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3 years ago

7

Electromagnetic waves all move at the same wave speed. which of these offer solutions to slowing them down select all that apply

1 answer:

Lelechka [254]3 years ago

6 0

option c...have the waves move thru the vacuum or space

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A 10 gauge copper wire carries a current of 20 A. Assuming one free electron per copper atom, calculate the magnitude of the dri

nordsb [41]

Complete Question

A 10 gauge copper wire carries a current of 20 A. Assuming one free electron per copper atom, calculate the drift velocity of the electrons. (The cross-sectional area of a 10-gauge wire is 5.261 mm2.) mm/s

Answer:

The drift velocity is $v = 0.0002808 \ m/s$

Explanation:

From the question we are told that

The current on the copper is $I = 20 \ A$

The cross-sectional area is $A = 5.261 \ mm^2 = 5.261 *10^{-6} \ m^2$

The number of copper atom in the wire is mathematically evaluated

$n = \frac{\rho * N_a}Z}$

Where $\rho$ is the density of copper with a value $\rho = 8.93 \ g/m^3$

$N_a$ is the Avogadro's number with a value $N_a = 6.02 *10^{23}\ atom/mol$

Z is the molar mass of copper with a value $Z = 63.55 \ g/mol$

So

$n = \frac{8.93 * 6.02 *10^{23}}{63.55}$

$n = 8.46 * 10^{28} \ atoms /m^3$

Given the 1 atom is equivalent to 1 free electron then the number of free electron is

$N = 8.46 * 10^{28} \ electrons$

The current through the wire is mathematically represented as

$I = N * e * v * A$

substituting values

$20 = 8.46 *10^{28} * (1.60*10^{-19}) * v * 5.261 *10^{-6}$

=> $v = 0.0002808 \ m/s$

8 0

3 years ago

What are the characteristics of high energy waves?

Temka [501]

Amplitude and frequency

4 0

3 years ago

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Which free body diagram is in equilibrium?

bearhunter [10]

A. because everything is balanced.

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3 years ago

If the ball shown in the figure lands in 0.5 s, about what height was it thrown from?

Ede4ka [16]

Answer: 1.22 m

Explanation:

The equation of motion in this situation is:

$y=y_{o}+V_{oy}t-\frac{g}{2}t^{2}$ (1)

Where:

$y=0$ is the final height of the ball

$y_{o}=h$ is the initial height of the ball

$V_{oy}=V_{o}sin(0\°)=0$ is the vertical component of the initial velocity (assuming the ball was thrown vertically and there is no horizontal velocity)

$t=0.5 s$ is the time at which the ball lands

$g=9.8 m/s^{2}$ is the acceleration due gravity

So, with these conditions the equation is rewritten as:

$h=\frac{g}{2}t^{2}$ (2)

$h=\frac{9.8 m/s^{2}}{2}(0.5 s)^{2}$ (3)

Finally:

$h=1.22 m$

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3 years ago

How could you record the number 4000 and report 2 significant figures?

yawa3891 [41]

Explanation:

Write in scientific notation.

4000 = 4.0×10³

5 0

3 years ago