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3 years ago

Why do concave lenses always form virtual images?

Physics

2 answers:

Dima020 [189]3 years ago

8 0

The rays diverge. Hope this helps :D

Licemer1 [7]3 years ago

5 0

Answer:

The extension of the refracted rays will intersect at a point. This point is known as the focal point. Notice that a diverging lens such as this double concave lens does not really focus the incident light rays that are parallel to the principal axis; rather, it diverges these light rays.

Explanation:

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A ball with a mass of 2000 g is floating on the surface of a pool of water. What is the minimum volume that the ball could have

Doss [256]

Answer:

$2000\; {\rm cm^{3}}$ .

Explanation:

When the ball is placed in this pool of water, part of the ball would be beneath the surface of the pool. The volume of the water that this ball displaced is equal to the volume of the ball that is beneath the water surface.

The buoyancy force on this ball would be equal in magnitude to the weight of water that this ball has displaced.

Let $m(\text{ball})$ denote the mass of this ball. Let $m(\text{water})$ denote the mass of water that this ball has displaced.

Let $g$ denote the gravitational field strength. The weight of this ball would be $m(\text{ball}) \, g$ . Likewise, the weight of water displaced would be $m(\text{water})\, g$ .

For this ball to stay afloat, the buoyancy force on this ball should be greater than or equal to the weight of this ball. In other words:

$\text{buoyancy} \ge m(\text{ball})\, g$ .

At the same time, buoyancy is equal in magnitude the the weight of water displaced. Thus:

$\text{buoyancy} = m(\text{water}) \, g$ .

Therefore:

$m(\text{water})\, g = \text{buoyancy} \ge m(\text{ball})\, g$ .

$m(\text{water}) \ge m(\text{ball})$ .

In other words, the mass of water that this ball displaced should be greater than or equal to the mass of of the ball. Let $\rho(\text{water})$ denote the density of water. The volume of water that this ball should displace would be:

$\begin{aligned}V(\text{water}) &= \frac{m(\text{water})}{\rho(\text{water})} \\ &\ge \frac{m(\text{ball}))}{\rho(\text{water})} \end{aligned}$ .

Given that $m(\text{ball}) = 2000\; {\rm g}$ while $\rho = 1.00\; {\rm g\cdot cm^{-3}}$ :

$\begin{aligned}V(\text{water}) &\ge \frac{m(\text{ball}))}{\rho(\text{water})} \\ &= \frac{2000\; {\rm g}}{1.00\; {\rm g\cdot cm^{-3}}} \\ &= 2000\; {\rm cm^{3}}\end{aligned}$ .

In other words, for this ball to stay afloat, at least $2000\; {\rm cm^{3}}$ of the volume of this ball should be under water. Therefore, the volume of this ball should be at least $2000\; {\rm cm^{3}}\!$ .

3 0

2 years ago

Pweese help luvs qwq pweese look at the image?

hjlf

Answer:

12 m/s

Explanation:

divide distance over time

72/6 = 12

6 0

3 years ago

a system that uses reflected radio waves to detect objects and to measure their distance and speed is called

viktelen [127]

Such system is called RADAR (RAdio Detection And Ranging).

The Radar emits radio waves, that are reflected back by the object. Since the speed of the radio waves is known (their speed is equal to the speed of light), by measuring the time the waves take to come back to the source it is possible to infer the distance they covered, and so the distance of the object.

3 0

3 years ago

Two wires are made out of the same metal, but one wire is twice as long as the other wire. which wire will have the greatest ela

lesya692 [45]

Elastic modulus only depends on the type of the material which is vary material to material based on its physical properties. elastic modulus does not depend on the length or the cross-section area of the wire or the force applied on the wire. so the lengths of wire does not decide the elastic modulus. since the two wires are made out of same metal ,hence their elastic modulus will be same.

7 0

3 years ago

Explain, in terms of waves, what is meant by a compression and a rarefaction.

Rom4ik [11]

Answer:

Compression is that part of longitudinal wave in which the medium of particles are closer and there is momentary decrease in volume of medium.

Rarefaction is that part of longitudinal wave in which the medium of particles apart and there is momentary increase in volume of medium.

Compression is that part in which particles vibrating in a medium come closer to each other and rarefaction is that part in which particles vibrating in a medium move apart from each other .

5 0

3 years ago