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Tems11 [23]
3 years ago
12

Which statement is true about a farsighted (hyperopic) eye?

Physics
1 answer:
Rufina [12.5K]3 years ago
5 0

The correct answer is C. I leave an attached image to have a clearer idea of the light process that uses the Light when entering the eye. When you have farsightedness, the image is formed behind the eye so the defective effect is generated.

What is sought with the lenses is to correct this condition and put the image back at the point of the retina.

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Gravity on the moon is about 1/6 th the gravity felt on the earth. This is because A) the moon is so far away from the earth. B)
ankoles [38]
<span>So we want to know why is there a difference between the force of gravity on the Moon and the force of gravity of the Earth. So the gravitational force between two objects depends on the masses of both objects. That can be seen from Newtons universal law of gravity. F=G*m1*m2*(1/r^2). So lets say we are holding an object of mass m=1kg on a height r=1m on the Moon and we are holding the same object on the Earth also on the same height of r=1m. The Gravitational force on the Earth will be Fg=G*M*m*(r^2) where M is the mass of the Earth. The force between the moon and that object will be Fg=G*n*m*(r^2), where n is the mass of the moon. Since mass of the Moon is much smaller than mass of the Earth, The gravitational force between the Moon and that body will be almost 6 times smaller than the gravitational force between the Earth and that body. So the correct answer is B. </span>
4 0
3 years ago
Read 2 more answers
Suppose that the angular separation of two stars is 0.1 arcseconds, and you photograph them with a telescope that has an angular
Crank

Answer: the photograph will likely show only one star.

Explanation:

Since their angular separation is smaller than the telescope's angular resolution, the picture will apparently show only one star rather than two.

4 0
3 years ago
A glass jar which is kept in freezing mixture but is in broken because​
Ann [662]

Answer:

the glass contains air bubbles that expands and contracts as the glass is heated or froze. when they expand they may cause the glass to break or even explode

8 0
3 years ago
The electric flux through a square-shaped area of side 5 cm near a very large, thin, uniformly-charged sheet is found to be 3\ti
deff fn [24]

Answer:

Explanation:

Given

side of square shape a=5\ cm

Electric flux \phi =3\times 10^{-5}\ N.m^2/C

Permittivity of free space \epsilon_0=8.85\times 10^{-12} \frac{C^2}{N.m^2}

Flux is given by

\phi =EA\cos \theta

where E=electric field strength

A=area

\theta=Angle between Electric field and area vector

E=\frac{\phi }{A\cos (0)}

E=\frac{3\times 10^{-5}}{25\times 10^{-4}\times \cos(0)}

E=0.012\ N/C

and Electric field  by a uniformly charged sheet is given by

E=\frac{\sigma }{2\epsilon_0}

where \sigma=charge density

=\frac{\sigma }{\epsilon_0}

\sigma =0.012\times 8.85\times 10^{-12}

\sigma =2.12\times 10^{-13}\ C/m^2    

5 0
3 years ago
An important diagnostic tool for heart disease is the pressure difference between blood pressure in the heart and in the aorta l
butalik [34]

Answer:

a)   f ’’ = f₀ \frac{1 + \frac{v}{c} }{1- \frac{v}{c} } , b)   Δf = 2 f₀ \frac{v}{c}

Explanation:

a) This is a Doppler effect exercise, which we must solve in two parts in the first the emitter is fixed and in the second when the sound is reflected the emitter is mobile.

Let's look for the frequency (f ’) that the mobile aorta receives, the blood is leaving the aorta or is moving towards the source

                    f ’= fo\frac{c+v}{c}

This sound wave is reflected by the blood that becomes the emitter, mobile and the receiver is fixed.

                   f ’’ = f’ \frac{c}{ c-v}

where c represents the sound velocity in stationary blood

therefore the received frequency is

                 f ’’ = f₀   \frac{c}{c-v}

let's simplify the expression

                f ’’ = f₀ \frac{c+v}{c-v}

                f ’’ = f₀ \frac{1 + \frac{v}{c} }{1- \frac{v}{c} }

         

b) At the low speed limit v <c, we can expand the quantity

                 (1 -x)ⁿ = 1 - x + n (n-1) x² + ...

                 ( 1- \frac{v}{c} ) ^{-1} = 1 + \frac{v}{c}

 

                f ’’ = fo ( 1+ \frac{v}{c}) ( 1 + \frac{v}{c} )

                f ’’ = fo ( 1 + 2 \frac{v}{c} + \frac{v^2}{ c^2} )

leave the linear term

               f ’’ = f₀ + f₀ 2\frac{v}{c}

the sound difference

               f ’’ -f₀ = 2f₀ v/c

               Δf = 2 f₀ \frac{v}{c}

4 0
3 years ago
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