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3 years ago

A wave that travels through is called a _____ wave? PLZ be specific just one word answer

Physics

2 answers:

slega [8]3 years ago

6 0

Answer:

A longitudinal wave

Explanation:

Hope this helps!

OleMash [197]3 years ago

6 0

I think it’s a Longitudinal wave plz give brainliest

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When one end of iron poker is put into the edge of a fire the other end of the poker gets hot due to what the heat transfers?

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Answer:

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3 years ago

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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 $\frac{1}{4}\Delta \lambda_{max}$

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3 years ago

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Answer:

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3 years ago

A 615 N student standing on a scale in an elevator notices that the scale reads 645 N. From this information, the student knows

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Answer:

The elevator must be moving upward.

Explanation:

During the motion of an elevator, the weight of the person deviates from his or her actual weight. This temporary weight during the motion is referred to as "Apparent Weight". So, when the elevator is moving downward, the apparent weight of the person becomes less than his or her actual weight.

On the other hand, for the upward motion of the elevator, the apparent weight of the person becomes more than the actual weight of that person.

Since the apparent weight (645 N) of the student, in this case, is greater than the actual weight (615 N) of the student.

<u>Therefore, the elevator must be moving upward.</u>

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Answer:

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Explanation:

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