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2 years ago

What affect, if any, does increasing the speed of the plunger have on the wavelength of the waves

1 answer:

6 0

As the speed of the plunger increases, the wavelength of the waves decreases. The greater the frequency, the smaller the wavelength. The smaller the frequency, the greater the wavelength. When we increase the speed of the plunger, the frequency of the waves also increases, and just like with the size of the ball, it’s the speed of the plunger and the frequency of the waves are directly related.

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radio waves are electromagnetic waves that travel at the speed of light 300 000 kilometers per second what is the wave length of

Anarel [89]

<h2>Answer:</h2>

<h3>The wavelength of 100-MHz radio waves is 3 m, yet using the sensitivity of the resonant frequency to the magnetic field strength, details smaller than a millimeter can be imaged.</h3>

<h2>Hope this helps you ❤️</h2>

<h2>MaRk mE aS braiNliest ❤️</h2>

3 0

3 years ago

Calculate the amount of energy transferred when a 40w light bulb is left on for 30 minutes.

Alinara [238K]

Answer:

E=72000J or 72kj

Explanation:

The formula is E=pt you need to convert your t from minutes to seconds before proceeding

3 0

3 years ago

A rotating fan completes 1200 revolutions every minute. Consider the tip of a blade, at a radius of 0.15 m. (a) Through what dis

olga55 [171]

Answer:

(a) 0.942 m

(b) 18.84 m/s

(d) 0.05 s

Explanation:

(a) In one revolution, it travels through one circumference, 2πr = 2 × 3.14 × 0.15 m = 0.942 m.

(b) Its frequency, f, is 1200 rev/min = $\dfrac{1200}{60}$ rev/s = 20 rev/s.

Its angular frequency, ω = 2πf = 2π × 20 = 40π

The speed is given by

v = ωr = 40π × 0.15 = 6π = 18.84 m/s

(d) The period is the inverse of the frequency because it is the time taken to complete one revolution.

$T = \dfrac{1}{f}$

T = 1/20 = 0.05 s

6 0

3 years ago

a skier starts from rest and skis down a 82 meter tall hill labeled h1, into a valley and staught back up another 35 meter hill(

horrorfan [7]

Answer:

She is going at 30.4 m/s at the top of the 35-meter hill.

Explanation:

We can find the velocity of the skier by energy conservation:

$E_{1} = E_{2}$

On the top of the hill 1 (h₁), she has only potential energy since she starts from rest. Now, on the top of the hill 2 (h₂), she has potential energy and kinetic energy.

$mgh_{1} = mgh_{2} + \frac{1}{2}mv_{2}^{2}$ (1)

Where:

m: is the mass of the skier

h₁: is the height 1 = 82 m

h₂: is the height 2 = 35 m

g: is the acceleration due to gravity = 9.81 m/s²

v₂: is the speed of the skier at the top of h₂ =?

Now, by solving equation (1) for v₂ we have:

$v_{2}^{2} = \frac{2mg(h_{1} - h_{2})}{m}$

$v_{2} = \sqrt{2g(h_{1} - h_{2})} = \sqrt{2*9.81 m/s^{2}*(82 m - 35 m)} = 30.4 m/s$

Therefore, she is going at 30.4 m/s at the top of the 35-meter hill.

I hope it helps you!

6 0

2 years ago

A ray of light crosses a boundary between two transparent materials. The medium the ray enters has a larger index of refraction.

Liula [17]

The wavelength of the light decreases as it enters into the medium with the greater index of refraction. The wavelength of the light remains constant as it transitions between materials.

7 0

2 years ago