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3 years ago

What distance does electromagnetic radiation travel in 55.0 μs ?

1 answer:

4 0

Electromagnetic radiation is an energy that is known as light. so electromagnetic radiation will have the same speed as the speed of light which is 3 x 10^8 m/s. so the distance it travel at 55 x 10^-6 s is:
D = ( 3 x 10^8 m/s ) ( 55 x 10^-6 s )
D = 16500 m

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A trombone can produce pitches ranging from 85 Hz to 660 Hz approximately. When the trombone is producing a 562 Hz tone, what is

tester [92]

To solve this problem we will apply the concept of wavelength, which warns that this is equivalent to the relationship between the speed of the air (in this case in through the air) and the frequency of that wave. The air is in standard conditions so we have the relation,

Frequency $= f = 562Hz$

Speed of sound in air $= v = 331m/s$

The definition of wavelength is,

$\lambda = \frac{v}{f}$

Here,

v = Velocity

f = Frequency

Replacing,

$\lambda = \frac{331m/s}{562Hz}$

$\lambda = 0.589m$

Therefore the wavelength of that tone in air at standard conditions is 0.589m

3 0

3 years ago

How to find initial velocity

lesya [120]

Answer:

Velocity is a function of time and defined by both a magnitude and a direction. [1] Often in physics problems, you will need to calculate the initial velocity (speed and direction) at which an object in question began to travel. There are multiple equations that can be used to determine initial velocity. Using the information given in a problem, you can determine the proper equation to use and easily answer your question.

Explanation:

Hope this helps

7 0

3 years ago

An object is allowed to fall freely near the surface of an unknown planet. The object falls 72 meters from rest in 4.0 seconds.

dimaraw [331]

<span>it fairly is going to attain a speed of 24 m/s in a 2d, yet between t = 0 and t = a million, it fairly is not any longer vacationing at that speed, yet at slower speeds. it fairly is 12 meters. ?D = [ ( a?T^2 + 2?Tv_i ) ] / 2 the place: ?D = displacement a = acceleration ?T = elapsed time v_i = preliminary speed ?D = [ ( 24m/s^2 • 1s • 1s + 2 • 1s • 0m/s ) ] / 2 ?D = 24 / 2 ?D = 12m</span>

7 0

3 years ago

Read 2 more answers

An object is observed for a time interval of 20 seconds. From time 6.7 s the object experiences a Force of 106 N that lasts unti

shusha [124]

Answer:

1654 kg m/s

Explanation:

The impulse experienced by an object is equal to the product between the force exerted on the object and the time during which the force lasts:

$I=F\Delta t$

where:

I is the impulse

F is the force exerted on the object

$\Delta t$ is the time during which the force is applied

For the object in this problem, we have

$F=106 N$ (force applied)

$\Delta t= 15.6 s$ (time interval)

Therefore, the impulse experienced by the object is:

$I=(106)(15.6)=1654 kg m/s$

3 0

3 years ago

Two equal charges are 2m2m apart. If the charges and the distance are divided by two, how is the force between the charges affec

lara31 [8.8K]

Answer:

The force between the charges are not affected.

Explanation:

Given;

distance between two equal charges, R = 2m

The force between the charges is given by;

$F = \frac{kq^2}{R^2}\\\\F_1 = \frac{kq_1^2}{R_1^2}\\\\When\ the \ charges \ and \ the \ distance \ are \ divided \ by \ two \ (q_2 = \frac{q_1}{2}, \ R_2 = \frac{R_1}{2} )\\\\ F_2 = \frac{kq_2^2}{R_2^2}\\\\F_2 = \frac{k(q_1/2)^2}{(R_1/2)^2}\\\\F_2= \frac{4k*q_1^2}{4*R_1^2}\\\\F_2 = \frac{k*q_1^2}{R_1^2}\\\\F_2 = F_1$

Therefore, the force between the charges are not affected.

7 0

3 years ago