 110^2* [sin [120 \pi t])^2}{ 144}} \, dt](https://tex.z-dn.net/?f=U%20%20%3D%20%20%5Cint%5Climits%5ET_0%20%7B%20%5Cfrac%7B%20110%5E2%2A%20%20%5Bsin%20%5B120%20%5Cpi%20t%5D%29%5E2%7D%7B%20144%7D%7D%20%5C%2C%20dt)
=> ![U = \frac{110^2}{144} \int\limits^T_0 { ( sin^2 [120 \pi t]} \, dt](https://tex.z-dn.net/?f=U%20%20%3D%20%20%5Cfrac%7B110%5E2%7D%7B144%7D%20%5Cint%5Climits%5ET_0%20%7B%20%28%20%20%20sin%5E2%20%5B120%20%5Cpi%20t%5D%7D%20%5C%2C%20dt)
=> 
=> 
=> ![U = \frac{110^2}{144} [\frac{t}{2} - [\frac{1}{2} * \frac{sin(240 \pi t)}{240 \pi} ] ]\left | T} \atop {0}} \right.](https://tex.z-dn.net/?f=U%20%3D%20%20%5Cfrac%7B110%5E2%7D%7B144%7D%20%5B%5Cfrac%7Bt%7D%7B2%7D%20%20-%20%5B%5Cfrac%7B1%7D%7B2%7D%20%2A%20%20%5Cfrac%7Bsin%28240%20%5Cpi%20t%29%7D%7B240%20%5Cpi%7D%20%5D%20%5D%5Cleft%20%20%7C%20T%7D%20%5Catop%20%7B0%7D%7D%20%5Cright.)
=> ![U = \frac{110^2}{144} [\frac{t}{2} - [\frac{1}{2} * \frac{sin(240 \pi t)}{240 \pi} ] ]\left | 18000} \atop {0}} \right.](https://tex.z-dn.net/?f=U%20%3D%20%20%5Cfrac%7B110%5E2%7D%7B144%7D%20%5B%5Cfrac%7Bt%7D%7B2%7D%20%20-%20%5B%5Cfrac%7B1%7D%7B2%7D%20%2A%20%20%5Cfrac%7Bsin%28240%20%5Cpi%20t%29%7D%7B240%20%5Cpi%7D%20%5D%20%5D%5Cleft%20%20%7C%2018000%7D%20%5Catop%20%7B0%7D%7D%20%5Cright.)
![U = \frac{110^2}{144} [\frac{18000}{2} - [\frac{1}{2} * \frac{sin(240 \pi (18000))}{240 \pi} ] ]](https://tex.z-dn.net/?f=U%20%3D%20%20%5Cfrac%7B110%5E2%7D%7B144%7D%20%5B%5Cfrac%7B18000%7D%7B2%7D%20%20-%20%5B%5Cfrac%7B1%7D%7B2%7D%20%2A%20%20%5Cfrac%7Bsin%28240%20%5Cpi%20%2818000%29%29%7D%7B240%20%5Cpi%7D%20%5D%20%5D)
=>
Gravity is a force because it pulls down on objects.
When a swimmer pushes through water to swim they are propelled forward because of the water resistance against the hand and feet. It's A. The water doesn't automatically push the swimmer forward. It releases a reaction after the swimmer pushes through the water.
R1 + R4 = 1430 + 1350 = 2780 = R14 series combination of R1 & R4
R2 + R5 = 1350 + 1150 = 2500 = R25
The circuit has been reduced to 3 resistors in parallel
R314 = 2780 * 1100 / (2780 + 1100) = 788 this is the resistance of the parallel combination of R14 and R3
R31425 = 2500 * 788 / (2500 + 788) = 599 which is the equivalent of the circuit - you can also use the formula for 3 resistors in parallel but this seems simpler
Answer:
i. Cv =3R/2
ii. Cp = 5R/2
Explanation:
i. Cv = Molar heat capacity at constant volume
Since the internal energy of the ideal monoatomic gas is U = 3/2RT and Cv = dU/dT
Differentiating U with respect to T, we have
= d(3/2RT)/dT
= 3R/2
ii. Cp - Molar heat capacity at constant pressure
Cp = Cv + R
substituting Cv into the equation, we have
Cp = 3R/2 + R
taking L.C.M
Cp = (3R + 2R)/2
Cp = 5R/2
