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kipiarov [429]
3 years ago
11

An object is placed 5.00 cm beyond the focal point of a convex lens whose focal length is 10.0 cm. If the object height is 3.0 c

m, calculate the image height.
A. 150
B. -6.0
C. 1.5

What is the orientation of the image? Erect or inverted?
Physics
1 answer:
Aleks04 [339]3 years ago
4 0

Answer:

The height of the image is, h' = 6.0 cm

The image is erect.

Explanation:

Given data,

The object distance, u = -5 cm

The focal length of convex lens, f = 10 cm

The object height, h = 3 cm

The lens formula,

                      \frac{1}{f}=\frac{1}{v}-\frac{1}{u}

                      \frac{1}{10}=\frac{1}{v}-\frac{1}{-5}

                      \frac{1}{v}=\frac{1}{10}-\frac{1}{5}

                      v = -10 cm

The magnification factor of lens

                     m=\frac{-10}{-5}

                     m = 2

                     m=\frac{h'}{h}

                     h'=h\times m

                     h'=3\times 2

                     h' = 6 cm

The height of the image is, h' = 6 cm

The image is erect.

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Acceleration = (-28 m/s) / (13 sec)

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5 0
3 years ago
A power company charges its customers for electricity based upon which one of the following?a. powerb. energyc. power per unit o
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B. Energy

A power company charges its customers for electricity based upon B. Energy.

<h3>Explanation:</h3>

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P = \frac{E}{t} \\E= P\times t

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7 0
3 years ago
Points A (-5,6), B (2,-2), and C (-6,-3) are placed in three different quadrants of a Cartesian coordinate system. Convert each
AURORKA [14]

Answer: A (\sqrt{61},309.8°)

              B (2\sqrt{2}, 315°)

             C (3\sqrt{5}, 26.56°)

Explanation: To transform rectangular coordinates into polar coordinates use:

r=\sqrt{x^{2}+y^{2}} and \theta=tan^{-1}(\frac{y}{x})

For point A:

r=\sqrt{(-5)^{2}+6^{2}}

r=\sqrt{61}

\theta=tan^{-1}(\frac{6}{-5})

\theta=tan^{-1}(-1.2)

\theta=-50.2°

Point A is in the II quadrant, so we substract the angle for 360° since it is in degrees:

\theta=360-50.2

\theta= 309.8°

Polar coordinates for point A is (\sqrt{61}, 309.8°)

For point B:

r=\sqrt{2^{2}+(-2)^{2}}

r=\sqrt{8}

r=2\sqrt{2}

\theta=tan^{-1}(\frac{-2}{2} )

\theta=tan^{-1}(1)

\theta=-45°

Point B is in IV quadrant, so:

\theta=360-45

\theta= 315°

Polar coordinates for point B is (2\sqrt{2}, 315°)

For point C:

r=\sqrt{(-6)^{2}+(-3)^{2}}

r=\sqrt{45}

r=3\sqrt{5}

\theta=tan^{-1}(\frac{-3}{-6} )

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Polar coordinates for point C is (3\sqrt{5}, 26.56°)

3 0
3 years ago
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Answer:

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A neutron undergoes the decaying process to produce an electron, a proton, and energy.

The reaction of neutron decay:

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by the same way:-
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6 0
3 years ago
Read 2 more answers
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