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3 years ago

13

two plane mirrors inclined at an angle of 50 degree to each othe r a ray is incident on mirror1 then thedeviation x of ray after

two reflections is

1 answer:

Kipish [7]3 years ago

6 0

The collapsed answer of Penchalreddy Badepalli is correct. The composition of two reflections via two mirror making an angle \alpha is equivalent to a single rotation by an angle 2\alpha, hence 2 * 60 deg = 120 deg. And turns is independent of the absolute orientation of the two mirrors in space and/or the direction of incidence of the incoming ray.

One could use elementary geometry to prove this (if you presume the direction of incidence is irrelevant imagine hitting the first mirror at 90 deg, then going retro right back along the normal to the first mirror, and follow the directions).

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. Consider the equation =0+0+02/2+03/6+04/24+5/120, where s is a length and t is a time. What are the dimensions and SI units of

Olegator [25]

Answer:

See Explanation

Explanation:

Given

$s=s_0+v_0t+\frac{a_0t^2}{2}+ \frac{j_0t^3}{6}+\frac{S_0t^4}{24}+\frac{ct^5}{120}$

Solving (a): Units and dimension of $s_0$

From the question, we understand that:

$s \to L$ --- length

$t \to T$ --- time

Remove the other terms of the equation, we have:

$s=s_0$

Rewrite as:

$s_0=s$

This implies that $s_0$ has the same unit and dimension as $s$

Hence:

$s_0 \to L$ --- dimension

$s_o \to$ Length (meters, kilometers, etc.)

Solving (b): Units and dimension of $v_0$

Remove the other terms of the equation, we have:

$s=v_0t$

Rewrite as:

$v_0t = s$

Make $v_0$ the subject

$v_0 = \frac{s}{t}$

Replace s and t with their units

$v_0 = \frac{L}{T}$

$v_0 = LT^{-1}$

Hence:

$v_0 \to LT^{-1}$ --- dimension

$v_0 \to$ $m/s$ --- unit

Solving (c): Units and dimension of $a_0$

Remove the other terms of the equation, we have:

$s=\frac{a_0t^2}{2}$

Rewrite as:

$\frac{a_0t^2}{2} = s_0$

Make $a_0$ the subject

$a_0 = \frac{2s_0}{t^2}$

Replace s and t with their units [ignore all constants]

$a_0 = \frac{L}{T^2}\\$

$a_0 = LT^{-2$

Hence:

$a_0 = LT^{-2$ --- dimension

$a_0 \to$ $m/s^2$ --- acceleration

Solving (d): Units and dimension of $j_0$

Remove the other terms of the equation, we have:

$s=\frac{j_0t^3}{6}$

Rewrite as:

$\frac{j_0t^3}{6} = s$

Make $j_0$ the subject

$j_0 = \frac{6s}{t^3}$

Replace s and t with their units [Ignore all constants]

$j_0 = \frac{L}{T^3}$

$j_0 = LT^{-3}$

Hence:

$j_0 = LT^{-3}$ --- dimension

$j_0 \to$ $m/s^3$ --- unit

Solving (e): Units and dimension of $s_0$

Remove the other terms of the equation, we have:

$s=\frac{S_0t^4}{24}$

Rewrite as:

$\frac{S_0t^4}{24} = s$

Make $S_0$ the subject

$S_0 = \frac{24s}{t^4}$

Replace s and t with their units [ignore all constants]

$S_0 = \frac{L}{T^4}$

$S_0 = LT^{-4$

Hence:

$S_0 = LT^{-4$ --- dimension

$S_0 \to$ $m/s^4$ --- unit

Solving (e): Units and dimension of $c$

Ignore other terms of the equation, we have:

$s=\frac{ct^5}{120}$

Rewrite as:

$\frac{ct^5}{120} = s$

Make $c$ the subject

$c = \frac{120s}{t^5}$

Replace s and t with their units [Ignore all constants]

$c = \frac{L}{T^5}$

$c = LT^{-5}$

Hence:

$c \to LT^{-5}$ --- dimension

$c \to m/s^5$ --- units

4 0

3 years ago

A screwdriver is used to pry the lid off a paint can. The resistance arm is 0.5 cm long and the effort arm is 20 cm long. What i

Lunna [17]

Answer:

40

Explanation:

Mechanical advantage = effort arm / load arm

MA = 20 cm / 0.5 cm

MA = 40

4 0

3 years ago

What do you mean by population size explain

torisob [31]

Population size is the actual number of individuals in a population. Population density is a measurement of population size per unit area, i.e., population size divided by total land area. Abundance refers to the relative representation of a species in a particular ecosystem.

8 0

3 years ago

They realize there is a thin film of oil on the surface of the puddle. If the index of refraction of the oil is 1.81, and they o

Sphinxa [80]

Answer:

The right solution is "165.8 nm".

Explanation:

Given:

Index of refraction,

n = 1.81

Wavelength,

λ = 600 nm

We know that,

⇒ $t=\frac{\lambda}{2\times n}$

By putting the values, we get

$=\frac{600}{2\times 1.81}$

$=165.8 \ nm$

3 0

3 years ago

Apparent brightness depends on which two properties?

Serjik [45]

Answer:

I believe it is luminosity and distance

Explanation:

So B

6 0

3 years ago

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