All categories

3 years ago

Which, if any, of these mirrors can produce a real image?

2 answers:

4 0

Real and virtual images.<span> In a dark room one can form an image of a candle flame on a white screen by letting the rays from the candle pass through a small opening. See Fig. 1. This idea is used in the pinhole camera. The pinhole camera is simply a light-proof box with a tiny pinhole in the front and photographic film stretched across the rear wall. See Fig. 2. One uncovers the pinhole for a minute or so, then covers it back up, to take a picture. If it is held steady, it can take a good picture. If one replaces the pinhole with a lens, the intensity of the incident light is increased and a sharp image can be recorded in a fraction of a second. The image formed by a pinhole or a lens is formed by incident rays of light on a surface and stands in contrast to another kind of image — the kind of image formed by a mirror. The image formed by a mirror is not real, it is an illusion formed by the way the light reflects off the mirror. The image seen in a mirror looks three dimensional and real but we know there is nothing where the image appears to be. An image such as that formed by a pinhole or lens is called a </span>real image<span>. An image such as that formed by a mirror is called a </span>virtual<span> image</span>

andrezito [222]3 years ago

4 0

Penn Foster STudents: Concave Spherical

You might be interested in

If an atom contains 1 electron and 1 proton will it carry any charge or not?

Ipatiy [6.2K]

Answer：

It is a neutral atom and does not carry a charge

7 0

3 years ago

In your own words, please describe the difference between a complete ionic equation and a net ionic equation.

beks73 [17]

Explanation:

In the molecular equation for a reaction, all of the reactants and products are represented as neutral molecules (even soluble ionic compounds and strong acids). In the complete ionic equation, soluble ionic compounds and strong acids are rewritten as dissociated ions.

The net ionic equation is a chemical equation for a reaction that lists only those species participating in the reaction. The net ionic equation is commonly used in acid-base neutralization reactions, double displacement reactions, and redox reactions.

6 0

3 years ago

Calculate the pH of the following?

Gwar [14]

The correct answer is c

4 0

3 years ago

If you dispense 40 ml of hexane, but it turns out you only need 5 ml, what should you do with the remainder?

Natasha2012 [34]

After subtracting the volume needed from the volume dispensed, we got a remainder of 35ml

<h3>Subtraction of Numbers</h3>

Given Data

Volume of Hexane dispensed = 40ml

Volume needed = 5 ml

Let us compute the amount of excess hexane/ the volume that will remain

Remainder = The difference in volume dispensed and the volume needed

Remainder = 40-5

Remainder = 35 ml

The remainder is 35ml

Learn more about subtraction of numbers here:

brainly.com/question/4721701

7 0

2 years ago

In each case, calculate the appropriate ratio to show that the information given is consistent with the law of multiple proporti

Alchen [17]

Answer:

(a) 3:2; (b) 2:1

Explanation:

The Law of Multiple Proportions states that when two elements A and B combine to form two or more compounds, the masses of B that combine with a given mass of A are in the ratios of small whole numbers.

That is, if one compound has a ratio r₁ and the other has a ratio r₂, the ratio of the ratios r is in small whole numbers.

(a) Ammonia and hydrazine.

In ammonia, the mass ratio of H:N is r₁ = 0.2158/1

In hydrazine, the mass ratio of H:N is r₂ = 0.1439/1

The ratio of the ratios is:

$r = \dfrac{r_{1}}{r_{2}} = \dfrac{ 0.2158}{0.1439} = \dfrac{1.500 }{1} = \dfrac{2.999}{2} \approx \mathbf{\dfrac{3}{2}}\\\\\text{The relative amounts of H in the two compounds are in the ratio }\boxed{\mathbf{\dfrac{3}{2}}}$

(b) Nitrogen oxides

In nitrogen monoxide, the mass ratio of O:N is r₁ = 1.142/1

In dinitrogen monoxide, the mass ratio of O:N is r₂ = 0.571/1

The ratio of the ratios is:

$r = \dfrac{r_{1}}{r_{2}} = \dfrac{ 1.142}{0.571} = \dfrac{2.000 }{1} \approx \mathbf{\dfrac{2}{1}}\\\\\text{The relative amounts of O in the two compounds are in the ratio }\boxed{\mathbf{\dfrac{2}{1}}}$

8 0

4 years ago