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Maksim231197 [3]
3 years ago
10

A capacitance C and an inductance L are operated at the same angular frequency.

Physics
1 answer:
PIT_PIT [208]3 years ago
6 0

Answer:

(B) 7279.70 rad/sec (C) X_L=37.1265\ ohm\ and \ X_C=37.1265\ ohm

Explanation:

We have given L=5.10mH and C=3.70\mu F

(B) The angular frequency is denoted as \omega whose value is given by \omega =\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{5.10\times 10^{-3}\times 3.70\times 10^{-6}}}=7279.70\ rad/sec

(C) The inductive reactance X_L=\omega L=7279.70\times 5.10\times 10^{-3}=37.1265\ ohm

Capacitive reactance  X_C=\frac{1}{\omega C}=\frac{1}{7279.70\times 3.70\times 10^{-6}}=37.1265\ ohm

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Io, a satellite of Jupiter, is the most volcanically active moon or planet in the solar system. It has volcanoes that send plume
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Answer:

1331.84 m/s

Explanation:

t = Time taken

u = Initial velocity

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s = Displacement = 490 km

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g = Acceleration due to gravity = 1.81 m/s² = a

From equation of linear motion

v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow u=\sqrt{v^2-2as}\\\Rightarrow u=\sqrt{0^2-2\times -1.81\times 490000}\\\Rightarrow u=1331.84\ m/s

The speed of the material must be 1331.84 m/s in order to reach the height of 490 km

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4 years ago
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4 years ago
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3 years ago
A particular coaxial cable is comprised of inner and outer conductors having radii 1 mm and 3 mm respectively, separated by air.
noname [10]

Answer:

The value is  \rho_s  =  4.026 *10^{-6} \  C/m^2

Explanation:

From the question we are told that

   The radius of the inner conductor  is  r_1 = 1 \ mm =  0.001 \ m

    The radius of the outer conductor is  r_2 = 3 \ mm = 0.003 \  m

    The potential at the outer conductor is  V = 1.5 kV  =  1.5 *10^{3} \  V

Generally the capacitance per length of the capacitor like set up of the two conductors is

      C= \frac{2 * \pi * \epsilon_o }{ ln [\frac{r_2}{r_1} ]}

Here \epsilon_o is the permitivity of free space with value  \epsilon_o =  8.85*10^{-12} C/(V \cdot m)

=>   C= \frac{2 *  3.142  * 8.85*10^{-12}  }{ ln [\frac{0.003}{0.001} ]}

=>   C= 50.6 *10^{-12} \  F/m

Generally given that the potential  of the outer conductor with respect to the inner conductor is positive it then mean that the outer conductor is positively charge

Generally the line  charge density of the outer  conductor is mathematically represented as

      \rho_l  =  C *  V

=>   \rho_d  =  50.6*10^{-12} *  1.5*10^{3}

=>   \rho_d  =  7.59*10^{-8} \  C/m

Generally the surface charge density is mathematically represented as

        \rho_s  =  \frac{\rho_l }{2 \pi * r_2 }    here 2 \pi r = (circumference \ of \ outer \  conductor  )

=>    \rho_s  =  \frac{7.59 *10^{-8} }{2* 3.142 * 0.003 }

=>    \rho_s  =  4.026 *10^{-6} \  C/m^2

3 0
3 years ago
Which has more pressure, FIRST one gets all the points.
SSSSS [86.1K]

<u>Answer:</u>

area of point has more pressure.

<u>Explanation:</u>

formula's : pressure = force ÷ area

1st pressure,

200 ÷ 0.2 = 1000 Pa

2nd pressure,

200 ÷ 0.004 = 50000 Pa

<h3>Therefore the pressure on the area of the point is more.</h3>
8 0
3 years ago
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