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slavikrds [6]
3 years ago
5

A(n) ____________ image cannot be projected and forms where light rays appear to originate.

Physics
2 answers:
IrinaVladis [17]3 years ago
6 0
A virtual image is the answer.
Sedbober [7]3 years ago
6 0
<h3><u>Answer;</u></h3>

Virtual image

A<em><u> virtual image</u></em> cannot be projected and forms where light rays appear to originate.

<h3><u>Explanation</u>;</h3>
  • <em><u>A virtual image is an image that can not be formed on a screen. This type of an image is formed when the rays of light appear to meet at a point after reflection or refraction. Virtual images are always erect or upright.</u></em>
  • <em><u>A real image on the other hand, is an image that can be formed on a screen. Real image is formed when the rays of light meet at some point after reflection or refraction. Real images are always inverted.</u></em>
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Does the orbital period of a planet depend on the mass of the planet or on the mass of the star that it orbits?
jasenka [17]

Answer:

The orbital period of a planet depends on the mass of the planet.

Explanation:

A less massive planet will take longer to complete one period than a more massive planet.

8 0
2 years ago
A block of mass M is connected by a string and pulley to a hanging mass m. The coefficient of kinetic friction between block M a
aleksklad [387]

Answer:

a)  y = 0.98 t², t=1s y= 0.98 m,  

b) he two blocks must move the same distance

c) v = 1.96 m / s,  d)  a = -1.96 m / s², e)  x = 0.98 m

Explanation:

For this exercise we can use Newton's second law

Big Block

Y axis

             N-W = 0

             N = M g

X axis

             T- fr = Ma

the friction force has the expression

             fr = μ N

             fr = μ Mg

small block

             w- T = m a

             

we write the system of equations

             T - fr = M a

             mg - T = m a

we add and resolved

             mg-  μ Mg = (M + m) a

             a = g \ \frac{m - \mu M}{m+M}

             a = 9.8 \ \frac{10- 0.2 \ 20}{ 10 \ +\ 20}

             a = 9.8 (6/30)

             a = 1.96 m / s²

a) now we can use the kinematic relations

             y = v₀ t + ½ a t²

the blocks come out of rest so their initial velocity is zero

             y = ½ a t²

             y = ½ 1.96 t²

             y = 0.98 t²

for t = 1s y = 0.98 m

       t = 2s y = 1.96 m

b) Time is a scale that is the same for the entire system, the question should be oriented to how far the big block will move.

As the curda is in tension the two blocks must move the same distance

c) the velocity of the block M

           v = vo + a t

           v = 0 + 1.96 t

for t = 1 s v = 1.96 m / s

       t = 2 s v = 3.92 m / s

d) the deceleration if the chain is cut

when removing the chain the tension becomes zero

           -fr = M a

          - μ M g = M a

          a = - μ g

          a = - 0.2 9.8

          a = -1.96 m / s²

e) the distance to stop the block is

         v² = vo² - 2 a x

        0 = vo² - 2a x

        x = vo² / 2a

        x = 1.96² / 2 1.96

        x = 0.98 m

the time to travel this distance is

        v = vo - a t

        t = vo / a

        t = 1.96 /1.96

        t = 1 s

3 0
3 years ago
What were some major reasons for the american television from analog broadcast to digital broadcast?
Vika [28.1K]

Answer: It frees up valuable portions of the broadcast spectrum, it has better audio and picture quality, and there are more options on digital broadcasting

Explanation:

6 0
3 years ago
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svlad2 [7]
Force=mass*acceleration
F=ma
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You are given two unknown point charges, Q1, and Q2. At a point on the line joining them, one third of the way from Q1 to Q2 the
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3 0
3 years ago
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