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QveST [7]
2 years ago
10

Image caughton Screen is called​

Physics
2 answers:
natka813 [3]2 years ago
5 0

Answer:

Real Image

Explanation:

Images which are formed on the screen by the actual intersection of light rays are called real images.

bixtya [17]2 years ago
4 0

Answer:

<h2>Virtual image</h2>

Explanation:

<h3><em>Virtual</em><em> </em><em>image</em><em> </em><em>can</em><em> </em><em>be</em><em> </em><em>caught</em><em> </em><em>on</em><em> </em><em>a</em><em> </em><em>screen</em></h3>

<em>hope</em><em> </em><em>this</em><em> </em><em>helps</em><em> </em><em>you</em><em>.</em>

<em>will</em><em> </em><em>give</em><em> </em><em>the</em><em> </em><em>brainliest</em><em>!</em>

<em>follow</em><em> </em><em>~</em><em>H</em><em>i</em><em>1</em><em>3</em><em>1</em><em>5</em><em>~</em>

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Answer:

I think it's 3 c only the loudness of the sound

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3 years ago
Name the forces acting on a plastic bucket container water held above ground level in your hand. Discuss why the forces acting o
SashulF [63]

There are two forces at play:

  • The gravitational force acting downward due to the mass of the bucket and the water that it contains.
  • The upward force that your hand exerts on the bucket.

If the magnitude of the force your hand exerts on the bucket equals the magnitude of the gravitational force, the bucket is in static equilibrium. That means the bucket is not moving and the forces acting on it balance each other out, making the net force 0.

Having 0 net force means the bucket doesn't undergo any acceleration, or change in motion.

5 0
3 years ago
If you accelerate from rest at 12m/s^2 for 5 seconds, what is your finally velocity?
jeka94

Answer:

v=60m/s

Explanation:

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v=v_o+at

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v=60m/s

3 0
2 years ago
Molecular motion is a measure of the movement of the molecules in a substance. Molecular motion is substantially different betwe
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8 0
3 years ago
Read 2 more answers
Practical steam engines utilize 450ºC steam, which is later exhausted at 270ºC.
Naily [24]

(a) 0.249 (24.9 %)

The maximum efficiency of a heat engine is given by

\eta = 1-\frac{T_C}{T_H}

where

Tc is the low-temperature reservoir

Th is the high-temperature reservoir

For the engine in this problem,

T_C = 270^{\circ}C+273=543 K

T_H = 450^{\circ}C+273=723 K

Therefore the maximum efficiency is

\eta = 1-\frac{T_C}{T_H}=1-\frac{543}{723}=0.249

(b-c) 0.221 (22.1 %)

The second steam engine operates using the exhaust of the first. So we have:

T_H = 270^{\circ}C+273=543 K is the high-temperature reservoir

T_C = 150^{\circ}C+273=423 K is the low-temperature reservoir

If we apply again the formula of the efficiency

\eta = 1-\frac{T_C}{T_H}

The maximum efficiency of the second engine is

\eta = 1-\frac{T_C}{T_H}=1-\frac{423}{543}=0.221

8 0
3 years ago
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