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2 years ago

Can someone pls help I’m stuck!

Physics

1 answer:

Elan Coil [88]2 years ago

6 0

B because why not i was wondering how ppl are so smart and let me tell ya i don’t know

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soldier1979 [14.2K]

For the sound wave passing through regions of the ocean with varying density, longer wavelengths correspond to greater density of the water.

<h3>What is effect of density of a medium on wavelength of a wave?</h3>

The density of a medium is directly proportional to the wavelength of a wave.

The higher the density of the medium, the longer the wavelength of a wave.

Therefore, for a sound wave passing through regions of the ocean with varying density, longer wavelengths correspond to greater density of the water.

Learn more about density and wavelength at: brainly.com/question/9486264

#SPJ1

4 0

1 year ago

A total positive charge of 12.00 mC is evenly distributed on a straight thin rod of length 6.00 cm.

Gala2k [10]

It’s not in English can you change it please

3 0

2 years ago

The maximum Compton shift in wavelength occurs when a photon isscattered through 180^\circ .

vlabodo [156]

Answer: $90\°$

Explanation:

The Compton Shift $\Delta \lambda$ in wavelength when the photons are scattered is given by the following equation:

$\Delta \lambda=\lambda_{c}(1-cos\theta)$ (1)

Where:

$\lambda_{c}=2.43(10)^{-12} m$ is a constant whose value is given by $\frac{h}{m_{e}c}$ , being $h$ the Planck constant, $m_{e}$ the mass of the electron and $c$ the speed of light in vacuum.

$\theta)$ the angle between incident phhoton and the scatered photon.

We are told the maximum Compton shift in wavelength occurs when a photon isscattered through $180\°$ :

$\Delta \lambda_{max}=\lambda_{c}(1-cos(180\°))$ (2)

$\Delta \lambda_{max}=\lambda_{c}(1-(-1))$

$\Delta \lambda_{max}=2\lambda_{c}$ (3)

Now, let's find the angle that will produce a fourth of this maximum value found in (3):

$\frac{1}{4}\Delta \lambda_{max}=\frac{1}{4}2\lambda_{c}(1-cos\theta)$ (4)

$\frac{1}{4}\Delta \lambda_{max}=\frac{1}{2}\lambda_{c}(1-cos\theta)$ (5)

If we want $\frac{1}{4}\Delta \lambda_{max}=\frac{1}{2}\lambda_{c}$ , $1-cos\theta$ must be equal to 1:

$1-cos\theta=1$ (6)

Finding $\theta$ :

$1-1=cos\theta$

$0=cos\theta$

$\theta=cos^{-1} (0)$

Finally:

$\theta=90\°$ This is the scattering angle that will produce $\frac{1}{4}\Delta \lambda_{max}$

7 0

2 years ago

What relationship exists between the amplitude of a wave and amount of disturbance in the water?

guapka [62]

A wave is a result of the disturbance in the equilibrium state. There are two types of wave, transverse and longitudinal. Transverse wave affects amplitude while longitudinal wave affects the frequency of the wave. As for the transverse wave, the magnitude of the perpendicular disturbance of the wave is directly proportional to the amplitude of the wave. The higher the transverse disturbance the higher the amplitude.

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2 years ago

If a 50 kg student is standing on the edge of a cliff. Find the student’s gravitational potential energy if the cliff is 40 m hi

Oduvanchick [21]

you can find it using the equation: potential energy=mass*gravitational acceleration*height.

energy=50kg*9.8N/kg*40m=19600Nm=19600J or 19.6kJ

Sometimes they use 10 instead of 9.8 for the g constant.

Rember to make me Brainliest!!!

3 0

3 years ago