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mixas84 [53]
3 years ago
11

Which is a longitudinal wave that requires a medium to move

Physics
2 answers:
jeka57 [31]3 years ago
5 0

Answer:

Longitudinal wave are those in which the motion of the particles is along the direction of the medium. It always requires the medium to travel.We can say that a wave that oscillates back and forth that means moving  in same interval.

And sound is the longitudinal wave that requires a medium to move.


kifflom [539]3 years ago
5 0
Sound is a longitudinal wave which requires medium to move.
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What is the displacement of this time velocity graph i need help asap
maw [93]

Answer:

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Explanation:

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4 years ago
A photoelectric effect experiment finds a stopping potential of 1.93 V when light of wavelength 200 nm is used to illuminate the
GenaCL600 [577]

a) Zinc (work function: 4.3 eV)

The equation for the photoelectric effect is:

E=\phi + K (1)

where

E=\frac{hc}{\lambda} is the energy of the incident photon, with

h = Planck constant

c = speed of light

\lambda = wavelength

\phi = work function of the metal

K = maximum kinetic energy of the photoelectrons emitted

The stopping potential (V) is the potential needed to stop the photoelectrons with maximum kinetic energy: so, the corresponding electric potential energy must be equal to the maximum kinetic energy,

eV=K

So we can rewrite (1) as

E=\phi + eV

where we have:

\lambda=200 nm = 2\cdot 10^{-7} m

V = 1.93 V

e is the electron charge

First of all, let's find the energy of the incident photon:

E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{2\cdot 10^{-7}m}=9.95\cdot 10^{-19} J

Converting into electronvolts,

E=\frac{9.95\cdot 10^{-19}J}{1.6\cdot 10^{-19} J/eV}=6.22 eV

And now we can solve eq.(1) to find the work function of the metal:

\phi = E-eV=6.22 eV-1.93 eV=4.29 eV

so, the metal is most likely zinc, which has a work function of 4.3 eV.

b) The stopping potential is still 1.93 V

Explanation:

The intensity of the incident light is proportional to the number of photons hitting the surface of the metal. However, the energy of the photons depends only on their frequency, so it does not depend on the intensity of the light. This means that the term E in eq.(1) does not change.

Moreover, the work function of the metal is also constant, since it depends only on the properties of the material: so \phi is also constant in the equation. As a result, the term (eV) must also be constant, and therefore V, the stopping potential, is constant as well.

6 0
4 years ago
You have a piece of cork with a volume of 2 cm^3 and a density of 210kg/m^3. You hold it under water and release it.
insens350 [35]

I am sorry if it didn't helped

answers;

Calculate the buoyant force of a piece of cork of 8cm3 that floats in water. Density of cork is 207kg/m3. ?

I need the mass, in order to get the volume to apply t to the Buoyancy formula of: B=(W)object=(m)object(g)

Explanation:

From Archimedes Principle, any object partially or totally submerged in a fluid is buoyed upwards with a force equal to the weight of the displaced fluid.

∴

B

=

ρ

f

l

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f

l

g

=

1000

k

g

/

m

3

×

8

×

10

−

6

m

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N

(assuming the density of water is at standard temperature and pressure, and that the cork is totally submerged as it floats in the water

it's not the answer of your question ⁉️ but it is similar ........

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3 years ago
A positively charged light metal ball is suspended between two oppositely charged metal plates on
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Answer:

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Check Explanation for more.

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Let the ball be in square brackets, and the plates in normal brackets.

(+) [+] (-)

From the law that like charges repel and unlike charges attract.

The positive ball would go first to the negatively charged plate. After which, the ball would hold more negative charges overall than before.

Because the ball is now more negatively charged, it then travels towards the positive plate. In the same manner, the ball would transfer negative electrons to the positive plate.

So, when leaving the positive plate, the ball would be more positive and be drawn towards the negative plate once more. In doing so, it would make the negative plate more positive.

Then, the ball again holds more negative electrons and is drawn towards the positive plate once more.

This back and forth process continues until the once-positive and once-negative plates become neutral, that is, they are discharged.

The ball hanging on the insulated thread becomes neutral too at this point.

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