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)
=> 
Answer: The motion of the object will remain the same
Explanation:
Answer:
Image distance of apple=-6.7 cm
Magnification of apple=0.33
Explanation:
We are given that an apple is placed 20.cm in front of a diverging lens.
Object distance=u=-20 cm
Focal length=f=-10 cm
Because focal length of diverging lens is negative.
We have to find the image distance and magnification of the apple.
Lens formula

Substitute the values then we get




Image distance of apple=-6.7 cm
Magnification=m=
Magnification of apple=
Hence, the magnification of apple=0.33
<u>Answer:</u>
<em>To relate the type of box material to the warmth inside the box
</em>
<u>Explanation:</u>
3 boxes are made with three different materials glass, plastic and Aluminium. Thermal conductivity is different for different materials. thermal conductors allow easy flow of heat through them and insulators allow minimal or no flow of heat through them.
Thus the amount by which an object gets heated up depends on the value of its conductivity. In this experiment, glass and plastic are insulators and aluminium is a conductor. Among glass and plastic, plastic is a better insulator.
<em>Thus the heat contained in the boxes will be of the order </em>Aluminium>glass>plastic.