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taurus [48]
3 years ago
9

Light is an oscillating magnetic field alone. an oscillating electric field alone. electric and magnetic fields that oscillate p

arallel to each other. electric and magnetic fields that oscillate perpendicularly to each other. electric and magnetic fields that oscillate in a random direction relative to each other. Which of the statements concerning light are true? The speed of light is the same no matter what material it is traveling through. Its propagation direction is perpendicular to both the electric field and the magnetic field. Its propagation direction is parallel to both the electric field and the magnetic field. The speed of light in matter is less than the speed of light in a vacuum. It moves at a constant speed through a vacuum. The speed of light in matter is greater than the speed of light in a vacuum.
Physics
1 answer:
cupoosta [38]3 years ago
7 0

Explanation:

  • As light is an electromagnetic wave that causes sensation of sight.  
  • Every electromagnetic wave has electrical and magnetic components in it which are perpendicular to the direction of propagation of wave.  
  • The electrical and magnetic components are mutually perpendicular to each other in an electromagnetic wave.
  • The speed of light in vacuum is c=3\times 10^{8}\ m.s^{-1}, which decreases when it passes through a matter (precisely a transparent medium) which allows the passage of light through it. When the light passes through some transparent matter then its speed decreases as compared to the speed of light in vacuum.
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a parallel circuit is sometimes called a because the current splits up among all the resistors in the parallel circuit
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A parallel circuit is sometimes called a current divider because current splits up among all the resistors in the parallel circuit. In addition, the current through the branches is inversely proportional to the resistance of the branch. If the resistance in each branch is kept constant but the voltage is decreased, the current will decrease.
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3 years ago
series RC circuit is built with a 15 kΩ resistor and a parallel-plate capacitor with 18-cm-diameter electrodes. A 18 V, 36 kHz s
andre [41]

Answer:

d=1.84\ mm

Explanation:

<u>Capacitance</u>

A two parallel-plate capacitor has a capacitance of

\displaystyle C=\frac{\epsilon_o A}{d}

where

\epsilon_o=8.85\cdot 10^{-12}\ F/m

A = area of the plates = \pi r^2

d = separation of the plates

\displaystyle d=\frac{\epsilon_o A}{C}=\frac{\epsilon_o \pi r^2}{C}

We need to compute C. We'll use the circuit parameters for that. The reactance of a capacitor is given by

\displaystyle X_c=\frac{1}{wC}

where w is the angular frequency

w=2\pi f=2\pi \cdot 36000=226194.67\ rad/s

Solving for C

\displaystyle C=\frac{1}{wX_c}

The reactance can be found knowing the total impedance of the circuit:

Z^2=R^2+X_c^2

Where R is the resistance, R=15 K\Omega=15000\Omega. Solving for Xc

X_c^2=Z^2-R^2

The magnitude of the impedance is computed as the ratio of the rms voltage and rms current

\displaystyle Z=\frac{V}{I}

The rms current is the peak current Ip divided by \sqrt{2}, thus

\displaystyle Z=\frac{\sqrt{2}V}{I_p}

I_p=0.65\ mA/1000=0.00065\ A

Now collect formulas

\displaystyle X_c^2=Z^2-R^2=\left(\frac{\sqrt{2}V}{I_p}\right)^2-R^2

Or, equivalently

\displaystyle X_c=\sqrt{\frac{2V^2}{I_p^2}-R^2}

\displaystyle X_c=\sqrt{\frac{2\cdot 18^2}{0.00065^2}-15000^2}

X_c=36176.34\ \Omega

The capacitance is now

\displaystyle C=\frac{1}{226194.67\cdot 36176.34}=1.22\cdot 10^{-10}\ F

The radius of the plates is

r=18\ cm/2=9 \ cm = 0.09 \ m

The separation between the plates is

\displaystyle d=\frac{8.85\cdot 10^{-12} \cdot \pi\cdot 0.09^2}{1.22\cdot 10^{-10}}

d=0.00184\ m

\boxed{d=1.84\ mm}

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A 2-kg bowling ball sits on top of a building that is 40 meters tall.
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The bowling ball is at rest, so it only has gravitational potential energy.

Ug = mgy
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