3 years ago

HELP PLS DUE SOON DUE SOON NEED HELP PLS PLS I BEG U

Physics

1 answer:

Bumek [7]3 years ago

3 0

Answer:

The thirds option

Explanation:

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Given that the speed of sound in air is 330m/s ,calculate the frequency of a note which setup a wavelength of 5cm in air.Is this

Pie

Explanation:

v = wavelength x frequency

330 = 5 . 10-² m x f

f = 6600 Hz

the frequency that human can hear is about 20 Hz - 20000 Hz

so human can hear the note.

3 0

3 years ago

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What is the mass of an object that experiences a gravitational force of 510 N near Earth's surface?

sineoko [7]

Answer:The mass of an object is 52 kg.

Explanation:

Gravitational force on the object ,F=510 N

Acceleration due to gravity = g = $9.8 m/s^2$

Mass of the object = m

Force = mass × acceleration

$510 N=m\times 9.8 m/s^2$

$m=52.04 kg\approx 52 kg$

The mass of an object is 52 kg.

8 0

3 years ago

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FREE BRAINLIEST! if you can answer this correctly ill give you brainliest and answer some of the questions you have posted :) th

Temka [501]

Answer:3,600 Newtons

Explanation:

The net force acting on the car is

3×10^3squared

Newtons.

Force is defined as the product of the mass of the body and its aaceleration,⇒F=ma

Substituting the above given values we get,F=(1500kg) (2.0m /s^2 squared)=3000 N=3×10^3 squared N.

N=newtons

7 0

3 years ago

What were some benefits to the american television from analog broadcast to digital broadcast?

Elanso [62]

Answer:

nothing no no no

Explanation:

nothing cause no

8 0

2 years ago

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The wave function of a particle is exp(i(kx-omegat)), where x is distance, t is time, and k and co are positive real numbers. Th

Step2247 [10]

Answer:

Momentum, $p=\hbar k$

Explanation:

The wave function of a particle is given by :

$y=exp[i(kx-\omega t)]$ ...............(1)

Where

x is the distance travelled

t is the time taken

k is the propagation constant

$\omega$ is the angular frequency

The relation between the momentum and wavelength is given by :

$p=\dfrac{h}{\lambda}$ ............(2)

From equation (1),

$k=\dfrac{2\pi}{\lambda}$

$\lambda=\dfrac{2\pi}{k}$

Use above equation in equation (2) as :

$p=\dfrac{h k}{2\pi }$

Since, $\dfrac{h}{2\pi}=\hbar$

$p=\hbar k$

So, the x-component of the momentum of the particle is $\hbar k$ . Hence, this is the required solution.

8 0

3 years ago