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3 years ago

12

Scientists use this part of the electromagnetic spectrum to diagnose broken bones. X–rays infrared waves gamma rays ultraviolet

waves

2 answers:

Westkost [7]3 years ago

5 0

Answer: XRays (I answered this on my test, and got it correct!)

Snezhnost [94]3 years ago

4 0

X-rays are used to identify broken bones

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How do you write a letter to the editor of a newspaper in reference to bullying

suter [353]

How do you write a letter to the editor?

Open the letter with a simple salutation. ...

Grab the reader's attention. ...

Explain what the letter is about at the start. ...

Explain why the issue is important. ...

Give evidence for any praise or criticism. ...

State your opinion about what should be done. ...

Keep it brief. ...

Sign the letter.

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3 years ago

PLEASE HELPPP FASTTTTTTTTTTTTTTT!!!!!!!!!!!!!!!!!!!!!!!!!!!In a paragraph of no less than five sentences, explain how these Russ

kompoz [17]

<span>These Russian nesting dolls are a good analogy for the strongest magnet in the world, because the magnitis were put inside each other. The magnets would be fitted together. Because they are closer together, the coils of magnet would be stronger. They would be able to magnetize better.</span>

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3 years ago

When light of wavelength 240 nm falls on a cobalt surface, electrons having a maximum kinetic energy of 0.17 eV are emitted. Fin

dusya [7]

Answer:

(a) 5.04 eV (B) 248.14 nm (c) $1.21\times 10^{15}Hz$

Explanation:

We have given Wavelength of the light \lambda = 240 nm

According to plank's rule ,energy of light

$E = h\nu = \frac{hc}{}\lambda$

$E = h\nu = \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{ 240\times 10^{-9} m\times 1.6\times 10^{-19}J/eV}= 5.21 eV$

Maximum KE of emitted electron i= 0.17 eV

Part( A) Using Einstien's equation

$E = KE_{max}+\Phi _{0}$ , here $\Phi _0$ is work function.

$\Phi _{0}=E - KE_{max}$ = 5.21 eV-0.17 eV = 5.04 eV

Part( B) We have to find cutoff wavelength

$\Phi _{0} = \frac{hc}{\lambda_{cuttoff}}$

$\lambda_{cuttoff}= \frac{hc}{\Phi _{0} }$

$\lambda_{cuttoff}= \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{5.04 eV\times 1.6\times 10^{-19}J/eV }=248.14 nm$

Part (C) In this part we have to find the cutoff frequency

$\nu = \frac{c}{\lambda_{cuttoff}}= \frac{3\times 10^{8}m/s}{248.14 \times 10^{-19} m }= 1.21\times 10^{15} Hz$

5 0

3 years ago

A transverse wave on a string has an amplitude a. A tiny spot on the string is colored red. As one cycle of the wave passes by

aliya0001 [1]

Answer:

Option D) 4A

Explanation:

As the cycle of the wave passes by, the amplitude gives the longest journey when the spot travels from the undistributed position. During each cycle the spot travels "Four times" .

Considering one of this cycle, if it begins to travel from it's undistributed position , there would be four movements i.e

* Upward movement through distance A

*Downward movement through distance A

*Downward again through distance A

*Upward through distance A.

Then it would travel back to its undistributed position held

4 0

2 years ago

A student places her 490 g physics book on a frictionless table. She pushes the book against a spring, compressing the spring by

Paul [167]

Answer:

book speed is 3.99 m/s

Explanation:

given data

mass m = 490 g = 0.490 kg

compressing x = 7.10 cm = 0.0710 m

spring constant k = 1550 N/m

to find out

book speed

solution

we know energy is conserve so

we can say

loss in spring energy is equal to gain in kinetic energy

so

$\frac{1}{2}*k*x^2 = \frac{1}{2}*m*v^2$ ..................1

put here value

$\frac{1}{2}*1550*0.071^2 = \frac{1}{2}*0.490*v^2$

v = 3.99 m/s

so book speed is 3.99 m/s

5 0

3 years ago