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Sladkaya [172]
3 years ago
12

When you take pictures with a camera, the distance between the lens and the film or chip has to be adjusted, depending on the di

stance at which you want to focus. This is done by moving the lens. If you want to change your focus so that you can take a picture of something farther away, which way do you have to move the lens
Physics
1 answer:
Shalnov [3]3 years ago
8 0

Answer:

Explanation:

When a camera shifts focus from a faraway object to a nearby object, the lens-to-film distance must increase. Likewise, when it shifts focus from a nearby object to a distant object, there must be an increase in the lens to film distance (that is, the image distance).

Therefore, if the picture of an object that is far away, the lens must move towards the film.

The focal length cannot be changed because it is fixed for a lens. Nevertheless, in order to focus on an object, the image distance can be changed.

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A circuit with resistive elements of 220, 100, 57, and 43 produce what total resistance?
Korvikt [17]

The total resistance is 420 ohm.

A circuit with resistive elements of 220, 100, 57, and 43 produce what total resistance  

 R= 220+ 100+ 57+ 43

    = 420 Ω

What is resistance and its types?

Resistance is a measure of the opposition to current flow in an electrical circuit also known as ohmic resistance or electrical resistance. Ohms are measured as resistance, symbolized by the Greek letter omega (Ω). The ratio of the applied voltage to the current through the material is then known as resistance.

What causes resistance?

An electric current flows when electrons move through a conductor, such as a metal wire. The moving electrons can collide with the ions in the metal. This makes it more difficult for the current to flow, and causes resistance.

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A. microscopes !!!!!!!!!!
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How many balls are there in a ski ball machine?
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Answer:

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Explanation:

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2 years ago
Radiation from the Sun The intensity of the radiation from the Sun measured on Earth is 1360 W/m2 and frequency is f = 60 MHz. T
Zina [86]

a) Total power output: 3.845\cdot 10^{26} W

b) The relative percentage change of power output is 1.67%

c) The intensity of the radiation on Mars is 540 W/m^2

Explanation:

a)

The intensity of electromagnetic radiation is given by

I=\frac{P}{A}

where

P is the power output

A is the surface area considered

In this problem, we have

I=1360 W/m^2 is the intensity of the solar radiation at the Earth

The area to be considered is area of a sphere of radius

r=1.5\cdot 10^{11} m (distance Earth-Sun)

Therefore

A=4\pi r^2 = 4 \pi (1.5\cdot 10^{11})^2=2.8\cdot 10^{23}m^2

And now, using the first equation, we can find the total power output of the Sun:

P=IA=(1360)(2.8\cdot 10^{23})=3.845\cdot 10^{26} W

b)

The energy of the solar radiation is directly proportional to its frequency, given the relationship

E=hf

where E is the energy, h is the Planck's constant, f is the frequency.

Also, the power output of the Sun is directly proportional to the energy,

P=\frac{E}{t}

where t is the time.

This means that the power output is proportional to the frequency:

P\propto f

Here the frequency increases by 1 MHz: the original frequency was

f_0 = 60 MHz

so the relative percentage change in frequency is

\frac{\Delta f}{f_0}\cdot 100 = \frac{1}{60}\cdot 100 =1.67\%

And therefore, the power also increases by 1.67 %.

c)

In this second  case, we have to calculate the new power output of the Sun:

P' = P + \frac{1.67}{100}P =1.167P=1.0167(3.845\cdot 10^{26})=3.910\cdot 10^{26} W

Now we want to calculate the intensity of the radiation measured on Mars. Mars is 60% farther from the Sun than the Earth, so its distance from the Sun is

r'=(1+0.60)r=1.60r=1.60(1.5\cdot 10^{11})=2.4\cdot 10^{11}m

Now we can find the radiation intensity with the equation

I=\frac{P}{A}

Where the area is

A=4\pi r'^2 = 4\pi(2.4\cdot 10^{11})^2=7.24\cdot 10^{23} m^2

And substituting,

I=\frac{3.910\cdot 10^{26}}{7.24\cdot 10^{23}}=540 W/m^2

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