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 ![U = \int\limits^T_0 {P(t)} \, dt](https://tex.z-dn.net/?f=U%20%20%3D%20%20%5Cint%5Climits%5ET_0%20%7BP%28t%29%7D%20%5C%2C%20dt)
Compute U if the bulb remains on for 5h
Answer:
The value is ![U = 7.563 *10^{5} \ J](https://tex.z-dn.net/?f=U%20%20%3D%20%207.563%20%2A10%5E%7B5%7D%20%5C%20%20J)
Explanation:
From the question we are told that
The power rating of the bulb is
The resistance is ![R = 143 \ \Omega](https://tex.z-dn.net/?f=R%20%3D%20%20143%20%5C%20%5COmega)
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 ![U = \int\limits^T_0 {P(t)} \, dt](https://tex.z-dn.net/?f=U%20%20%3D%20%20%5Cint%5Climits%5ET_0%20%7BP%28t%29%7D%20%5C%2C%20dt)
The voltage ![V_o = 110 \ V](https://tex.z-dn.net/?f=V_o%20%20%3D%20%20110%20%5C%20%20V)
The frequency is ![f = 60 \ Hz](https://tex.z-dn.net/?f=f%20%3D%20%2060%20%5C%20%20Hz)
The time considered is ![t = 5 \ h = 18000 \ s](https://tex.z-dn.net/?f=t%20%3D%20%205%20%5C%20%20h%20%20%3D%20%2018000%20%5C%20%20s)
Generally power is mathematically represented as
![P = \frac{V^2}{ R}](https://tex.z-dn.net/?f=P%20%3D%20%20%5Cfrac%7BV%5E2%7D%7B%20R%7D)
=> ![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} \int\limits^T_0 { \frac{1 - cos 2 (120\pi t)}{2} } \, dt](https://tex.z-dn.net/?f=U%20%3D%20%20%5Cfrac%7B110%5E2%7D%7B144%7D%20%5Cint%5Climits%5ET_0%20%7B%20%5Cfrac%7B1%20-%20cos%202%20%28120%5Cpi%20t%29%7D%7B2%7D%20%7D%20%5C%2C%20dt)
=> ![U = \frac{110^2}{144} \int\limits^T_0 { \frac{1 - cos 240 \pi t)}{2} } \, dt](https://tex.z-dn.net/?f=U%20%3D%20%20%5Cfrac%7B110%5E2%7D%7B144%7D%20%5Cint%5Climits%5ET_0%20%7B%20%5Cfrac%7B1%20-%20cos%20240%20%5Cpi%20t%29%7D%7B2%7D%20%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)
=> ![U = 7.563 *10^{5} \ J](https://tex.z-dn.net/?f=U%20%20%3D%20%207.563%20%2A10%5E%7B5%7D%20%5C%20%20J)
Answer:The time interval is always positive quantity because time always moves forward. Time can not be reversed.
Explanation:
Velocity is defined as rate of change of position nothing but the displacement. Displacement can be positive or negative.
In case of time interval, time is always a positive quantity. it is impossible to reverse the time. To obtain any displacement we need to measure the time taken. which has to be positive. It always moves forward.
The atom is to small to see so we can't identify it.
Answer: The answer is (D)Reduced impact time will increase the impact force.
Answer:
In a series circuit, how does the voltage supplied by the battery compare to the voltage on each load? The voltage of the battery is equal to the voltage of each load added together. ... The voltage across the two resistors must both have the same voltage of the battery.
Explanation:
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