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3 years ago

Differences Between light year and astronomical unit in two points .

1 answer:

5 0

One astronomical unit (AU) is the average distance from the Earth to the Sun, which is about 93-million miles or about 8.5 light-minutes.

A light-year is the distance light travels in one year in a vacuum, which is about 6-trillion miles. One light-year is about 64,500 times longer than one AU.

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The law of reflection is quite useful for mirrors and other flat, shiny surfaces. (This sort of reflection is called specular re

Sever21 [200]

Answer:

Explanation:

Suppose initially the plane was horizontal and light was reflected back at some angle θ from the normal .

Now the reflecting surface is twisted so that is becomes inclined at angle alpha .

The reflected light will be deviated from its original direction by angle

2 x alpha .

Similarly when the reflecting surface is further twisted so that it becomes inclined at angle beta then again the reflected beam will deviated by angle

2 x beta

Hence angle between these two reflected beam

= 2 beta - 2 alpha

= 2 ( β - α )

So, angular separation between the rays reflected from the two surfaces

= 2 ( β - α ) .

5 0

3 years ago

Suppose the coefficient of static friction between a quarter and the back wall of a rocket car is 0.383. At what minimum rate wo

djverab [1.8K]

Answer:

25.59 m/s²

Explanation:

Using the formula for the force of static friction:

$f_s = \mu_s N$ --- (1)

where;

$f_s =$ static friction force

$\mu_s =$ coefficient of static friction

N = normal force

Also, recall that:

F = mass × acceleration

Similarly, N = mg

here, due to min. acceleration of the car;

$N = ma_{min}$

From equation (1)

$f_s = \mu_s ma_{min}$

However, there is a need to balance the frictional force by using the force due to the car's acceleration between the quarter and the wall of the rocket.

Thus,

$F = f_s$

$mg = \mu_s ma_{min}$

$a_{min} = \dfrac{mg }{ \mu_s m}$

$a_{min} = \dfrac{g }{ \mu_s }$

where;

$\mu_s = 0.383$ and g = 9.8 m/s²

$a_{min} = \dfrac{9.8 \ m/s^2 }{0.383 }$

$\mathbf{a_{min}= 25.59 \ m/s^2}$

3 0

3 years ago

A cyclist rides 6.2 km east, then 9.28 km in a direction 27.27 degrees west of north, then 7.99 km west. A. What is the magnitud

Pani-rosa [81]

Answer:

Explanation:

given,

cyclist ride 6.2 km east and then 9.28 km in the direction of 27.27° west of north and then 7.99 km west.

vertical component = 9.28 cos∅

= 9.28 cos 27.27°

= 8.24 km

horizontal axis component = 9.28 sin ∅

= 9.28 sin 27.27°

= 4.5 km

distance of the final point from the origin

= 7.99 -(6.2-4.5)

= 6.29 km

displacement

$d = \sqrt{6.29^2+8.24^2}$

d = 10.37 km

b) $tan \theta = \dfrac{6.29}{8.24}$

θ = 37.36°

4 0

3 years ago

Mass of water= 357g density water= 1.0g/cm3

aliya0001 [1]

$m=357g\\\\\rho=1.0\ \dfrac{g}{cm^3}\\\\\rho=\dfrac{m}{V}\to V=\dfrac{m}{\rho}\\\\\text{substitute}\\\\V=\dfrac{357g}{1.0\frac{g}{cm^3}}=375g\cdot1.0\dfrac{cm^3}{g}=375\ cm^3$

8 0

3 years ago

Why is a specimen smaller than 200 nm not visible with a light microscope?

kifflom [539]

Because the specimen is very small with a light microscope

8 0

3 years ago