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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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A block on a horizontal frictionless plane is attached to a spring, as shown below. The block oscillates along the x-axis with s
AleksandrR [38]

The question is about unclear since no picture provided. But from the question, it could be guessed that the box is moving back and forth on the frictionless plane at the amplitude of A in simple harmonic motion.

Answer:

D. At x=0, it's acceleration is at a maximum

Explanation:

As the box move forward, it reaches point A and than move backward. Theoretically, the box will move backwards, through its origin, to point -A and then going forward.

Point A is the maximum displacement of the box in this case. At this point, the box instantaneously stop to go backward. Therefore the velocity at that moment is zero.

From point -A, the box travel forward and keep building up speed due to the release in potential energy of the spring. And at point x=0, the velocity become maximum. After point x=0, the velocity of the box slows down due to the conversion of kinetic energy to potential energy of the spring. And as it reaches point A, it reaches zero velocity.

The same can be said as the box travels backward from point A to -A

8 0
3 years ago
In general, how do you find the average velocity of any object falling in a vacuum?
irina [24]
In general, how do you find the average velocity of any object falling in a vacuum? (Assume you know the final velocity.) Multiply the final velocity by final time. 3. Calculate : Distance, average velocity, and time are related by the equation, d = v • t A
5 0
2 years ago
What is the difference between heat and temperature
Tasya [4]
Temperature is the measurement system we use to measure the heat
7 0
2 years ago
Two boats leave the same port at the same time, with boat A traveling north at 15 knots (nautical miles per hour) and boat B tra
Mrrafil [7]

Answer:

The chance in distance is 25 knots

Explanation:

The distance between the two particles is given by:

s^2 = (x_A - x_B)^2+(y_A - y_B)^2  (1)

Since A is traveling north and B is traveling east we can say that their displacement vector are perpendicular and therefore (1) transformed as:

s^2 = x_B^2+y_A^2 (2)

Taking the differential with respect to time:

\displaystyle{2s\frac{ds}{dt}= 2x_B\frac{dx_B}{dt}+2y_A\frac{dy_A}{dt}}  (3)

where \displaystyle{\frac{dx_B}{dt}}=v_B and \displaystyle{\frac{dx_A}{dt}}=v_A are the respective given velocities of the boats. To find s and x_B we make use of the given position for A, y_A=30, the Pythagoras theorem and the relation between distance and velocity for a movement with constant velocity.

\displaystyle{y_A = v_A\cdot t\rightarrow t = \frac{y_A}{v_A}=\frac{30}{15}=2 h

with this time, we know can now calculate the distance at which B is:

\displaystyle{x_B = v_B\cdot t= 20 \cdot 2 = 40\ nmi

and applying Pythagoras:

\displaystyle{s = \sqrt{x_B^2+y_A^2}=\sqrt{30^2 + 40^2}=\sqrt{2500}=50}

Now substituting all the values in (3) and solving for  \displaystyle{\frac{ds}{dt} } we get:

\displaystyle{\frac{ds}{dt} = \frac{1}{2s}(2x_B\frac{dx_B}{dt}+2y_A\frac{dy_A}{dt})}\\\displaystyle{\frac{ds}{dt} = 25 \ knots}

4 0
3 years ago
A cheetah can run 113 km/h in short busts. How far can a cheetah run in 0.25 hours (15 minutes)?
Fofino [41]
The cheetah can run 28,25 km
4 0
2 years ago
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