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3 years ago

A sound wave of frequency 440 Hz is played by a musician. What is the

Physics

1 answer:

TiliK225 [7]3 years ago

7 0

Answer:

343/440

Explanation:

Recall that v=d/t

Now, this is the same thing.

Frequency is 1/T and wavelength is the distance travelled in one period.

So Vs=f*λ

(the greek letter is used as the symbol of wavelength; it's arbitrary)

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Which of the following is NOT an important physical property of matter?

Xelga [282]

Volume isn't a property of matter at all. You can have a big lump or a tiny lump of the same substance.

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3 years ago

A toy plane with a mass of 20 kg is flying at 10 m/s. What is its kinetic energy?

gladu [14]

Answer:

$\boxed {\boxed {\sf A. \ 1000 \ Joules}}$

Explanation:

Kinetic energy can be found using the following formula.

$E_k=\frac{1}{2}mv^2$

where m is the mass and v is the velocity.

The mass of the toy plane is 20 kilograms and the velocity is 10 meters per second.

$m= 20 \ kg \\v= 10 \ m/s \\$

Substitute the values into the formula.

$E_k=\frac{1}{2}(20 \ kg) (10 \ m/s)^2$

Solve the exponent first.

(10 m/s)²= 10 m/s *10 m/s= 100 m²/s²

$E_k=\frac{1}{2} (20 \ kg) (100 \ m^2/s^2)$

Multiply the last two numbers together.

$E_k= \frac{1}{2}(2000 \ kg*m^2/s^2)$

Multiply by 1/2 or divide by 2.

$E_k= 1000 \ kg*m^2/s^2$

1 kilogram square meter per square second is equal to 1 Joule.
Our answer of 1000 kg*m²/s² is equal to 1000 Joules

$E_k= 1000 \ J$

The kinetic energy of the toy plane is 1000 Joules

4 0

3 years ago

Read 2 more answers

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 mass is suspended from the roof of a lift (elevator) by means of a spring balance. The lift (elevator) is

larisa [96]

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.

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3 years ago

Elementary physics in a real automobile accident

Fudgin [204]

Is this even a question

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3 years ago