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3 years ago

If angle between two plane mirrors is 120 degrees then how many images were formed

Physics

1 answer:

Goryan [66]3 years ago

5 0

<h2>Answer:3</h2>

Explanation:

The number of images formed by a pair mirrors kept at an angle $\alpha$ is

$n=\frac{360}{\alpha } -1$ if $\frac{360}{\alpha }$ comes even.

$n=\frac{360}{\alpha }$ if $\frac{360}{\alpha }$ comes odd.

where $n$ is the number of images formed.

In our question,

$\alpha =120^{0}$

$\frac{360}{\alpha } =\frac{360}{120}=3$

It is odd,so $n=\frac{360}{120}=3$

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The units we use to measure the sound intensity level are _____.

Sergio039 [100]

Answer:

decibels (dB)

Explanation:

The sound intensity level is a quantity derived from the sound intensity.

The intensity of a wave is defined as the power of the source of the wave divided by the area through which the power of the wave is spread, mathematically:

$I=\frac{P}{A}$

where

P is the power of the source

$A=4\pi r^2$ is the surface area over which the wave spreads (assuming that the wave propagates in all directions, it corresponds to the surface area of a sphere of radius $r$ , where r is the distance between the source of the wave and the observer)

For sound waves, the intensity is often expressed using another unit, called decibel (dB), defined as follows:

$\beta(dB)=10Log_{10}(\frac{I}{I_0})$

where

$\beta$ is the sound intensity level in decibels

I is the intensity of the sound wave

$I_0=1\cdot 10^{-12} W/m^2$ is the threshold intensity of a sound that a person can normally hear.

3 0

3 years ago

A test rocket is launched vertically from ground level (y=0), at time t=0.0s. The rocket engine provides constant upward acceler

Luden [163]

From the definition of average velocity,

$\bar v=\dfrac{\Delta y}{\Delta t}=\dfrac{49\,\mathrm m}t$ ,

and the fact that constant acceleration means

$\bar v=\dfrac{v_f+v_0}2=\dfrac{30\,\frac{\mathrm m}{\mathrm s}}2=15\,\dfrac{\mathrm m}{\mathrm s}$

we can solve for the time $t$ :

$15\,\dfrac{\mathrm m}{\mathrm s}=\dfrac{49\,\mathrm m}t\implies t=3.3\,\mathrm s$

7 0

3 years ago

A bucket that has a mass of 20 kg when filled with sand needs to be lifted to the top of a 15 meter tall building. You have a ro

prohojiy [21]

Answer:

work done lifting the bucket (sand and rope) to the top of the building,

W=67.46 Nm

Explanation:

in this question we have given

mass of bucket=20kg

mass of rope= $.2\frac{kg}{m}$

height of building= 15 meter

We have to find the work done lifting the bucket (sand and rope) to the building =work done in lifting the rope + work done in lifting the sand

work done in lifting the rope is given as, $W_{1}=Force \times displacement$

= $\int\limits^{15}_0 {.2x} \, dx$ ..............(1)

= $.1\times 15^2$

=22.5 Nm

work done in lifting the sand is given as, $W_{2}=Force \times displacement$

$W_{2}=\int\limits^{15}_0 F \, dx$ .................(2)

Here,

F=mx+c

here,

c=20-18

c=2

m= $\frac{20-18}{15-0}$

m=.133

Therefore,

$F=.133x+2$

Put value of F in equation 2

$W_{2}=\int\limits^{15}_0 (.133x+2) \, dx$

$W_{2}=.133 \times 112.5+2\times15\\W_{2}=14.96+30\\W_{2}=44.96 Nm$

Therefore,

work done lifting the bucket (sand and rope) to the top of the building, $W=W_{1}+W_{2}$

W=22.5 Nm+44.96 Nm

W=67.46 Nm

4 0

4 years ago

What is your interpretation of 'nothing'? (Talking astrophysics) More answers the better!

Sladkaya [172]

MY personal interpretation of nothing is no atoms or particles of anything. but keep in mind im 11 <span />

5 0

4 years ago

Plz i need help for the 5 problems. plz show the work!!!

Artemon [7]

Answer:

1. 3 $m/s^{2}$

2. 1.5 $m/s^{2}$

3. 3 seconds

4. 0 $m/s^{2}$

5. 2.2 seconds

Explanation:

(1)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making a the subject we have

$a=\frac {v-u}{t}$

Substituting u=0 since it’s at rest, v=30m/s and t=10 seconds

a = $\frac {30-0}{10}=3 m/s^{2}$

(2)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making a the subject we have

$a=\frac {v-u}{t}$

Substituting u=10m/s, v=22m/s and t=8 seconds

a = $\frac {22-10}{8}=1.5 m/s^{2}$

(3)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making t the subject we have

$t=\frac {v-u}{a}$

Substituting u=0m/s since at rest, v=15m/s and a=5 $\frac {m}{s^{2}}$

= $\frac {15-0}{5}=3s$

(4)

When initial and final velocity are constant, there’s no acceleration as proven below

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making a the subject we have

$a=\frac {v-u}{t}$

Substituting u=20 since it’s at rest, v=20m/s and t=10 seconds

a = $\frac {20-20}{10}=0 m/s^{2}$

(5)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making t the subject we have

$t=\frac {v-u}{a}$

Substituting u=9m/s since at rest, v=0m/s and a=-4.1 $\frac {m}{s^{2}}$

= $\frac {0-9}{-4.1}=2.2s$

8 0

3 years ago