All categories

3 years ago

Describe what we may observe from Earth regarding the moon’s revolution around the earth and its rotation

1 answer:

4 0

When we look at the moon from the Earth, we always see the same light spots, dark spots, and shapes. It never changes. There could be two possible reasons for this:

-- The moon is a flat disk with some markings on it, and one side of it always faces the Earth.

-- The moon is a round ball with some markings on it, and one side of it always faces the Earth.

Either way, since the same side always faces the Earth, the only way that can happen is if the moon's revolution around the Earth and rotation on its axis both take EXACTLY the same length of time.

Even if they were only one second different, then we would see the moon's whole surface over a long period of time. But we don't. So the moon's rotation and revolution must be EXACTLY locked to the same period of time.

You might be interested in

find the time taken, if the speed of a train increased from 72 km/hr to 90 km/hr for 234 km. leave your answer in seconds

Airida [17]

Answer:

Time taken = 10400 s

Explanation:

Given:

Initial speed of the train, $u=72\textrm{ km/h}=72\times \frac{5}{18}=20\textrm{ m/s}$

Final speed of the train, $v=90\textrm{ km/h}=90\times \frac{5}{18}=25\textrm{ m/s}$

Displacement of the train, $S=234\textrm{ km}=234\times 1000=234000\textrm{ m}$

Using Newton's equation of motion,

$v - u = at\\a=\frac{v-u}{t}$

Now, using Newton's equation of motion for displacement,

$v^{2}-u^{2}=2aS$

Now, plug in the value of $a=\frac{v-u}{t}$ in the above equation. This gives,

$v^{2}-u^{2}=2\times \frac{v-u}{t}\times S\\(v+u)(v-u)=\frac{2(v-u)S}{t}\\t=\frac{2(v-u)S}{(v+u)(v-u)}\\t=\frac{2S}{v+u}$

Now, plug in 234000 m for $S$ , 25 m/s for $v$ and 20 m/s for $u$ . Solve for $t$ .

$t=\frac{2S}{v+u}\\t=\frac{2\times 234000}{25+20}\\t=\frac{468000}{45}=10400\textrm{ s}$

Therefore, the time taken by the train is 10400 s.

3 0

4 years ago

The hubbles telescopes orbit is 5.6 x10 ^5 meters above earths suface. the telescope has a mass os 1.1 x10^4 kilograms. earth ex

Andru [333]

(3) 8.3 N/kg. The gravitational field strength at a point is the force per unit mass exerted on a mass placed at that point. So at the point where the Hubble telescope is, it is (9.1 x 10^4)N/(1.1 x 10^4 kg) = 8.3 N/kg

Fam

7 0

3 years ago

A block is released to slide down a frictionless incline of 15∘ and then it encounters a frictional surface with a coefficient o

Elodia [21]

The block's potential energy at the top of the incline (at a height h from the horizontal surface) is equal to its kinetic energy at the bottom of the incline, so that

mgh = 1/2 mv²

where v is its speed at the bottom of the incline. It follows that

v = √(2gh)

If the incline is 20.4 m long, that means the block has a starting height of

sin(15°) = h/(20.4 m) ⇒ h = (20.4 m) sin(15°) ≈ 5.2799 m

and so the block attains a speed of

v = √(2gh) ≈ 10.1728 m/s

The block then slides to a rest over a distance d. Kinetic friction exerts a magnitude F over this distance and performs an amount of work equal to Fd. By the work-energy theorem, this quantity is equal to the block's change in kinetic energy, so that

Fd = 0 - 1/2 mv² ⇒ d = (-1293.58 J)/F

By Newton's second law, the net vertical force on the block as it slides is

∑ F [vertical] = n - mg = 0

where n is the magnitude of the normal force, so that

n = mg = (25 kg) g = 245 N

and thus the magnitude of friction is

F = -0.16 (245 N) = -39.2 N

(negative since it opposes the block's motion)

Then the block slides a distance of

d = (-1293.58 J) / (-39.2 N) ≈ 32.9994 m ≈ 33 m

5 0

3 years ago