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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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Determine the ratio of the flow rate through capillary tubes A and B (that is, Qa/Qb).
Oliga [24]

To solve this problem we can use the concepts related to the change of flow of a fluid within a tube, which is without a rubuleous movement and therefore has a laminar fluid.

It is sometimes called Poiseuille’s law for laminar flow, or simply Poiseuille’s law.

The mathematical equation that expresses this concept is

\dot{Q} = \frac{\pi r^4 (P_2-P_1)}{8\eta L}

Where

P = Pressure at each point

r = Radius

\eta = Viscosity

l = Length

Of all these variables we have so much that the change in pressure and viscosity remains constant so the ratio between the two flows would be

\frac{\dot{Q_A}}{\dot{Q_B}} = \frac{r_A^4}{r_B^4}\frac{L_B}{L_A}

From the problem two terms are given

R_A = \frac{R_B}{2}

L_A = 2L_B

Replacing we have to

\frac{\dot{Q_A}}{\dot{Q_B}} = \frac{r_A^4}{r_B^4}\frac{L_B}{L_A}

\frac{\dot{Q_A}}{\dot{Q_B}} = \frac{r_B^4}{16*r_B^4}\frac{L_B}{2*L_B}

\frac{\dot{Q_A}}{\dot{Q_B}} = \frac{1}{32}

Therefore the ratio of the flow rate through capillary tubes A and B is 1/32

6 0
3 years ago
A rocket, weighing 4.36 x 10^4 N, has an engine that provides an upward force of 8.90 x 10^5 N. It reaches a maximum speed of 86
Pavel [41]

The net force on the rocket is  846400 N.

Answer:

Explanation:

It is known that weight is the influence of gravitational force acting on any mass of the object. So in the present case, the weighing force given is equal to the gravitational force acting on the rocket. Thus, the gravitational force will be acting towards downward direction. But an upward force is required by the rocket for thrusting purpose and that force is given as upward force. So the net force acting on the rocket is the vector addition of all the forces acting on the rocket. As in this case, only upward and downward force is acting on the rocket. The vector addition will be equal to subtraction of downward acting gravitational force from upward force or force provided by engine.

Net force = Engine force - Gravitational force = 890000-43600=846400 N

So the net force acting on the rocket is 846400 N.

8 0
3 years ago
List three devices that use electric current from batteries and three that use regular house current.
Bond [772]
Charger
counsels
TV

Fan
light
car
4 0
3 years ago
Which of the following is correct definition of electrical energy
tatyana61 [14]

Answer:

Electrical energy. Jump to navigation Jump to search. Electrical energy is energy derived from electric potential energy or kinetic energy. When used loosely, electrical energy refers to energy that has been converted from electric potential energy

Explanation:

6 0
3 years ago
Assume the earth to be a nonrotating sphere with mass MEME and radius RERE. If an astronaut weights WW on the ground, what is hi
Solnce55 [7]

Answer:

The weight at a distance 2 RE from surface of earth is <em>W/9</em>

Explanation:

For the value of acceleration due to gravity (g), we have a formula, that is:

g = (G)(ME)/(RE)²    ----- equation (1)

where,

G = Gravitational Constant

ME = Mass of Earth

RE = Radius of Earth

g = Acceleration due to gravity on surface of earth = 9.8 ms²

When the person goes 2RE, distance above earth's surface. Then the total distance from center of earth becomes: 2RE + RE = 3RE.

Therefore, equation (1) becomes:

gh = (G)(ME)/(3RE)²

where,

gh = acceleration due to gravity at height

gh = (G)(ME)/(RE)²9

using equation (1), we get:

gh = g/9

Now, he weight is given by formula:

W = mg   ------- equation (2)

At height 2RE

Wh = (m)(gh)

where,

Wh = Weight at height = ?

m = mass of astronaut

Therefore, using vale of gh, we get:

Wh = mg/9

Using equation (2), we get:

<u>Wh = W/9</u>

6 0
3 years ago
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