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3 years ago

Determine the capacitive reactance for a 20 uF capacitor that is across a 20 volt, 60 Hz source

Physics

1 answer:

aleksandr82 [10.1K]3 years ago

4 0

Answer:

Capacitive reactance is 132.6 Ω.

Explanation:

It is given that,

Capacitance, $C=20\ \mu F=20\times 10^{-6}\ F=2\times 10^{-5}\ F$

Voltage source, V = 20 volt

Frequency of source, f = 60 Hz

We need to find the capacitive reactance. It is defined as the reactance for a capacitor. It is given by :

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

$X_C=\dfrac{1}{2\pi \times 60\ Hz\times 2\times 10^{-5}\ F}$

$X_C=132.6\ \Omega$

So, the capacitive reactance of the capacitor is 132.6 Ω. Hence, this is the required solution.

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What is the momentum of an object with 5.22 kg and 1.7 m/s

BlackZzzverrR [31]

Answer:

8.874

Explanation:

You need to times 5.22 kg and 1.7 m/s to get 8.874.

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3 years ago

A 2.0-kg object moving with a velocity of 5.0 m/s in the positive x direction strikes and sticks to a 3.0-kg object moving with

Andrej [43]

Answer:

5.4 J.

Explanation:

Given,

mass of the object, m = 2 Kg

initial speed, u = 5 m/s

mass of another object,m' = 3 kg

initial speed of another orbit,u' = 2 m/s

KE lost after collusion = ?

Final velocity of the system

Using conservation of momentum

m u + m'u' = (m + m') V

2 x 5 + 3 x 2 = ( 2 + 3 )V

16 = 5 V

V = 3.2 m/s

Initial KE = $\dfrac{1}{2}mu^2 + \dfrac{1}{2}m'u'^2$

= $\dfrac{1}{2}\times 2\times 5^2 + \dfrac{1}{2}\times 3 \times 2^2$

= 31 J

Final KE = $\dfrac{1}{2} (m+m')V^2 = \dfrac{1}{2}\times 5 \times 3.2^2 = 25.6 J$

Loss in KE = 31 J - 25.6 J = 5.4 J.

4 0

3 years ago

A 15 kg bucket of water is suspended by a very light ropewrapped around a solid uniform cylinder 0.300 m in diamter withmass 12.

matrenka [14]

Answer:

Part a)

$T = 42 N$

Part b)

$v_f = 11.8 m/s$

Part c)

$t = 1.7 s$

Part d)

$F = 159.7 N$

Explanation:

Part a)

While bucket is falling downwards we have force equation of the bucket given as

$mg - T = ma$

for uniform cylinder we will have

$TR = I\alpha$

so we have

$T = \frac{1}{2}MR^2(\frac{a}{R^2})$

$T = \frac{1}{2}Ma$

now we have

$mg = (\frac{M}{2} + m)a$

$a = \frac{mg}{(\frac{M}{2} + m)}$

$a = \frac{15 \times 9.81}{(6 + 15)}$

$a = 7 m/s^2$

now we have

$T = \frac{12 \times 7}{2}$

$T = 42 N$

Part b)

speed of the bucket can be found using kinematics

so we have

$v_f^2 - v_i^2 = 2 a d$

$v_f^2 - 0 = 2(7)(10)$

$v_f = 11.8 m/s$

Part c)

now in order to find the time of fall we can use another equation

$v_f - v_i = at$

$11.8 - 0 = 7 t$

$t = 1.7 s$

Part d)

as we know that cylinder is at rest and not moving downwards

so here we can use force balance

$F = T + Mg$

$F = 42 + (12 \times 9.81)$

$F = 159.7 N$

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3 years ago

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Answer:

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Explanation:

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3 years ago

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3 years ago