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2 years ago

Why would adding oil to the moving parts of a machine increase its efficiency?

Physics

1 answer:

luda_lava [24]2 years ago

5 0

Adding oil reduces friction between the moving parts
Less energy is used

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How do solar panels create electricity

xenn [34]

Answer:

They create electricity from the sun's rays

Explanation:

By the use of "solar cells" or "solar panel", you can trap solar energy in silicon/germanium rods, then you can convert in into electricity too. Solar powered panels convert the sun's rays into electricity by exciting electrons in silicon cells using the photons of light from the sun. This electricity can then be used to supply renewable energy to your home or business.

5 0

2 years ago

You and your friend are going bungee jumping! You wait directly below them with a camera. When they leap from the bridge they be

Alex

Answer:

The amplitude is $A = 90.2 \ m$

Explanation:

From the question we are told that

The frequency of when sound is approaching observer is $f = 392 Hz$

The frequency as the move away from observer is $f_ a = 330 \ Hz$

The time between the pitch are $t = 10 \ s$

Here you are the observer and your friends are the source of the sound

The period is mathematically evaluated as

$T = 2 t$

as it is the time to complete one oscillation which from on highest pitch to the next highest pitch

Now T can also be mathematically represented as

$T = \frac{2 \pi}{w}$

Where $w$ is the angular velocity

=> $\frac{2 \pi}{w} = 2 * 10$

=> $w = 0.314 \ rad/sec$

Now using Doppler Effect,

The source of the sound is approaching the observer

The

$f = f_o (\frac{v}{v- wA} )$

$392 = f_o (\frac{v}{v- wA} )$

Where A is the amplitude

So when the source is moving away from the observer

$f_a = f_o (\frac{v}{v+ wA} )$

$330 = f_o (\frac{v}{v+ wA} )$

Here $f_o$ is the fundamental frequency

Dividing the both equation we have

$\frac{392}{330} = \frac{f_o(\frac{v}{v-wA} )}{f_o(\frac{v}{v+wA}}$

$1.1878 = \frac{v+wA}{v-wA}$

$1.1878 v - 1.1878 wA = v+wA$

$1.1878 v = 2.1878 wA$

=> $A = \frac{(0.1878 * (330))}{(2.1878)* (0.314)}$

$A = 90.2 \ m$

7 0

3 years ago

A trombone plays a C3 note. If the speed of sound in air is 343 m/s and the wavelength of this note is

Umnica [9.8K]

The frequency of note C3 is 131 $s^{-1}$ .

<u>Explanation:</u>

Frequency is the measure of repetition of same thing a certain number of times. So frequency is inversely proportional to the wavelength. As wavelength is distance between two successive crests or troughs in a sound wave.

And frequency is the completion of number of cycles in a given time in sound waves. The frequency and wavelength are inversely proportional to each other with velocity of sound being the proportionality constant.

Thus, here the speed of sound is given as 343 m/s, the wavelength of the note is also given as 2.62 m, then frequency will be as follows:

$Frequency=\frac{speed of sound}{Wavelength of note}$