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2 years ago

What is the largest-aperture Earth-based telescope currently in use at visible wavelengths?

Physics

1 answer:

Svetlanka [38]2 years ago

3 0

Answer:

Gran Telescopio Canarias

Explanation:

There are numerous giant telescopes all over the world which give us amazingly details data of the Universe. These work in various wavelengths. Talking about the telescopes which operates in visible wavelengths, the largest operational telescope will be Gran Telescopio Canarias. This telescope has an aperture of 10.4m. It is situated in Canary Island, Spain.

In future larger telescopes such as Thirty Meter Telescope and EELT will will make it look like a small telescope.

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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

A 12 N force is used to pull a wagon 10 m to the left in 15 seconds. Which of the following represents the magnitude of the forc

Lana71 [14]

1. - 12N

2.- 2 hours

3. - 0.5 m/s/s

7 0

3 years ago

Read 2 more answers

Why is MA always less than VR?

jarptica [38.1K]

Answer:

$\boxed{ \bold{ \boxed{ \sf{see \: below}}}}$

Explanation:

MA ( Mechanical Advantage ) is always less than VR ( Velocity Ratio ) because MA is reduced by friction but VR is not affected by friction.

Hope I helped!

Best regards! :D

6 0

2 years ago

Calculate the specific heat at constant volume of water vapor, assuming the nonlinear triatomic molecule has three translational

vampirchik [111]

Answer:

I) $c=1385.667\frac{J}{kg K}$

II)The difference from the value obtained on part I is: 2000-1385.67 =614.33 $\frac{J}{Kg K}$

The possible reason of this difference is that the vibrational motion can increase the value, since if we take in count this factor we will have a higher heat capacity, because molecules with vibrational motion require more heat to vibrate and necessary higher specific heat capacity.

Explanation:

From the problem we have the molar mass given $M=18\frac{gr}{mol}$ of water vapor and at constant volume condition. It's important to say that the vapour molecules have 3 transitionsl and 3 rotational degrees of freedom and the rotational motion no contribution.

Part I

Calculate the specific heat at constant volume of water vapor, assuming the nonlinear triatomic molecule has three translational and three rotational degrees of freedom and that vibrational motion does not contribute. The molar mass of water is 18.0 g/mol=0.018kg/mol.

Let $C_v (\frac{J}{Kg K})$ the molar heat capacity at constant volume and this amount represent the quantity of heat absorbed by mole.

Let $C (\frac{J}{Kg K})$ the specific heat capcity this value represent the heat capacity aboserbed by mass.

For the problem we have a total of 6 degrees of freedom and from the thoery we know that for each degree of freedom the molar heat capacity at constant volume is given by $C_v =\frac{R}{2}$ so the total for the 6 degrees of freedom would be:

$C_v =6*\frac{R}{2}=3R=3x8.314\frac{J}{mol K}=24.942\frac{J}{mol K}$

And by definition we know that the specific heat capacity is defined:

$c=\frac{C_V}{M}$

If we replace all the values we have:

$c=\frac{24.942\frac{J}{mol K}}{0.018\frac{kg}{mol}}=1385.667\frac{J}{kg K}$

So on this case the specific heat capacity with constant volume and with three translational and three rotational degrees of freedom is $c=1385.667\frac{J}{kg K}$

Part II

The actual specific heat of water vapor at low pressures is about 2000 J/(kg * K). Compare this with your calculation.

The difference from the value obtained on part I is: 2000-1385.67 =614.33 $\frac{J}{Kg K}$

4 0

3 years ago

If the caffeine concentration in a particular brand of soda is 2.97 mg/oz, 2.97 mg/oz, drinking how many cans of soda would be l

Ray Of Light [21]

Explanation:

The given data is as follows.

Concentration of caffeine = 2.97 mg/oz

Number of oz in a can = 12 oz

Therefore, the concentration of caffeine in one can is calculated as follows.