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Lynna [10]
3 years ago
12

A star moving away emits each wave farther,stretching them. longer waves are ____.

Physics
1 answer:
Snezhnost [94]3 years ago
6 0
I think the answer you're looking for is answer B) Not there. Good luck and hoped it helped!
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As Earth moves around the Sun, its _____ is approximately perpendicular to the force of gravity exerted by the Sun.
Bogdan [553]

SORRY i forgot but i think its A because the sun has a really strong gravitational pull


sorry and hope it helps

4 0
3 years ago
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Plate tectonics is the scientific theory explaining the movement of the earths crust. The movement of these tectonic plates is l
zzz [600]

Answer:

Continental drift theory describes the long term effect of plate tectonics.

Explanation:

The long term result of plate tectonic movement is the continental drift. The continents of Earth lay on tectonic plates, that are in motion and interaction via plate tectonics. The drift of the Earths continent is an ongoing process evident in the rift valleys and seafloor spreading zones.

The theory that  the Earth's continents are dynamic and have drifted relative to each other is known as continental drift which correlates with the theory of plate tectonics.

Every year, the Earth's outer shell plates are displaced by a small amount due to the heat coming from the Earths interior via convection currents.

7 0
4 years ago
Which picture represents 2NO2?
wariber [46]
2NO2 means that there is 2 oxygen atoms and one nitrogen with two sets of that. So its the third one

3 0
4 years ago
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An engineer is designing a runway. She knows that a plane, starting at rest, needs to reach a speed of 180mph at take-off. If th
kvv77 [185]

Answer:

The plane would need to travel at least 8,\!580\; {\rm ft} (8.58 \times 10^{3}\; {\rm ft}.)

The 10,\!000\; {\rm ft} runway should be sufficient.

Explanation:

Convert unit of the the take-off velocity of this plane to \rm ft\cdot s^{-1}:

\begin{aligned}v &= 180\; {\rm mph} \\ &= 180\; {\rm mi \cdot hrs^{-1}} \times \frac{1\; {\rm hrs}}{3600\; {\rm s}} \times \frac{5280\; {\rm ft}}{1\; {\rm mi}} \\ &= 264\; {\rm ft \cdot s^{-1}}\end{aligned}.

Initial velocity of the plane: u = 0\; {\rm ft \cdot s^{-1}}.

Take-off velocity of the plane v =264\; {\rm ft\cdot s^{-1}}.

Let x denote the distance that the plane travelled along the runway. Since acceleration is constant but unknown, make use of the SUVAT equation x = ((u + v) / 2) \, t.

Notice that this equation does not require the value of acceleration. Rather, this equation make use of the fact that the distance travelled (under constant acceleration) is equal to duration t times average velocity (u + v) / 2.

The distance that the plane need to cover would be:

\begin{aligned}x &= \left(\frac{u + v}{2}\right)\, t \\ &= \frac{0\; {\rm ft \cdot s^{-1}} + 264\; {\rm ft \cdot s^{-1}}}{2} \times 65.0\; {\rm s} \\ &= 8.58\times 10^{3}\; {\rm ft}\end{aligned}.

4 0
3 years ago
14. Convert 22 degrees celsius to fahrenheit.<br> Please show your work
forsale [732]
(22°C × 9/5) + 32 = 71.6°F
6 0
3 years ago
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