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4 years ago

How do you think altering factors like wavelength and frequency affect they way we see and hear?

1 answer:

4 0

<span>Well think about how Pitch often depends on the frequency of a sound wave and that Frequency is the number of wavelengths . Don't forget that a wavelength can equal to 1 compression and 1 rarefaction.</span>

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10 facts on how old your ultrasounds are electromagnetic

Katyanochek1 [597]

Answer:

1.Ultrasound is a mechanical wave

2.Ultrasounds are sound waves having a frequency above 20,000Hz

3.ltrasound is not included in the electromagnetic spectrum

4.An ultrasound scan uses high-frequency sound waves to create images of the inside of the body

5.Therapeutic ultrasound has no known harmful effects when done right by your therapist

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2 years ago

Where is the best place to put pfds while you are out on your boat?

Alexeev081 [22]

The best place to put PFDs while you are out on your boat is at the top deck of your boat. PFDs must be placed at an area where it is easily accessible, especially in times of accidents, without any unnecessary gear or objects covering the PFDs, but it is best to wear a life jacket during the whole duration of the boat trip.

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4 years ago

at 25°c three moles of gas occupy a volume of 4.0L at 1.9 kpa. if the volume increases to 8.0L what will the new pressure be

ki77a [65]

Answer:0.95kPa

Explanation:

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3 years ago

Objects with masses of 255 kg and a 555 kg are separated by 0.390 m.

eimsori [14]

Answer:

(a). The net gravitational force is $4.20\times10^{-6}\ N$

(b). The position is at 0.232 m.

Explanation:

Given that,

Mass of one object M = 255 kg

Mass of another object M'= 555 kg

Separation = 0.390 m

(a). We need to calculate the net gravitational force

Using formula of force

$F_{net}=\dfrac{GM'm}{r^2}-\dfrac{GMm}{r^2}$

$F_{net}=\dfrac{Gm(M'-M)}{r^2}$

Put the value into the formula

$F_{net}=\dfrac{6.67\times10^{-11}\times32.0(555-255)}{(0.390)^2}$

$F_{net}=4.20\times10^{-6}\ N$

The net gravitational force is $4.20\times10^{-6}\ N$

(b). We need to calculate the position

Force from 555 mass = Force from 255 mass

$\dfrac{Gm\times555}{x^2}=\dfrac{Gm\times255}{(0.390-x)^2}$

$555(0.390-x)^2=255x^2$

$300x^2-0.78x+84.4155=0$

$x=0.232\ m$

The position is at 0.232 m.

Hence, (a). The net gravitational force is $4.20\times10^{-6}\ N$

(b). The position is at 0.232 m.

5 0

3 years ago

Read 2 more answers

Which of these help create radio waves?

inysia [295]

The one that help create radio waves is :
Changing electric and magnetic fields applied at right angles

Radio waves are transverse wave, which means that the oscillations occurring perpendicular to the direction of energy transfer

hope this helps

3 0

3 years ago