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3 years ago

The term angle of deviation is used in reference to

1 answer:

4 0

B. a prism.

This is because the term "angle of deviation" is used to describe the angle at which the light bends compared to the normal of the ray of light

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The escape velocity is defined to be the minimum speed with which an object of mass m must move to escape from the gravitational

s344n2d4d5 [400]

Answer:

v = √2G $M_{earth}$ / R

Explanation:

For this problem we use energy conservation, the energy initiated is potential and kinetic and the final energy is only potential (infinite r)

Eo = K + U = ½ m1 v² - G m1 m2 / r1

Ef = - G m1 m2 / r2

When the body is at a distance R> Re, for the furthest point (r2) let's call it Rinf

Eo = Ef

½ m1v² - G m1 $M_{earth}$ / R = - G m1 $M_{earth}$ / R

v² = 2G $M_{earth}$ (1 / R - 1 / Rinf)

If we do Rinf = infinity 1 / Rinf = 0

v = √2G $M_{earth}$ / R

Ef = = - G m1 m2 / R

The mechanical energy is conserved

Em = -G m1 $M_{earth}$ / R

Em = - G m1 $M_{earth}$ / R

R = int ⇒ Em = 0

6 0

2 years ago

The gravitational force,F, on a rocket at a distance,r, from the center of the earth isgiven byF=kr2wherek= 1013N·km2. (Newton·k

Brrunno [24]

Answer:

The gravitational force changing velocity is

$\frac{dF}{dt}=-8\frac{N}{s}$

Explanation:

The expression for the gravitational force is

$F=\frac{k}{r^{2}}\\\\k=10x10^{13} N*km^{2}\\\\r=10x10^{4} km\\\\V=0.4 \frac{km}{s}$

Differentiate the above equation

$\frac{dF}{dt}=\frac{k}{r^{2}}\\\frac{dF}{dt}=k*r^{-2}\\\frac{dF}{dt}=-2*k*r^{-3} \frac{dr}{dt}\\\frac{dF}{dt}=\frac{-2k}{r^{3}}\frac{dr}{dt}$

The velocity is the distance in at time so

$V=\frac{dr}{dt}=0.4 \frac{km}{s}$

$\frac{dF}{dt}=\frac{-2*k}{r^{3}}*0.4\\\frac{dF}{dt}=\frac{-8*10x^{13}N*km^{2} }{(10x10^{4}) ^{3}} \\\frac{dF}{dt}=\frac{-8x10^{12} }{1x10^{12}}$

$\frac{dF}{dt}=-8\frac{N}{s}$

8 0

3 years ago

Use the drop-down menu to complete the statement. Based on the field lines, the electric charges indicated by the question marks

zheka24 [161]

Answer: The same

Explanation: I just did it on Edg

5 0

3 years ago

Read 2 more answers

Can you plz help me in on 11 12 and 14?

anygoal [31]

I’m pretty sure 14 is mutations

3 0

3 years ago

The speed of an electric locomotive is 90kmh .express this speed in m/s with method

Maksim231197 [3]

Answer:

24.3 m/s

Explanation:

1 kmh = 0.27 m/s, that makes a conversion ratio of 0.27/1kmh

$\frac{90 kmh}{1}$ x $\frac{0.27 m/s}{1 kmh}$

The "kmh" n the top and bottom cancel out. And then you just multiply the top 90 x 0.27 and the bottom 1 x 1 to get

$\frac{24.3 m/s}{1}$

and since its over 1 its just 24.3 m/s

8 0

3 years ago