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3 years ago

What are the characteristics of the image based on the values?Check all that apply

Physics

2 answers:

Ahat [919]3 years ago

8 0

inverted, real, behind the lens is the answer

kifflom [539]3 years ago

7 0

Answer:

Inverted

Real

Behind the lens

Explanation:

As we know that

$d_o = 16 cm$

$d_i = 16 cm$

also we know that

$h_o = 4 cm$

$h_i = -4 cm$

now we know that magnification is given by

$M = \frac{d_i}{d_o} = \frac{h_i}{h_o}$

so we can say here that

M = 1

so here image size will be same as object size and it must be real and inverted as magnification is negative in sign

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A projector is placed on the ground 22 ft. away from a projector screen. A 5.2 ft. tall person is walking toward the screen at a

Stella [2.4K]

Answer:

y = 67.6 feet, y = 114.4/ (22 - 3t)

Explanation:

For this exercise let's use that light travels in a straight line and some trigonometric relationships, the symbols are in the attached diagram

Large triangle Projector up to the screen

tan θ = y / L

For the small triangle. Projector up to the person

tan θ = y₀ / (L-d)

The angle is the same, so we equate the two equations

y₀ / (L -d) = y / L

y = y₀ L / (L-d)

The distance from the screen (d), we look for it with kinematics

v = d / t

d = v t

we replace

y = y₀ L / (L - v t)

y = 5.2 22 / (22 - 3 t)

y = 114.4 (22 - 3t)⁻¹

This is the equation of the shadow height change as a function of time

For the suggested distance the shadow has a height of

y = 114.4 / (22-13)

y = 67.6 feet

7 0

3 years ago

Show that rigid body rotation near the Galactic center is consistent with a spherically symmetric mass distribution of constant

irakobra [83]

To solve this problem we will use the concepts related to gravitational acceleration and centripetal acceleration. The equality between these two forces that maintains the balance will allow to determine how the rigid body is consistent with a spherically symmetric mass distribution of constant density. Let's start with the gravitational acceleration of the Star, which is

$a_g = \frac{GM}{R^2}$