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2 years ago

11

Why are the largest visible-light telescopes in the world made with mirrors rather than lenses?

1 answer:

malfutka [58]2 years ago

8 0

Light distorts through lenses more then it does on mirrors. Images are harder to make out and mirrors are cheaper then all the glass they need for the lenses especially because of contorting the lenses.

Hope this helps!

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When given a compound, adding up all of the _______ of all of the individual atoms within one molecule of a compound will determ

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Atomic masses should be about right

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2 years ago

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Suppose that a comet that was seen in 550 A.D. by Chinese astronomers was spotted again in year 1941. Assume the time between ob

pickupchik [31]

To solve the problem it is necessary to apply the concepts related to Kepler's third law as well as the calculation of distances in orbits with eccentricities.

Kepler's third law tells us that

$T^2 = \frac{4\pi^2}{GM}a^3$

Where

T= Period

G= Gravitational constant

M = Mass of the sun

a= The semimajor axis of the comet's orbit

The period in years would be given by

$T= 1941-550\\T= 1391y(\frac{31536000s}{1y})\\T=4.3866*10^{10}s$

PART A) Replacing the values to find a, we have

$a^3= \frac{T^2 GM}{4\pi^2}$

$a^3 = \frac{(4.3866*10^{10})^2(6.67*10^{-11})(1.989*10^{30})}{4\pi^2}$

$a^3 = 6.46632*10^{39}$

$a = 1.86303*10^{13}m$

Therefore the semimajor axis is $1.86303*10^{13}m$

PART B) If the semi-major axis a and the eccentricity e of an orbit are known, then the periapsis and apoapsis distances can be calculated by

$R = a(1-e)$

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$R= 5.58*10^{10}m$

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3 years ago

As a sample's temperature increases, which two factors also increase?

mote1985 [20]

Particle kinetic energy and particle speed

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2 years ago

Jake lifted a 10-kg bell by a distance of 0.50 meters in 0.50 seconds. How much power did Jake use?

Reil [10]

Answer:

10 watts

Explanation:

first calculate work.

Work =force×distance cos thita

10Kg×0.50M cos 0= 5joules

Therefore, Power=Work÷ Time

Therefore, 5joules÷0.50s=10 watts.

8 0

2 years ago

When a car of mass 1167 kg accelerates from 10.0 m/s to some final speed, 4.00  10 5J of work are done. Find this final speed.

Akimi4 [234]

Mass=1167kg
Initial velocity=u=10m/s
Acceleration=a=4m/s^2
Work done=105J=W
Final velocity=v=?
Force=F
Distance=d

Apply Newton's second law

$\\ \tt\hookrightarrow F=ma$

$\\ \tt\hookrightarrow F=1167(4)=4668N$

Now

$\\ \tt\hookrightarrow W=Fd$

$\\ \tt\hookrightarrow d=\dfrac{W}{F}$

$\\ \tt\hookrightarrow d=\dfrac{105}{4668}$

$\\ \tt\hookrightarrow d=0.022m$

Now

d be s

According to third equation of kinematics

$\\ \tt\hookrightarrow v^2=u^2+2as$

$\\ \tt\hookrightarrow v^2=10^2+2(4)(0.022)$

$\\ \tt\hookrightarrow v^2=100+8(0.022)$

$\\ \tt\hookrightarrow v^2=100+0.176$

$\\ \tt\hookrightarrow v^2=100.176$

$\\ \tt\hookrightarrow v=10.001m/s$

8 0

2 years ago