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Volgvan
3 years ago
7

The Capacitive reactance of a 0.094uF capacitor when a frequency of 25kHz is applied is lus 68 Ohms 0.68 Ohms luf

Physics
1 answer:
Nana76 [90]3 years ago
8 0

Answer:

Capacitive reactance of the capacitor is 68 ohms

Explanation:

It is given that,

Capacitance, C=0.094\ \mu F=0.094\times 10^{-6}\ F

Frequency, f=25\ kHz=25\times 10^3\ Hz

Capacitive reactance is given by :

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

X_C=\dfrac{1}{2\pi 25\times 10^3\times 0.094\times 10^{-6}}

X_C=67.72\ \Omega

or

X_C=68\ \Omega

So, the capacitive reactance of the capacitor is 68 ohms. Hence, this is the required solution.

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Your answer is 311.29271 lbs
5 0
3 years ago
A bug is 12 cm from the center of a turntable that is rotating with a frequency of 45 rev/min . What minimum coefficient frictio
Agata [3.3K]

Answer:

The minimum coefficient of friction is 0.27.

Explanation:

To solve this problem, start with identifying the forces at play here. First, the bug staying on the rotating turntable will be subject to the centripetal force constantly acting toward the center of the turntable (in absence of which the bug would leave the turntable in a straight line). Second, there is the force of friction due to which the bug can stick to the table. The friction force acts as an intermediary to enable the centripetal acceleration to happen.

Centripetal force is written as

F_c = m\frac{v^2}{r}

with v the linear velocity and r the radius of the turntable. We are not given v, but we can write it as

v = r\omega

with ω denoting the angular velocity, which we are given. With that, the above becomes:

F_c = m\frac{v^2}{r}=m\omega^2 r

Now, the friction force must be at least as much (in magnitude) as Fc. The coefficient (static) of friction μ must be large enough. How large?

F_r=\mu mg \geq m\omega^2 r = F_c\implies\\\mu \geq \frac{\omega^2 r}{g}

Let's plug in the numbers. The angular velocity should be in radians per second. We are given rev/min, which can be easily transformed by a factor 2pi/60:

\frac{1 rev}{1 min}\cdot\frac{\frac{2\pi rad}{rev}}{\frac{60s}{1 min}}=\frac{2\pi}{60}\frac{rad}{s}

and so 45 rev/min = 4.71 rad/s.

\mu \geq \frac{\omega^2 r}{g}=\frac{4.71^2\frac{1}{s^2}\cdot 0.12m}{9.8\frac{m}{s^2}}=0.27

A static coefficient of friction of at least be 0.27 must be present for the bug to continue enjoying the ride on the turntable.



3 0
3 years ago
Using planck's constant and a wavelength of 656e-9 what is the energy of the photon
lbvjy [14]

Answer:

3.03e-19 J

Explanation:

Use the formula E = hc/λ

Where:

h (Planck's constant) = 6.626e-34 J*s

c (speed of light, constant) = 3.00e8 m/s

λ (wavelength) = 656e-9 m

E = energy (in Joules)

E = (6.626e-34 * 3.00e8) / 656e-9 = 3.03018293e-19 = 3.03e-19 J

4 0
2 years ago
a transformer changes 95 v acorss the primary to 875 V acorss the secondary. If the primmary coil has 450 turns how many turns d
natita [175]

Answer:

The number of turns in the secondary coil is 4145 turns

Explanation:

Given;

the induced emf on the primary coil, E_p = 95 V

the induced emf on the secondary coil, E_s = 875 V

the number of turns in the primary coil, N_p = 450 turns

the number of turns in the secondary coil, N_s = ?

The number of turns in the secondary coil is calculated as;

\frac{N_p}{N_s} = \frac{E_p}{E_s}

N_s = \frac{N_pE_s}{E_p} \\\\N_s = \frac{450*875}{95} \\\\N_s = 4145 \ turns

Therefore, the number of turns in the secondary coil is 4145 turns.

8 0
3 years ago
2 / 3
kiruha [24]

Answer:

Explanation:

   

3 0
3 years ago
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