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Yakvenalex [24]
3 years ago
11

Which of the following represents an image that is located behind a mirror?

Physics
2 answers:
9966 [12]3 years ago
3 0

Answer:

i think its -di........

Marina CMI [18]3 years ago
3 0

Explanation:

There are some conventions while solving the problems based on mirrors. All parameters are taken like x-y coordinate system.

All the measurements are done from the optical centre of the mirror.

d_i is the image distance

d_o is the object distance

-d_i is the image distance which is formed behind the mirror.

-d_o is the object distance when an object is placed behind the mirror.

So, -d_i shows the image that is located behind the mirror. Hence, this is the required solution.

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julia-pushkina [17]

Answer:

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4 0
3 years ago
Read 2 more answers
Dejamos caer un objeto desde lo alto de una torre y medimos el tiempo que tarda en llegar al suelo que resulta ser de 0,02 minut
Harman [31]

Answer: a) 11.76 m/s  b) 7.056 m

Explanation:

The described situation is as follows:

An object is dropped from the top of a tower and when measuring the time it takes to reach the ground that turns out to be 0.02 minutes.

This situation is related to free fall, this also means we have constant acceleration, hence the equations we will use are:

V_{f}=V_{o}+at (1)  

{V_{f}}^{2}={V_{o}}^{2}+2ad (2)  

Where:  

V_{f} Is the final velocity of the object

V_{o}=0 Is the initial velocity of the object (it was dropped)

a=9.8 m/s^{2} is the acceleration due gravity

d is the height of the tower

t=0.02min=1.2 s is the time it takes to the object to reach the ground

b) Begining with (1):

V_{f}=0+at (3)  

V_{f}=at=(9.8 m/s^{2})(1.2 s) (4)  

V_{f}=11.76 m/s (5)  This is the final velocity of the object

a) Substituting (5) in (2):

(11.76 m/s)^{2}=0+2(9.8 m/s^{2})d (6)  

Clearing d:

d=\frac{(11.76 m/s)^{2}}{2(9.8 m/s^{2})} (7)  

d=7.056 m (8)  This is the height of the tower

4 0
3 years ago
A charged cloud system produces an electric field in the air near earth's surface. a particle of charge -2.0 × 10-9 c is acted o
defon

Part a)

Magnitude of electric field is given by force per unit charge

E = \frac{F}{q}

E = \frac{4.3 * 10^{-6}}{2 * 10^{-9}}

E = 2150 N/C

Part b)

Electrostatic force on the proton is given as

F = qE

F = 1.6 * 10^{-19} * 2150

F = 3.44 * 10^{-16} N

PART C)

Gravitational force is given by

F_g = mg

F_g = 1.6 * 10^{-27}*9.8

F_g = 1.57 * 10^{-26} N

PART d)

Ratio of electric force to weight

\frac{F_e}{F_g} = \frac{3.44 * 10^{-16}}{1.57*10^{-26}}

\frac{F_e}{F_g} = 2.2 * 10^{10}

7 0
3 years ago
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