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ser-zykov [4K]
3 years ago
12

An object located behind the focal point of a convex lens forms a(an)

Physics
2 answers:
fredd [130]3 years ago
8 0

Answer:

When object placed behind the focal length of lens (convex) forms enlarge/ magnified image.

Explanation:

When object placed behind the focal length of lens (convex) forms enlarge/ magnified image.

convex lens are the combination of 2 spherical surface which are bulging outwards. These are said to be convex as they are directing all light into one point. therefore when object placed behind focal length it form real, inverted and enlarge image

Tomtit [17]3 years ago
7 0
Enlarged real image.

A converging lens always has a positive focal point so the light rays always converge on the real side of the lens, creating a real image.

It will also be enlarged because when you draw a ray diagram you can see that the image will be about two times or so larger depending on where the position of the object actually is. Usually when an image is reduced, it is when it is a diverging / concave lens.
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Two marbles are launched at t = 0 in the experiment illustrated in the figure below. Marble 1 is launched horizontally with a sp
Alex787 [66]

Answer:

Two marbles are launched at t = 0 in the experiment illustrated in the figure below. Marble 1 is launched horizontally with a speed of 4.20 m/s from a height h = 0.950 m. Marble 2 is launched from ground level with a speed of 5.94 m/s at an angle above the horizontal. (a) Where would the marbles collide in the absence of gravity? Give the x and y coordinates of the collision point. (b) Where do the marbles collide given that gravity produces a downward acceleration of g = 9.81 m/s2? Give the x and y coordinates.

Explanation:

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3 years ago
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a student conducts an experiment to determine how the additional of salt to water affects the density of the water. the student
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How does friction affect a machine efficiency?
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How does displacement, acceleration, time, and velocity affect motion?
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X-rays with an energy of 301 keV undergo Compton scattering with a target. If the scattered X-rays are detected at 77.5^{\circ}
Alex73 [517]

Answer:

6.03\cdot 10^{-12} m

Explanation:

First of all, we need to find the initial wavelength of the photon.

We know that its energy is

E=301 keV = 4.82\cdot 10^{-14}J

So its wavelength is given by:

\lambda = \frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{4.82\cdot 10^{-14} J}=4.13\cdot 10^{-12}m

The formula for the Compton scattering is:

\lambda' = \lambda +\frac{h}{mc}(1-cos \theta)

where

\lambda is the original wavelength

h is the Planck constant

m is the electron mass

c is the speed of light

\theta=77.5^{\circ} is the angle of the scattered photon

Substituting, we find

\lambda' = 4.13\cdot 10^{-12} m +\frac{6.63\cdot 10^{-34} Js)}{(9.11\cdot 10^{-31}kg)(3\cdot 10^8 m/s)}(1-cos 77.5^{\circ})=6.03\cdot 10^{-12} m

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