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3 years ago

Which equations could be used as is, or rearranged to calculate for frequency of a wave? Check all that apply.

2 answers:

amm18123 years ago

7 0

-- Equations #2 and #6 are both the same equation,
and are both correct.

-- If you divide each side by 'wavelength', you get Equation #4,
which is also correct.

-- If you divide each side by 'frequency', you get Equation #3,
which is also correct.
With some work, you can rearrange this one and use it to calculate
frequency.

Summary:

-- Equations #2, #3, #4, and #6 are all correct statements,
and can be used to find frequency.

-- Equations #1 and #5 are incorrect statements.

alexira [117]3 years ago

3 0

The correct answers to the question are 2,3,4 and 6 options respectively.

EXPLANATION:

Let us consider a wave which is moving with a speed v in a medium .

Let f and $\lambda$ is the frequency and wavelength of the wave.

The relation between frequency,wavelength and speed is given as -

speed = $=\ frequency\times wavelength$

= $f\times \lambda$

The wavelength of the wave is calculated as -

$wavelength=\ \frac{speed}{frequency}$

Similarly frequency of the wave can be calculated as-

$frequency=\ \frac{speed}{wavelength}$

The option 1 is a wrong relation. So, we can not calculate the wavelength through this equation.

The option 2 is a right relation. By rearranging it, we can calculate the frequency of the wave.

The option 3 is a correct relation. So, we can calculate the wavelength through it.

The option 4 is also a correct relation. By rearranging it, we can calculate the wavelength.

The option 5 is a wrong relation.

The option 6 is also a right relation. By rearranging it, we can calculate the wavelength of the wave.

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I have this sheet and I don’t understand any of it at all

katen-ka-za [31]

I know this probably is not what you are looking for, but this is what I would suggest.

Write down on the side what values they are giving you and the corresponding variable used for that value.

Like for the first question, they give you mass is 3000kg.

On the side write

m = 3000kg

m is the variable used for mass

Then list any other variables they give you in the question and then what you are trying to look for.

a = 2m/s^2

F = ?

a is the variable for acceleration and F is the variable for force.

So now you have:

m = 3000kg

a = 2m/s^2

F = ?

You have a and m, and you are looking for F (force), you know that because it asks "with what force".

Also note, that you have to know what the variable is based on the units and you likely have some type of formula sheet with all of them listed.

You can use the variables and what is unknown to match it to a formula that will solve for the unknown variables. Also note, that it might not always be simply a=b+c, you may know what "a" and "b" are, but they may ask for c so you will have to rearrange the formula with some simple algebra.

This question however is simple. You can see you are going to use the formula

F=ma.

So now just substitute in your variables to solve for F.

(3000kg)(2m/s^2) = 6000N

Make sure to write the units which are probably on your formula sheet as well. Force is measured in Newtons, so you use the unit N.

Also note, if they gave you mass in grams (another unit), you cannot use that in the formula (your formula sheet would tell you the unit it must be in), so you would have to convert it to kg.

Hope this helps as a starting point. I don't intend to do your homework for you, but explain a process you can use to help you do it on your own.

3 0

3 years ago

A thin metallic spherical shell of radius 0.357 m has a total charge of 5.03 times 10^-6 C placed on it. At the center of the sh

Artyom0805 [142]

Answer:

The electric field is $5.623\times10^{4}\ N/C$

Explanation:

Given that,

Radius = 0.357 m

Charge $Q=5.03\times10^{-6}\ C$

Point charge $q=4.15\times10^{-6}\ C$

Distance = 0.815 m

We need to calculate the total electric field

Using formula of electric field

$E=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{q}{r^2}$

Where, q = point charge

r = distance

Put the value into the formula

$E=\dfrac{9\times10^{9}\times4.15\times10^{-6}}{(0.815)^2}$

$E=5.623\times10^{4}\ N/C$

Hence, The electric field is $5.623\times10^{4}\ N/C$

4 0

3 years ago

Three noise sources produce volume (loudness) levels of 70, 73, and 80 dB when acting separately. When the sources act together,

Mashutka [201]

Sound at 70 dB is 70 dB louder than the human reference level. That's 10⁷ times as much as the reference sound power.

Sound at 73 dB is 73 dB louder than the human reference level. That's 10⁷.³ or 2 x 10⁷ times as much as the reference sound power.

Sound at 80 dB is 80 dB louder than the human reference level. That's 10⁸ or 10 x 10⁷ times as much as the reference sound power.

Now we can adumup:

Intensity of all 3 sources = (10⁷) + (2 x 10⁷) + (10 x 10⁷)

Intensity = (13 x 10⁷) times the sound power reference intensity.

Intensity in dB = 10 log (13 x 10⁷) = 10 (7 + log(13)

Intensity = 70 + 10 log(13)

Intensity = 70 + 10 (1.114)

Intensity = 70 + 11.14

Intensity = 81.14 dB

______________________________________

Looking at the questioner's profile, I seriously wonder whether I'll ever get a comment in return from this creature, and how I'll ever find out if my solution is correct. For that matter, I'm also seriously questioning how and whether my solution will ever be used for anything.

8 0

3 years ago

Air enters a nozzle steadily at 2.21 kg/m3 and 20 m/s and leaves at 0.762 kg/m3 and 150 m/s. If the inlet area of the nozzle is

saveliy_v [14]

Answer:

a) The mass flow rate through the nozzle is 0.27 kg/s.

b) The exit area of the nozzle is 23.6 cm².

Explanation:

a) The mass flow rate through the nozzle can be calculated with the following equation:

$\dot{m_{i}} = \rho_{i} v_{i}A_{i}$