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alexandr1967 [171]
4 years ago
13

A 250 - ω resistor is connected in series with a 4.80 - μf capacitor. the voltage across the capacitor is vc=(7.60v)⋅sin[(120rad

/s) t ]. part a determine the capacitive reactance of the capacitor.

Physics
2 answers:
nata0808 [166]4 years ago
8 0

The capacitive reactance of the capacitor is about 1740 Ω

\texttt{ }

<h3>Further explanation</h3>

Let's recall Capacitive Reactance and Impedance formula as follows:

\boxed{ X_c = \frac{1}{ \omega C } }

<em>where:</em>

<em>Xc = capacitive reactance ( Ohm )</em>

<em>ω = angular frequency ( rad/s )</em>

<em>C = capacitance ( F )</em>

\texttt{ }

\boxed{ Z^2 = R^2 + (X_L - X_c)^2}

<em>where:</em>

<em>Xc = capacitive reactance ( Ohm )</em>

<em>XL = inductive reactance ( Ohm )</em>

<em>R = resistance ( Ohm )</em>

<em>Z = impedance ( Ohm )</em>

Let us now tackle the problem!

\texttt{ }

<u>Given:</u>

resistance = R = 250 Ω

capacitance = C = 4.80 μF = 4.80 × 10⁻⁶ F

angular frequency = ω = 120 rad/s

maximum voltage across the capacitor = Vc_max = 7.60 V

<u>Asked:</u>

capacitive reactance of the capacitor = Xc = ?

<u>Solution:</u>

X_c = \frac{1}{ \omega C }

X_c = \frac{1}{ 120 \times 4.80 \times 10^{-6} }

\boxed {X_c \approx 1740\ \Omega}

\texttt{ }

<h3>Conclusion :</h3>

The capacitive reactance of the capacitor is about 1740 Ω

\texttt{ }

<h3>Learn more</h3>
  • The three resistors : brainly.com/question/9503202
  • A series circuit : brainly.com/question/1518810
  • Compare and contrast a series and parallel circuit : brainly.com/question/539204

\texttt{ }

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Alternating Current

galben [10]4 years ago
6 0
<span>553 ohms The Capacitive reactance of a capacitor is dependent upon the frequency. The lower the frequency, the higher the reactance, the higher the frequency, the lower the reactance. The equation is Xc = 1/(2*pi*f*C) where Xc = Reactance in ohms pi = 3.1415926535..... f = frequency in hertz. C = capacitance in farads. I'm assuming that the voltage and resistor mentioned in the question are for later parts that are not mentioned in this question. Reason is that they have no effect on the reactance, but would have an effect if a question about current draw is made in a later part. With that said, let's calculate the reactance. The 120 rad/s frequency is better known as 60 Hz. Substitute known values into the formula. Xc = 1/(2*pi* 60 * 0.00000480) Xc = 1/0.001809557 Xc = 552.6213302 Rounding to 3 significant figures gives 553 ohms.</span>
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2 years ago
A worker exerts a pulling force on a box. The worker exerts this force by attaching a rope to the box and pulling on the rope so
shusha [124]

Answer:

<em>The magnitude of the pulling force is 20.66% of the gravitational force acting on the box</em>

Explanation:

<u>Accelerated Motion</u>

The net force exerted on a body is the (vector) sum of all forces applied to the body. The net force can be decomposed in its rectangular components and the dynamics of the body can be studied in each direction x,y separately.

Let's start off by calculating the acceleration the worker gives to the box when pulling it. The distance traveled by the box initially at rest in a time t at an acceleration a is given by

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Now we analyze the geometric of the forces applied to the box. Please refer to the free body diagram provided below.

The forces in the y-axis must be in equilibrium since no movement takes place there, thus, being g the acceleration of gravity:

T_y+N=m.g

There Ty is the vertical component of the tension of the rope, N is the normal force, and m is the mass of the box

The decomposition of T gives us

T_y=Tsin\theta

T_x=Tcos\theta

Solving the above equation for N

Tsin\theta+N=m.g

N=m.g-Tsin\theta\text{..........[1]}

Now for the x-axis, there are two forces acting on the box, the x-component of the tension and the friction force Fr. Those forces are not equilibrated, thus acceleration is produced:

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Recalling that

F_r=\mu N

Tcos\theta-\mu N=m.a

Replacing N from [1]

Tcos\theta-\mu (m.g-Tsin\theta)=m.a

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T(cos\theta+\mu sin\theta)=m.a+\mu m.g

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\displaystyle T=m\frac{a+\mu g}{cos\theta+\mu sin\theta}

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\displaystyle T=m\frac{0.763+0.10\cdot 9.8}{cos36.8^o+0.10 sin36.8^o}

T=2.0252m

Dividing by the weight of the box m.g

T/(m.g)=2.0252/9.8=0.2066

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7 0
4 years ago
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Damm [24]

Answer:

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3 years ago
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