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Vesnalui [34]
2 years ago
12

PLEASE HELP!

Physics
1 answer:
Kobotan [32]2 years ago
3 0

Answer:

No

Explanation:

Cause a monster truck don

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A 100-W (watt) light bulb has resistance R=143Ω (ohms) when attached to household current, where voltage varies as V=V0sin(2πft)
Phantasy [73]

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

Compute U if the bulb remains on for 5h

Answer:

The value is  U  =  7.563 *10^{5} \  J

Explanation:

From the question we are told that

   The power rating of the bulb is P  =  100 \  W

   The resistance is   R =  143 \ \Omega

   The  voltage is  V  =  V_o  sin [2 \pi ft]

   The  energy expanded is U  =  \int\limits^T_0 {P(t)} \, dt

   The  voltage  V_o  =  110 \  V

   The frequency is  f =  60 \  Hz

    The  time considered is  t =  5 \  h  =  18000 \  s

Generally power is mathematically represented as

             P =  \frac{V^2}{ R}

=>          P =  \frac{( 110  sin [2 \pi * 60t])^2}{ 144}

=>           P =  \frac{ 110^2 [ sin [120 \pi t])^2}{ 144}

So  

     U  =  \int\limits^T_0 { \frac{ 110^2*  [sin [120 \pi t])^2}{ 144}} \, dt

=>  U  =  \frac{110^2}{144} \int\limits^T_0 { (   sin^2 [120 \pi t]} \, dt

=>  U =  \frac{110^2}{144} \int\limits^T_0 { \frac{1 - cos 2 (120\pi t)}{2} } \, dt

=>  U =  \frac{110^2}{144} \int\limits^T_0 { \frac{1 - cos 240 \pi t)}{2} } \, dt

=>  U =  \frac{110^2}{144} [\frac{t}{2}  - [\frac{1}{2} *  \frac{sin(240 \pi t)}{240 \pi} ] ]\left  | T} \atop {0}} \right.

=>  U =  \frac{110^2}{144} [\frac{t}{2}  - [\frac{1}{2} *  \frac{sin(240 \pi t)}{240 \pi} ] ]\left  | 18000} \atop {0}} \right.

U =  \frac{110^2}{144} [\frac{18000}{2}  - [\frac{1}{2} *  \frac{sin(240 \pi (18000))}{240 \pi} ] ]

=>   U  =  7.563 *10^{5} \  J

7 0
3 years ago
Which is the MOST accurate statement about the heating curve and energy?
V125BC [204]

The answer is A). Moving from A to C the temperature and the kinetic energy increases.

3 0
3 years ago
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Choose either eclipses or lunar phases and make a model of the event. Your model can take any form, providing it can explain how
babymother [125]

Answer: See the explanation below.

Explanation: For this assignment, I chose to display how eclipses are created.

My model was made utilizing a 3D displaying device program for all intents and purposes. The items utilized are three models I made for this presentation, Earth, the moon, and the sun. These three models will be utilized for the showcase.  

The light that shines from the sun would create a shadow on the moon. The moon would then catch the light that should've arrived on Earth, making the shadow we call an eclipse. Earth gets a shadow of the moon and the remainder of Earth is lit up from the rest of the light, making an eclipse.  

The individual I demonstrated my project to was [<em>Someone you know</em>], [<em>Pronoun</em>] said it precisely took after the occasion of an eclipse. The light from the sun being shined on to the moon rather than the Earth, creating the shadow we call an eclipse.

5 0
3 years ago
Lol i'm going to fail please help
Novosadov [1.4K]

Answer:

Explanation:

The frequency is 16.0 Hz. That means that 16 of these waves can pass a single point in 1 second. We are given frequency and wavelength. The equation that relates them is

f=\frac{v}{\lambda} where f is frequency, v is velocity, and λ is wavelength. Putting all this together:

16.0=\frac{v}{97.5} and solving for velocity,

v = 16.0(97.5) so

v = 1560 m/s. This wave can travel 1560 meters in a single second!!! Now that we know this velocity, we can use it in a proportion to find our unknown, which is how long, t, it will take to hear this sound 11000m away. (11 km is 11000m):

\frac{1560m}{1s}=\frac{11000}{t} and cross multiply to get

1560t = 11000 so

t = 7.1 seconds

6 0
3 years ago
On level ground a shell is fired with an initial velocity of 49.0 m/s at 70.0 โ above the horizontal and feels no appreciable ai
xenn [34]

Answer:

Horizontal component = 16.8 m/s

Vertical component = 46.0 m/s

Explanation:

If we denote the initial velocity by <em>v</em> and the angle above the horizontal by <em>θ</em>,

the horizontal component of this initial velocity is given by

v_x = v\cos \theta

v_x = (49.0\text{ m/s})\cos\,(70.0^\circ) = 16.8\text{ m/s}

The vertical component is given by

v_y = v\sin\theta

v_x = (49.0\text{ m/s})\sin\,(70.0^\circ) = 46.0\text{ m/s}

3 0
3 years ago
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