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3 years ago

Which statement describes the speed of electromagnetic waves?

Physics

2 answers:

goldfiish [28.3K]3 years ago

7 0

Answer:

Their speed in a vacuum is a constant value.

Explanation:

Electromagnetic waves consits of oscillations of electric field and magnetic field. The oscillations of these fields occur in a direction perpendicular to the direction of propagation of the waves, so they are transverse wave. Electromagnetic waves, contrary to mechanical waves, do not need a medium to propagate, so they can also travel through a vacuum. In a vacuum, their speed is constant and has always the same value, the speed of light:

$c=3\cdot 10^8 m/s$

Marianna [84]3 years ago

3 0

Answer:

Their speed in a vacuum is a constant value.

Explanation:

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And observer reviews for identical objects that are traveling at speed close to the speed of light in their legs be used to rank

soldier1979 [14.2K]

Answer:

C :)

Explanation:

5 0

3 years ago

Can a boat moving eastward accelerate to the west? What happens to the boatâs velocity? Name other examples of objects accelerat

Tatiana [17]

Answer: yes

Explanation: Assuming a it's moving 20m/s in the west direction and slowing down at a rate of -5m/s² (deceleration), the negative sign would imply the westward direction because -5 west is the same thing as +5 east. So, the boat is moving 20m/s west direction and accelerating at 5m/s² east direction.

Other examples of these are: Cars and and tricycles.

4 0

3 years ago

A 2 kg mass connected to a spring with spring constant k = 10 N/m oscillates in simple harmonic motion with an amplitude of A =

insens350 [35]

Answer:

Kinetic energy at 0.05 m is 0.037 J

Explanation:

Given:

Mass, m = 2 kg

Spring constant, k = 10 N/m

Amplitude, A = 0.1 m

Angular frequency, ω = √k/m

Substitute the suitable values in the above equation.

$\omega = \sqrt\frac{10}{2}$

ω = 2.24 s⁻¹

Simple harmonic equation is represent by the equation:

x = A cos ωt

Substitute 0.05 m for x, 0.1 m for A and 2.24 s⁻¹ for ω in the above equation.

$0.05 = 0.1\cos(2.24t)$

t = 0.47 s

Kinetic energy at x = 0.05 is determine by the relation:

$E=\frac{1}{2} kA^{2}\sin^{2}(\omega t)$

Substitute the suitable values in the above equation.

$E=\frac{1}{2} \times 10 \times 0.1^{2} \sin^{2}(2.24\times0.47)$

E = 0.037 J

5 0

3 years ago

An automatic dryer spins wet clothes at an angular speed of 4.3 rad/s. Starting from rest, the dryer reaches its operating speed

JulsSmile [24]

Answer:

The time taken by dryer to come up to speed is 1.075 seconds.

Explanation:

Initial angular speed, $\omega_i=0$

Final angular speed, $\omega_f=4.3\ rad/s$

The dryer reaches its operating speed with an average angular acceleration of, $a=4\ rad/s^2$

Let t is the time taken by dryer to come up to speed. Using first equation of motion as :

$\omega_f=\omega_i+\alpha t\\\\t=\dfrac{\omega_f}{\alpha }\\\\t=\dfrac{4.3}{4}\\\\t=1.075\ s$

So, the time taken by dryer to come up to speed is 1.075 seconds.

5 0

3 years ago

How can you connect four of them to produce an equivalent resistance of 0.25 kω?

Charra [1.4K]

Answer:

Question incomplete

This is the complete question

You have a collection of 1.0 kΩ resistors.

How can you connect four of them to produce an equivalent resistance of 0.25 kΩ?

Explanation:

Given that,

We have collection of resistor

R = 1.0 kΩ.

And we want to design an equivalent resistance of

R = 0.25kΩ

We know that,

Series connection of add up resistance and this increases the value of resistance, I.e

Req = R1 + R2 + R3 +.....

While

Parallel resistance add up the reciprocal of resistance and this reduces resistance

So, 1/Req = 1/R1 + 1/R2 + 1/R3 +......

Now, we want to design 0.25kΩ

Connect four of the resistor in parallel will give a equivalent resistance of 0.25 kΩ

1/Req = 1/R1 + 1/R2 + 1/R3 + 1/R4

R1 = R2 = R3 = R4 = 1.0 kΩ

1/Req = 1/1 + 1/1 + 1/1 + 1/1

1/Req = 1+1+1+1

1/Req = 4

Taking reciprocal of both sides..

Req = ¼ kΩ

This is the required equivalent resistance

So, connect the four of the resistor in parallel will give the required 0.25 kΩ resistance

4 0

3 years ago