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3 years ago

THE RIGHT ANSWER WILL RECEIVE A BRAINLESS AND POINTS AND THANKS!!!

Physics

1 answer:

poizon [28]3 years ago

4 0

Answer:

391.67Hz

Explanation:

The fundamental frequency formula in string is expressed as;

Fo = V/2L

V is the velocity of the wave = 329m/s

L is the length of the string = 42cm = 0.42m

Substitute

Fo = 329/2(0.42)

Fo = 329/0.84

Fo = 391.67Hertz

Hence the fundamental frequency of a mandolin string is 391.67Hz

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Look at the photo and answer the question

Naddik [55]

Answer:

I think D. maybe C

Explanation:

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3 years ago

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A scientist compares two samples of white powder. One powder was present at the beginning of an experiment. The other powder was

Jet001 [13]

Density of powder 1 = 0.5 g / 45 cm^3 = 1/90 g/cm^3
Density of powder 2 = 1.3 g / 65 cm^3 = 1/50 g/cm^3

Therefore the densities of the two powders are different, hence chemical reaction has occurred.

(note: none of the other choices make sense. In fact, a different density does not necessarily indicate a chemical change, see paragraph below).

Density of powders are not definitive unless they are each of the same size and texture. For example, granular sugar, rock sugar, and icing sugar all have different densities. I would conclude that this experiment does not lend to a reliable answer.

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3 years ago

What is the (magnitude of the) centripetal acceleration (as a multiple of g=9.8~\mathrm{m/s^2}g=9.8 m/s 2 ) towards the Eart

Wittaler [7]

Answer:

The centripetal acceleration as a multiple of $g=9.8 m/s^{2}$ is $1.020x10^{-3}m/s^{2}$

Explanation:

The centripetal acceleration is defined as:

$a = \frac{v^{2}}{r}$ (1)

Where v is the velocity and r is the radius

Since the person is standing in the Earth surfaces, their velocity will be the same of the Earth. That one can be determined by means of the orbital velocity:

$v = \frac{2 \pi r}{T}$ (2)

Where r is the radius and T is the period.

For this case the person is standing at a latitude $71.9^{\circ}$ . Remember that the latitude is given from the equator. The configuration of this system is shown in the image below.

It is necessary to use the radius at the latitude given. That radius can be found by means of trigonometric.

$\cos \theta = \frac{adjacent}{hypotenuse}$

$\cos \theta = \frac{r_{71.9^{\circ}}}{r_{e}}$ (3)

Where $r_{71.9^{\circ}}$ is the radius at the latitude of $71.9^{\circ}$ and $r_{e}$ is the radius at the equator ( $6.37x10^{6}m$ ).

$r_{71.9^{\circ}}$ can be isolated from equation 3:

$r_{71.9^{\circ}} = r_{e} \cos \theta$ (4)

$r_{71.9^{\circ}} = (6.37x10^{6}m) \cos (71.9^{\circ})$

$r_{71.9^{\circ}} = 1.97x10^{6} m$

Then, equation 2 can be used

$v = \frac{2 \pi (1.97x10^{6} m)}{24h}$

Notice that the period is the time that the Earth takes to give a complete revolution (24 hours), this period will be expressed in seconds for a better representation of the velocity.

$T = 24h . \frac{3600s}{1h}$ ⇒ $84600s$

$v = \frac{2 \pi (1.97x10^{6} m)}{84600s}$

$v = 146.31m/s$

Finally, equation 1 can be used:

$a = \frac{(146.31m/s)^{2}}{(1.97x10^{6}m)}$

$a = 0.010m/s^{2}$

Hence, the centripetal acceleration is $0.010m/s^{2}$

To given the centripetal acceleration as a multiple of $g=9.8 m/s^{2}$ it is gotten:

$\frac{0.010m/s^{2}}{9.8 m/s^{2}} = 1.020x10^{-3}m/s^{2}$

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3 years ago

A whistle you use to call your hunting dog has a frequency of 21 kHz, but your dog is ignoring it. You suspect the whistle may n

jeka94

The whistle you use to call has a frequency of 21 kHz, so you do not recognize it also because your threshold of hearing is only 20 kHz. Since you are unable to hear sounds above this value, you request your friend to blow the whistle and move farther away. The frequency will decrease due to Doppler effect.

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4 years ago

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Who was the first man to go into the orbit?

kkurt [141]

D. Yuri Gagarin

(Typing this to circumnavigate the 20 character minimum requirement)

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4 years ago