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worty [1.4K]
3 years ago
15

HELP PLEASE

Physics
1 answer:
yulyashka [42]3 years ago
5 0

Answer:

I can msg it toyou

------->------>---->

Explanation:

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What is sound wave and examples
Nezavi [6.7K]

<em>★</em><em> </em><em>«</em><em> </em><em><u>what is sound wave and examples</u></em><em><u> </u></em><em>»</em><em> </em><em>★</em>

  • <em>A sound wave is the pattern of disturbance caused by the movement of energy traveling through a medium (such as air, water, or any other liquid or solid matter) as it propagates away from the source of the sound. The source is some object that causes a vibration, such as a ringing telephone, or a person's vocal chords.</em>

<em>hope </em><em>it</em><em> helps</em>

3 0
3 years ago
To understand thermal linear expansion in solid materials. Most materials expand when their temperatures increase. Such thermal
Marat540 [252]

Answer:

Check the explanation

Explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

5 0
3 years ago
Help pls, see picture. Will mark Brainliest
Leno4ka [110]

Answer:

B

Explanation:

the graph shows the line going up (accelerating) and it isn't curving like d so it doesn't stop accelerating

Hope this helps :)

4 0
3 years ago
A huge rotating cloud of particles in space gravitate together to form an increasingly dense ball. As ir shrinks in size, the cl
Trava [24]

Answer:

rotates faster

Explanation:

A huge rotating cloud of particles in space gravitate together to form an increasingly dense ball As it shrinks in size, the cloud rotates faster. Because Angular momentum is conserved, so when it shrinks the moment of inertia decreases, then angular speed must increase. So it rotates fast.

4 0
3 years ago
The work function for tungsten metal is 4.52eV a. What is the cutoff (threshold) wavelength for tungsten? b. What is the maximum
Tanya [424]

Answer: a) 274.34 nm; b) 1.74 eV c) 1.74 V

Explanation: In order to solve this problem we have to consider the energy balance for the photoelectric effect on tungsten:

h*ν = Ek+W ; where h is the Planck constant, ek the kinetic energy of electrons and W the work funcion of the metal catode.

In order to calculate the cutoff wavelength we have to consider that Ek=0

in this case  h*ν=W

(h*c)/λ=4.52 eV

λ= (h*c)/4.52 eV

λ= (1240 eV*nm)/(4.52 eV)=274.34 nm

From this h*ν = Ek+W;  we can calculate the kinetic energy for a radiation wavelength of 198 nm

then we have

(h*c)/(λ)-W= Ek

Ek=(1240 eV*nm)/(198 nm)-4.52 eV=1.74 eV

Finally, if we want to stop these electrons we have to applied a stop potental equal to 1.74 V . At this potential the photo-current drop to zero. This potential is lower to the catode, so this  acts to slow down the ejected electrons from the catode.

5 0
3 years ago
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