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SOVA2 [1]
4 years ago
7

At what frequency will a 3.0 μF capacitor have a reactance of 7.0 kΩ?

Physics
1 answer:
blagie [28]4 years ago
5 0

Answer:

Frequency, f = 7.57 Hz

Explanation:

It is given that,

Capacitance, C=3\ \mu F=3\times 10^{-6}\ F

Capacitive reactance, X_C=7\ k\Omega=7\times 10^3\ \Omega

We need to find the frequency. The capacitive reactance of the capacitor is given by :

X_C=\dfrac{1}{2\pi fC}

f is the frequency

f=\dfrac{1}{2\pi CX_C}

f=\dfrac{1}{2\pi \times 3\times 10^{-6}\ F\times 7\times 10^3\ \Omega}

f = 7.57 Hz

Hence, this is the required solution.

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A child sits on a toboggan and slides down a hill with a particular acceleration. If another child joins the first one on the to
Troyanec [42]

Answer:

it is airpod

Explanation:

becue  is gay

3 0
4 years ago
By means of a rope whose mass is negligible, two blocks are suspended over a pulley, as the drawing shows, with m1 = 12.1 kg and
puteri [66]

Answer:

14.8 kg

Explanation:

We are given that

m_1=43.7 kg

m_2=12.1 kg

g=9.8 m/s^2

a=\frac{1}{2}(9.8)=4.9 m/s^2

We have to find the mass of the pulley.

According to question

T_2-m_2 g=m_2 a

T_2=m_2a+m_2g=m_2(a+g)=12.1(9.8+4.9)=177.87 N

T_1=m_1(g-a)=43.7(9.8-4.9)=214.13 N

Moment of inertia of pulley=I=\frac{1}{2}Mr^2

(T_2-T_1)r=I(-\alpha)=\frac{1}{2}Mr^2(\frac{-a}{r})=\frac{1}{2}Mr(-4.9)

Where \alpha=\frac{a}{r}

(177.87-214.13)=-\frac{1}{2}(4.9)M

-36.26=-\frac{1}{2}(4.9)M

M=\frac{36.26\times 2}{4.9}=14.8 kg

Hence, the mass of the pulley=14.8 kg

6 0
3 years ago
Total energy input is 1400 and my efficiency is 80% so what is the useful energy output in jules?
Ierofanga [76]

Answer

1400 * 0.8 = 1,120

the useful energy output is 1,120 joules

4 0
4 years ago
In the attached position versus time graph what is the magnitude of average velocity for the entire 19 seconds of the motion. An
evablogger [386]

According to the plot, the positions at time <em>t</em> = 0 s and <em>t</em> = 19 s are -1 m and -2 m, respectively. So the average velocity for the 19-s interval is

v_{\rm ave} = \dfrac{-2\,\mathrm m - (-1\,\mathrm m)}{19\,\mathrm s} = -\dfrac1{19}\dfrac{\rm m}{\rm s}\approx \boxed{-0.05\dfrac{\rm m}{\rm s}}

8 0
3 years ago
A 6.5 x 104 W engine exerts a constant force on of 5.5 x 103 N on a car, the resulting velocity is?
jek_recluse [69]

Answer:

12m/s

Explanation:

Given parameters:

Power = 6.5 x 10⁴W

Force  = 5.5 x 10³N

Unknown:

The resulting velocity  = ?

Solution:

 The velocity of a body is related to force and power using the expression below;

            Power  = Force x velocity

Insert the parameters and solve for velocity

         6.5 x 10⁴  = 5.5 x 10³ x velocity

          velocity  = \frac{6.5 x 10^{4} }{5.5 x 10^{3} }   = 12m/s

4 0
3 years ago
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