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erma4kov [3.2K]
3 years ago
8

Ground infrared telescopes Versailles at high altitude and dry climate. What explains the location for those infrared telescopes

Physics
1 answer:
Natali5045456 [20]3 years ago
3 0

Answer:

High altitude

Explanation:

Infrared rays are absorbed by water vapor in the lower parts of our atmosphere such as near the sea level.Infrared telescopes positioned on high mountains can observe the cosmos at a wavelength near-infrared. In this case, this telescope is positioned on a mountain top where the altitude is high.

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How energy and power are different ???
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Explanation:

Energy is what makes change happen and can be transferred form one object to another. ... Power is the rate at which energy is transferred. It is not energy but is often confused with energy. The watt is the most commonly used unit of measure for power.

6 0
3 years ago
Which image illustrates the bouncing of a light wave off of a surface?
Naily [24]

Answer:

The answer is A good luck :P

8 0
3 years ago
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Over a period of operation, the useful work output of the fluorescent bulb was
Nadya [2.5K]

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199.0521 Will be the answer

5 0
3 years ago
A shuttle on Earth has a mass of 4.5 E 5 kg. Compare its weight on Earth to its weight while in orbit at a height of 6.3 E 5 met
faltersainse [42]

Answer:

83%

Explanation:

On the surface, the weight is:

W = GMm / R²

where G is the gravitational constant, M is the mass of the Earth, m is the mass of the shuttle, and R is the radius of the Earth.

In orbit, the weight is:

w = GMm / (R+h)²

where h is the height of the shuttle above the surface of the Earth.

The ratio is:

w/W = R² / (R+h)²

w/W = (R / (R+h))²

Given that R = 6.4×10⁶ m and h = 6.3×10⁵ m:

w/W = (6.4×10⁶ / 7.03×10⁶)²

w/W = 0.83

The shuttle in orbit retains 83% of its weight on Earth.

4 0
4 years ago
The total charge a battery can supply is rated in mA⋅h , the product of the current (in mA ) and the time (in h ) that the batte
natita [175]

Answer: 0.2  hours

Explanation: In order to solve this question we have to considerer that a recargeable battery can supply 1800 mA  in one hour then we have to determine how long could this battery drive current through a long, thin wire of resistance 34 Ω .

Besides, this battery has a voltage of 12 V

so by using the Ohm law we also know that V=R*I,

Fron this we can obtain:

I= V/R= 12 V/ 34 Ω=0.35 A= 350 mA

then considering that this battery can supply 1800 mA in one hour we have this battery can supply 350 mA  in x time in the form:

1hour------- 1800 mA

x hour--------350 mA

time= 350/1800= 0.2 hour

4 0
3 years ago
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