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USPshnik [31]
4 years ago
15

the planet neptune is approximately 4.5*10^9 kilometers from the sun. The planet Venus is approximately 1.1*10^8 kilometers from

the sun. Which is the best estimate of how many times as far from the sun Neptune is as Venus?
Physics
1 answer:
Firlakuza [10]4 years ago
7 0

Answer:

Neptune is approximately 41 times as far from the sun as Venus

Explanation:

Estimate = distance of Neptune from the sun ÷ distance of Venus from the sun = 4.5×10^9 ÷ 1.18×10^8 = 40.9 (approximately 41)

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two clear, colorless liquids are poured together. A bright yellow solid forms. What physical properties have changed?
jek_recluse [69]
Change in state(from liquid to solid) and change in colour I believe.
3 0
3 years ago
What is the equation of preassure <br><br><br>​
oksano4ka [1.4K]
If u mean pressure, pressure = Force/Area
Correct me if I am wrong :D
8 0
3 years ago
In 1976, the SR-71A, flying at 20 km altitude (T = –56 0C), set the official jet-powered aircraft speed record of 3530 km/hr (21
Lapatulllka [165]

To solve this problem we will apply the concepts related to the calculation of the speed of sound, the calculation of the Mach number and finally the calculation of the temperature at the front stagnation point. We will calculate the speed in international units as well as the temperature. With these values we will calculate the speed of the sound and the number of Mach. Finally we will calculate the temperature at the front stagnation point.

The altitude is,

z = 20km

And the velocity can be written as,

V = 3530km/h (\frac{1000m}{1km})(\frac{1h}{3600s})

V = 980.55m/s

From the properties of standard atmosphere at altitude z = 20km temperature is

T = 216.66K

k = 1.4

R = 287 J/kg

Velocity of sound at this altitude is

a = \sqrt{kRT}

a = \sqrt{(1.4)(287)(216.66)}

a = 295.049m/s

Then the Mach number

Ma = \frac{V}{a}

Ma = \frac{980.55}{296.049}

Ma = 3.312

So front stagnation temperature

T_0 = T(1+\frac{k-1}{2}Ma^2)

T_0 = (216.66)(1+\frac{1.4-1}{2}*3.312^2)

T_0 = 689.87K

Therefore the temperature at its front stagnation point is 689.87K

6 0
4 years ago
A snail crawls at a speed of 0.0004 m/s. How long will ot take to climb a garden stick 1.6m high?​
tekilochka [14]

4000 seconds

Explanation:

speed = distance / time

0.0004m/s = 1.6m / time

Subject time

time = 1.6 / 0.0004

time = 4000 seconds.

Hope this helps. Mark as brainliest if possible. tks

5 0
3 years ago
USE THE PICTURE THANKS WILL GIVE B WRONG ANSWER=report lol love ya
KIM [24]

Answer:

I would have to say B THe su would rise in the west and set in the east But this is just a guess

8 0
3 years ago
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