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AlladinOne [14]
3 years ago
15

an object that is 8 cm tall is located 2.5 m in front of a plane mirror. The image formed by the mirror appears to be. Find di a

nd whether behind or in front of mirror​
Physics
1 answer:
UkoKoshka [18]3 years ago
6 0

The image is at 2.5 m behind the mirror (virtual image)

Explanation:

We can solve the problem by using the mirror equation:

\frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i}

where

f is the focal length of the mirror

d_o is the distance between the object and the mirror

d_i is the distance between the image and the mirror

For a plane mirror, the focal length is infinite:

f=\infty

And in this problem, the distance of the object from the mirror is

d_o = 2.5 m

Substituting and solving for d_i, we find the location of the image:

0=\frac{1}{d_o}+\frac{1}{d_i}\\d_i = -d_o = -2.5 m

And the negative sign means that the image is located behind the mirror, so it is virtual.

Learn more about reflection and refraction:

brainly.com/question/3183125

brainly.com/question/12370040

#LearnwithBrainly

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The charges of two particles are as follows: Q1=2 x 10 -8 C and Q2 = 3 x 10 -7 C. Find the magnitude of the force between these
Fantom [35]

Answer:

F = 3.86 x 10⁻⁶ N

Explanation:

First, we will find the distance between the two particles:

r = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2+(z_{2}-z_{1})^2}\\

where,

r = distance between the particles = ?

(x₁, y₁, z₁) = (2, 5, 1)

(x₂, y₂, z₂) = (3, 2, 3)

Therefore,

r = \sqrt{(3-2)^2+(2-5)^2+(3-1)^2}\\r = 3.741\ m\\

Now, we will calculate the magnitude of the force between the charges by using Coulomb's Law:

F = \frac{kq_{1}q_{2}}{r^2}\\

where,

F = magnitude of force = ?

k = Coulomb's Constant = 9 x 10⁹ Nm²/C²

q₁ = magnitude of first charge = 2 x 10⁻⁸ C

q₂ = magnitude of second charge = 3 x 10⁻⁷ C

r = distance between the charges = 3.741 m

Therefore,

F = \frac{(9\ x\ 10^9\ Nm^2/C^2)(2\ x\ 10^{-8}\ C)(3\ x\ 10^{-7}\ C)}{(3.741\ m)^2}\\

<u>F = 3.86 x 10⁻⁶ N</u>

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3 years ago
Two asteroids identical to those above collide at right angles and stick together; i.e, their initial velocities were perpendicu
11111nata11111 [884]

Answer:

velocity = 62.89 m/s  in 58 degree measured from the x-axis

Explanation:

Relevant information:

Before the collision, asteroid A of mass 1,000 kg moved at 100 m/s, and asteroid B of mass 2,000 kg moved at 80 m/s.

Two asteroids moving with velocities collide at right angles and stick together. Asteroid A initially moving to right direction and asteroid B initially move in the upward direction.

Before collision Momentum of A = 1000 x 100 = $ 10^5$ kg - m/s in the right direction.

Before collision Momentum of B = 2000 x 80 = 1.6 x $ 10^5$  kg - m/s in upward direction.

Mass of System of after collision = 1000 + 2000 = 3000 kg

Now applying the Momentum Conservation, we get

Initial momentum in right direction = final momentum in right direction = $ 10^5$

And, Initial momentum in upward direction = Final momentum in upward direction = 1.6 x $ 10^5$

So, $ V_x = \frac{10^5}{3000} $  = $ \frac{100}{3} $  m/s

and $ V_y=\frac{160}{3}$  m/s

Therefore, velocity is = $ \sqrt{V_x^2 + V_y^2} $

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And direction is

tan θ = $ \frac{V_y}{V_x}$     = 1.6

therefore, $ \theta = \tan^{-1}1.6 $

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