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EastWind [94]
3 years ago
8

What would earth be like without the moon?

Physics
2 answers:
sleet_krkn [62]3 years ago
8 0
<span>Earth without its moon would be a very different world indeed. No eclipses. Smaller tides. But the biggest change would be in the length of Earth’s day.</span>
maxonik [38]3 years ago
6 0
1.       Nights would be much, much darker. The next brightest object in the night sky is Venus. But it still wouldn't be enough to light up the sky. A full moon is nearly two thousand times brighter than Venus is at its brightest.

2.       Without the moon, a day on earth would only last six to twelve hours. There could be more than a thousand days in one year! That's because the Earth's rotation slows down over time thanks to the gravitational force -- or pull of the moon -- and without it, days would go by in a blink.

3.       A moonless earth would also change the size of ocean tides -- making them about one-third as high as they are now. 

4.       Forget about seeing any lunar eclipses -- or any solar eclipses -- without the moon, there would be nothing to block the sun.

5.       Without a moon the tilt of our earth's axis would vary over time. This could create some very wild weather. Right now, thanks to our moon, our axis stays tilted at twenty-three point five degrees. But without the moon the earth might tilt too far over or hardly tilt at all leading to no seasons or even extreme seasons. 

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According to Kepler, what do all bodies in orbit around another have in common?
dolphi86 [110]
D.
All follow an elliptical path that has two foci, rather than a circular path.

Kepler found paths are elliptical, not circular
3 0
3 years ago
An airplane engine starts from rest; and 2 seconds later, it is rotating with an angular speed of 300 rev/min. if the angular ac
elena-s [515]

First of all, we need to convert the angular speed from rev/min into rev/s:

\omega_f=300 rev/min=5 rev/s

The angular acceleration is the variation of angular speed divided by the time:

\alpha=\frac{\omega_f-\omega_i}{t}=\frac{5 rev/s-0}{2 s}=2.5 rev/s^2

And this is constant, so we can use the following equation to calculate the angle through which the engine has rotated:

\theta(t)=\frac{1}{2}\alpha t^2 =\frac{1}{2}(2.5 rev/s^2)(2 s)^2=5 rev

so, 5 revolutions.

3 0
3 years ago
Our Sun shines bright with a luminosity of 3.828 x 1025 Watt. Her energies
kifflom [539]

Answer:

a)   E = 1.58 10²¹ J , b) Oil = 4,236 107 liter ,  e)   T = 54.3 C

Explanation:

a) To calculate the energy that reaches Earth, let us combine that the power emitted by the Sun is distributed uniformly on a spherical surface

     I = P / A

     A = 4π r²

in this case the radius of the sphere is the distance from the Sun to Earth r = 1.5 10¹¹ m

     I = P / A

     I = P / 4π r²

let's calculate

     I = 3,828 10²⁵/4 pi (1.5 10¹¹)²

     I = 1.3539 10²W / m² = 135.4 W / m2

the energy that reaches the disk of the Earth is

    E = I A

the area of ​​a disc

    A = π r²

    E = I π r²

where r is the radius of the Earth 6.37 10⁶ m

     E = 135.4 π(6.37 10⁶)

     E = 1,726 10¹⁶ W

This is the energy per unit of time that reaches Earth

    t = 1 dai (24h / 1day) (3600s / 1h) = 86400 s

     

    E = 1,826 10¹⁶ 86400

     E = 1.58 10²¹ J

b) for this part we can use a direct proportions rule

      Oil = 1.58 10²¹ (1 / 37.3 10⁶)

      Oil = 4,236 10⁷ liter

c) to silence the surface temperature of the Earth we use the Stefan-Bolztman Law

       P = σ A e T⁴

       T = \sqrt[4]{P/Ae}

nos indicate the refect, therefore the amount of absorbencies

       P_absorbed = 0.7 P

let's calculate

       T = REA (0.7 1.58 1021 / [pi (6.37 106) 2 1)

       T = RER (8,676 106)

       T = 54.3 C

b) Among the other factors that must be taken into account is the greenhouse effect, due to the absorption of gases from the atmosphere

4 0
3 years ago
Which planet weighs twice as much as all other planets.
Bumek [7]

Answer:

Jupiter. Although it is a gas planet, it has a ton of mass. Mass can determine the weight of an object. We can thereby assume that Jupiter weighs way more than most planets.

8 0
2 years ago
The force needed to keep a car from skidding on a curve varies inversely as the radius of the curve and jointly as the weight of
motikmotik

Explanation:

It is given that, the force needed to keep a car from skidding on a curve varies inversely as the radius of the curve and jointly as the weight of the car and the square of the car's speed such that,

F\propto \dfrac{mgv^2}{r}

F=\dfrac{kmgv^2}{r}

mg is the weight of the car

r is the radius of the curve

v is the speed of the car

Case 1.

F = 640 pounds

Weight of the car, W = mg = 2600 pound

Radius of the curve, r = 650 ft

Speed of the car, v = 40 mph

640=\dfrac{k(2600)(40)^2}{650}

k = 0.1

Case 2.

Radius of the curve, r = 750 ft

Speed of the car, v = 30 mph

F=\dfrac{0.1\times 2600\times (30)^2}{750}

F = 312 N

Hence, this is the required solution.

6 0
3 years ago
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