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In-s [12.5K]
3 years ago
13

Plz help An X-ray technician will place a lead cloth over parts of your body when taking an X-ray to protect your cells from bei

ng damaged. Why do X-rays have the potential to harm your body but radio waves do not?
Question 6 options:

a.X-rays have a taller wavelength than radio waves as shown in the diagram, so they are more harmful.


c.Radio waves have longer wavelengths than X-rays and therefore carry more energy than X-rays.


c.Radio waves are shorter than X-rays and therefore weaker in strength..


d.X-rays have a much shorter wavelength than radio waves, and therefore carry more energy that can be transferred to your body.

Physics
2 answers:
gayaneshka [121]3 years ago
6 0
Letter d. Hope it helps.
dalvyx [7]3 years ago
6 0

Answer: The correct option is d.

Explanation:

The wavelength and the frequency have an inverse relationship. The shorter the wavelength of the wave, higher the frequency of the wave. Then the photon energy of the waves will be high because the energy of the photon is directly proportional to the frequency.

X-rays have a much shorter wavelength than radio waves. It carries more energy. It can ionize the matter. When it is incident on the biological material then it can disrupt the molecular bond.

This is the reason that an X-ray technician places a a lead cloth over parts of your body when taking an X-ray to protect your cells from being damaged.

Therefore, the correct option is d.

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There is a repulsive force between two charged objects when they are of like charges/ they are likely charged (like charges repel each other)
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A technician wishes to determine the wavelength of the light in a laser beam. To do so, she directs the beam toward a partition
Nana76 [90]

Answer:

λ = 5.656 x 10⁻⁷ m = 565.6 nm

Explanation:

Using the formula of fringe spacing from the Young's Double Slit experiment, which is given as follows:

\Delta x = \frac{\lambda L}{d}\\\\\lambda = \frac{\Delta x\ d}{L}

where,

λ = wavelength = ?

Δx = fringe spacing = 1.6 cm = 0.016 m

L = Distance between slits and screen = 4.95 m

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Therefore,

\lambda = \frac{(0.016\ m)(0.000175\ m)}{4.95\ m}\\\\

<u>λ = 5.656 x 10⁻⁷ m = 565.6 nm</u>

6 0
3 years ago
Air as an ideal gas enters a diffuser operating at steady state at 5 bar, 280 K with a velocity of 510 m/s. The exit velocity is
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Answer:

Explanation:

Calculating the exit temperature for K = 1.4

The value of c_p is determined via the expression:

c_p = \frac{KR}{K_1}

where ;

R = universal gas constant = \frac{8.314 \ J}{28.97 \ kg.K}

k = constant = 1.4

c_p = \frac{1.4(\frac{8.314}{28.97} )}{1.4 -1}

c_p= 1.004 \ kJ/kg.K

The derived expression from mass and energy rate balances reduce for the isothermal process of ideal gas is :

0=(h_1-h_2)+\frac{(v_1^2-v_2^2)}{2}     ------ equation(1)

we can rewrite the above equation as :

0 = c_p(T_1-T_2)+ \frac{(v_1^2-v_2^2)}{2}

T_2 =T_1+ \frac{(v_1^2-v_2^2)}{2 c_p}

where:

T_1  = 280 K \\ \\ v_1 = 510 m/s \\ \\ v_2 = 120 m/s \\ \\c_p = 1.0004 \ kJ/kg.K

T_2= 280+\frac{((510)^2-(120)^2)}{2(1.004)} *\frac{1}{10^3}

T_2 = 402.36 \ K

Thus, the exit temperature = 402.36 K

The exit pressure is determined by using the relation:\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{k}{k-1}

P_2=P_1(\frac{T_2}{T_1})^\frac{k}{k-1}

P_2 = 5 (\frac{402.36}{280} )^\frac{1.4}{1.4-1}

P_2 = 17.79 \ bar

Therefore, the exit pressure is 17.79 bar

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