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Ivan
3 years ago
5

A _____________ is any visual, audio, audiovisual, or other sensory material used to enhance a verbal message.

Physics
1 answer:
Ber [7]3 years ago
8 0

Answer:

Presentational aid

Explanation:

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Increase the slit width to 1050 nm. What happens to the width of the central bright fringe? (Always remember to wait a few secon
djyliett [7]

Answer:

b.

Explanation:

In case of Single Slit, diffraction will occur.

Then In Single slit Diffraction, width of central fringe is

x_c= 2D\lambda/a


where D= distance b/w screen and slit

a= slit width

\lambda = wavelength

Thus if Screen width increases keeping other factors same then width of central fringe becomes narrower as

x_c\propto 1/a

On increasing the slit width the central bright fringe width The width of the central bright fringe becomes narrower.

3 0
3 years ago
A 1000-kg whale swims horizontally to the right at a speed of 6.0 m/s. It suddenly collides directly with a stationary seal of m
anzhelika [568]

Answer:

Momentum after collision will be 6000 kgm/sec

Explanation:

We have given mass of the whale = 1000

Initial velocity v = 6 m/sec

It collides with other mass of 200 kg which is at stationary

Initial momentum of the whale = 1000×6 = 6000 kgm/sec

We have to find the momentum after collision

From conservation of momentum

Initial momentum = final momentum

So final momentum = 6000 kgm/sec

5 0
3 years ago
The once-ler used what to make thneeds
Bogdan [553]
He used truffula trees to make thneeds
6 0
3 years ago
There are stars located in the center bulge of the Milky Way and the spiral arms of the Milky Way. What is the difference betwee
nadya68 [22]

Answer:

The stars at the center bulge are bigger and brighter than the stars in the arms.

Explanation:

5 0
3 years ago
An infinitely long cylindrical insulating shell of inner radius a and outer radius b has a uniform volume charge density p. Dete
xenn [34]

Answer: The electric field is: a) r<a , E0=; b) a<r<b E=ρ (r-a)/εo;

c) r>b E=ρ b (b-a)/r*εo

Explanation: In order to solve this problem we have to use the Gaussian law in diffrengios regions.

As we know,

∫E.dr= Qinside/εo

For r<a --->Qinside=0 then E=0

for a<r<b er have

E*2π*r*L= Q inside/εo       in this case Qinside= ρ.Vol=ρ*2*π*r*(r-a)*L

E*2π*r*L =ρ*2*π*r* (r-a)*L/εo

E=ρ*(r-a)/εo

Finally for r>b

E*2π*r*L =ρ*2*π*b* (b-a)*L/εo

E=ρ*b* (b-a)*/r*εo

3 0
3 years ago
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