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3 years ago

10

What type of electromagnetic wave does a light bulb and a radio antenna release?

1 answer:

andrew-mc [135]3 years ago

4 0

I believe that a light bulb releases visible light and a radio antenna releases a radio waves

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How to find aceleration

lions [1.4K]

Answer:

$a=\frac{v_{f}-v_{i}}{t_{f}-t_{i}}$

Explanation:

Formula for AVERAGE acceleration

$a=\frac{v_{f}-v_{i}}{t_{f}-t_{i}}$

$v_{f}$ is Final Velocity

$v_{i}$ is Initial Velocity

$t_{f}$ is Final Time

$t_{i}$ is Initial Time

Most of the time, the initial time will be 0, so you won't have to subtract.

5 0

3 years ago

Byron bought a keyless entry door lock that has the digits 0 through 9 on the keypad. He wants to choose a three-digit entry cod

DedPeter [7]

Answer:

27

Explanation:

8 0

3 years ago

Read 2 more answers

In Example 2.12, two circus performers rehearse a trick in which a ball and a dart collide. Horatio stands on a platform 6.4 m a

pickupchik [31]

Answer:

time of collision is

t = 0.395 s

$h = 5.63 m$

so they will collide at height of 5.63 m from ground

Explanation:

initial speed of the ball when it is dropped down is

$v_1 = 0$

similarly initial speed of the object which is projected by spring is given as

$v_2 = 16.2 m/s$

now relative velocity of object with respect to ball

$v_r = 16.2 m/s$

now since we know that both are moving under gravity so their relative acceleration is ZERO and the relative distance between them is 6.4 m

$d = v_r t$

$6.4 = 16.2 t$

$t = 0.395 m$

Now the height attained by the object in the same time is given as

$h = v_2 t - \frac{1}{2}gt^2$

$h = 16.2(0.395) - \frac{1}{2}(9.81).395^2$

$h = 5.63 m$

so they will collide at height of 5.63 m from ground

4 0

3 years ago

A charged particle is moving in a uniform magnetic field at a speed of 8.2Ã10^3 m/s in a direction 87Â° from the direction of th

77julia77 [94]

Answer:

Magnetic field, B = 0.042 T

Explanation:

It is given that,

Speed of charged particle, $v=8.2\times 10^3\ m/s$

Angle between velocity and the magnetic field, $\theta=87$

Charge, $q=5.7\ \mu C=5.7\times 10^{-6}\ C$

Magnetic force, F = 0.002 N

The magnetic force is given by :

$F=qvB\ sin\theta$

B is the magnetic field

$B=\dfrac{F}{qv\ sin\theta}$

$B=\dfrac{0.002}{5.7\times 10^{-6}\times 8.2\times 10^3\times sin(87)}$

B = 0.042 T

So, the strength of the magnetic field is 0.042 Tesla. Hence, this is the required solution.

5 0

3 years ago

If the frequency of the motion of a simple harmonic oscillator is doubled, by what factor does the maximum speed of the oscillat

Maksim231197 [3]

Answer:

If the frequency of the motion of a simple harmonic oscillator is doubled , then maximum speed of the oscillator changes by the factor 2

Explanation:

We know that in a simple harmonic oscillator the maximum speed is given by

$v_{max}$ = $Aw$

Here A is amplitude which is constant , so from above equation we see that maximum speed is directly proportional to $w\\$ of the oscillation .

Since $w = 2 \pi f$

$v_{max}^{|}/v_{max} = 2f/f$ = 2

Where $v_{max}^{|}$ is the maximum speed when frequency is doubled .

6 0

3 years ago