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Gekata [30.6K]
3 years ago
11

A 0.74 mF capacitor is connected to a standard outlet (rms voltage 82 V, frequency 49 Hz ). Determine the magnitude of the curre

nt in the capacitor at t
Physics
1 answer:
disa [49]3 years ago
7 0

Answer:

I =  26.36 cosω t A

Explanation:

Given that

C=0.74 mF

Vrms= 82 V

Frequency ,f= 49 Hz

We know that ω = 2 π f

ω = 2 x  π x 49

ω = 307.72 rad/s

As we know that voltage given as

V= Vo sinω t

V_o=\sqrt2\ V_{rms}

V_o=\sqrt2\ \times 82\

Vo=115.96 V

V=115.96 sinω t

The current given as

I=C\dfrac{dV}{dt}

I=0.74\times \dfrac{dV}{dt}\ mA

\dfrac{dV}{dt}=115.96\omega cos\omega t

I=0.74\times 115.96\times 307.22 cos\omega t\ mA

I = 26362.67 cosω t mA

I =  26.36 cosω t A

This is the current at time ant time t.

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PLEASE HELP ASAP!!!! I need to know some machines that put force where you want it
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Answer:

A wheel and axle, force multipliers and a lever.

Explanation:

6 0
3 years ago
Aliens come blasting into our solar system and wipe out everything but the Sun, the Earth, and Jupiter. Discuss (conceptually) w
KATRIN_1 [288]

Answer:

There is no short answer.

Explanation:

In theory, the gravitational pull between the planets depends on their total mass and the distance between them. The sun has a mass of 1.989x10^{30} kg, Jupiter has a mass of 1.898x10^{27} kg and Earth has a mass of 5.972x10^{24} kg.

Earth is closer to Jupiter than it is to the sun. The other planets that are behind jupiter are gone so that means the sun's pulling force's effect on jupiter is increased drastically since there is no force the balance that and that means that jupiter is going to get closer to earth.

The new center of mass of the new solar system is roughly located between earth and the sun, closer to earth.

I hope this answer helps.

4 0
3 years ago
¿cual es la velocidad de un haz de electrones que marchan sin desviarse cuando pasan a traves de un campo magnetico perpendicula
Elina [12.6K]

Answer:

La velocidad del haz de electrones es 1.78x10⁵ m/s. Este valor se obtuvo asumiendo que el campo magnético dado (3500007) estaba en tesla y que la fuerza venía dada en nN.

Explanation:

Podemos encontrar la velocidad del haz de electrones usando la Ley de Lorentz:

F = |q|vBsin(\theta)     (1)

En donde:

F: es la fuerza magnética = 100 nN

q: es el módulo de la carga del electron = 1.6x10⁻¹⁹ C

v: es la velocidad del haz de electrones =?

B: es el campo magnético = 3500007 T

θ: es el ángulo entre el vector velocidad y el campo magnético = 90°

Introduciendo los valores en la ecuación (1) y resolviendo para "v" tenemos:

v = \frac{F}{qBsin(\theta)} = \frac{100 \cdot 10^{-9} N}{1.6 \cdot 10^{-19} C*3500007 T*sin(90)} = 1.78 \cdot 10^{5} m/s            

Este valor se calculó asumiendo que el campo magnético está dado en tesla (no tiene unidades en el enunciado). De igual manera se asumió que la fuerza indicada viene dada en nN.

Entonces, la velocidad del haz de electrones es 1.78x10⁵ m/s.  

Espero que te sea de utilidad!                                        

7 0
3 years ago
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
Fission is the process of creating energy by
Colt1911 [192]
<h2>Answer: The separation of the components of the nucleous of the atom  </h2><h2> </h2>

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It should be noted that in the process, the emitted neutrons can interact with new fissionable nuclei that will emit new neutrons and so on. Effect better known as chain reaction.

3 0
3 years ago
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