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Korolek [52]
3 years ago
12

Can a falling object reach terminal velocity in outer space?

Physics
2 answers:
yan [13]3 years ago
6 0

Answer: lth air resistance acting on an object that has been dropped, the object will eventually reach a terminal velocity, which is around 53 m/s (195 km/h or 122 mph) for a human skydiver (on the moon, the gravitational acceleration is much less on earth, approximately 1.6 m/s2.)

Explanation:

Triss [41]3 years ago
4 0

Answer:

There is no gravity in outer space so how will any object fall in the first place

Also the principle of terminal velocity is limited to free fall where the acceleration is equal to the gravity of the planet

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g The speed of light in a vacuum is larger for short wavelength electromagnetic waves. larger for high frequency electromagnetic
Orlov [11]

Answer:

A constant value everywhere in the universe.

Explanation:

The speed of light in a vacuum is a constant value. It is not affected by change in frequency or wavelength of the light.

Mathematically the speed of light is given as:

c = λf

where λ = wavelength and f - frequency

The speed of light is the constant of proportionality between frequency and wavelength. In order words, wavelength and frequency are inversely proportional. As the wavelength increases, frequency decreases and vice versa.

While the change in wavelength and frequency of light affect the energy of the light, its speed is a constant value as long as the medium is a vacuum.

The speed of light is also not dependent on the manner with which the light wave is moving.

4 0
3 years ago
What is responsible for all changes in motion? acceleration force changing direction energy
FinnZ [79.3K]
Force is responsible for all changes in motion.
6 0
4 years ago
Read 2 more answers
Consider three force vectors F~ 1 with magnitude 43 N and direction 38◦ , F~ 2 with magnitude 26 N and direction −140◦ , and F~
Rama09 [41]

Answer:

34.70 N

Explanation:

Given :

F~ 1 = 43 N in direction 38◦

F~ 2 = 26 N in direction −140◦

F~ 3 = 27 N in direction 110◦

Therefore,

F~x = 43 cos (38) + 26 cos (-140) + 27 cos (110)

      = 43  (0.7) + 26  (-0.7) + 27  (-0.3)

      =  3.8

F~y = 43 sin (38) + 26 sin (-140) + 27 sin (110)

      = 43  (0.6) + 26  (-0.6) + 27  (0.9)

      = 34.5

so, F~ = $ \sqrt{3.8^2 + 34.5^2}$

          = 34.70 N

6 0
3 years ago
Can you solve a simple physics problem
mart [117]

Answer:

0.41

Explanation:

A 49 N block has a mass of 5 kg:

49/9.8

A horizontal force of 50 N giving it an acceleration of 6 m/s^2 means that 30 N is taken up by the acceleration

This leaves 20 N to take care for friction. so the coefficient of friction is:

20/49 = 0.41.

6 0
3 years ago
PLEASE HELP!! 9TH K12 GCA LIGHT TEST REVIEW SHEET FOR PHYSICS! 100 POINTS!!
Svetllana [295]

I got a 100 on this. I took a bunch of snips of it for you. Mine is a little different though. I only have 6 questions and most of yours are the same as mine. And some of yours that I have are in a different order. I hope I helped.

6 0
4 years ago
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