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2 years ago

If there are 20 waves in a 5-second time frame, what is the frequency?

Physics

2 answers:

mestny [16]2 years ago

6 0

Answer:

The frequency is four waves per second.

Explanation:

You make it so the time is one second.

To do that you divide 5 seconds by 5.

Since you divide the seconds by 5 you now have to divide the amount of waves by 5.

20/5=4 5/5=1

Now your answer is four waves per second.

creativ13 [48]2 years ago

3 0

Your answer would be 0.20 Hz

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Enzo investigated the effect of different types of fertilizers on the growth of bean plants. He hypothesized that fertilizer B w

Mama L [17]

Answer:

The answer is "In this information, Enzo will be confident that if any fertilizer is required only for plants".

Explanation:

The Beans plants were members of the group of legumes. Legume flowering plant under which the roots form nodules. These nodules require a specific bacterium called bacteria, which fix nitrogen. Its bacterium transforms nitrogen from the atmosphere into the soil's nitrate. Consequently, for some of its growth, beans do not require fertilizer.

4 0

3 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

Cindy runs 500 meters every morning. It takes her 2 hours to complet

marta [7]

Answer:

250 / 2.5*10^2

Explanation:

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1 year ago

The Hubble Space Telescope orbits the Earth at approximately 612,000m altitude. Its mass is 11,100 kg and the mass of earth is 5

nexus9112 [7]

Answer:

7.55 km/s

Explanation:

The force of gravity between the Earth and the Hubble Telescope corresponds to the centripetal force that keeps the telescope in uniform circular motion around the Earth:

$G\frac{mM}{R^2}=m\frac{v^2}{R}$

where

$G=6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2}$ is the gravitational constant

$m=11,100 kg$ is the mass of the telescope

$M=5.97\cdot 10^{24} kg$ is the mass of the Earth

$R=r+h=6.38\cdot 10^6 m+612,000 m=6.99\cdot 10^6 m$ is the distance between the telescope and the Earth's centre (given by the sum of the Earth's radius, r, and the telescope altitude, h)

v = ? is the orbital velocity of the Hubble telescope

Re-arranging the equation and substituting numbers, we find the orbital velocity:

$v=\sqrt{\frac{GM}{R}}=\sqrt{\frac{(6.67\cdot 10^{-11})(5.97\cdot 10^{24} kg)}{6.99\cdot 10^6 m}}=7548 m/s=7.55 km/s$

6 0

3 years ago

Solenoid is another word for

Artyom0805 [142]

Answer: coil of wire

5 0

3 years ago