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:
Fx1 (6 m) sin 60 = 300 (3 m) cos 60 balancing torques about floor
Fx1 = 900 * 1/2 / 5.20 = 86.6 N this is the horizontal force that must be supplied by the wall to balance torques about the floor
This is also equal to the static force of friction that must be applied at the point of contact with the floor to balance forces in the x-direction.
Fx1 = Fx2 = 86.6 N
Answer i dont even know im just putting this cus id ont care
Explanation:
Hey! So referring to the data the thing we can clearly see is that in a vacuum, everything, regardless of its mass, falls at the same speed.
Acceleration is often confused with speed, or velocity, but the difference is, acceleration by definition is the rate of which an object falls with respect to its mass and time.
Every single thing in the world falls at the same acceleration, this is because of gravity. The difference is the speed of which it falls. In space, there is not any gravity, and so, the objects are able to fall at the same speed regardless of their mass.
Answer:
Explanation:
If an elements atomic mass and number are known you can also determine the number of protons and neutrons it has. You can also determine where it is on the periodic table. For example, if a element had a atomic mass of 10 then it would be higher up on the table then one with a mass of 20. Btw I do apex too.