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2 years ago

If planet A is three times as far from planet C, then the period of its orbit will be __ times as long

1 answer:

6 0

I may be wrong, but I think you're trying to say that Planet-A is
<em>3 times as far from the sun</em> as Planet-C is.

If that's the real question, then the answer is that the period of Orbit-A
is about<em> 5.2</em> times as long as the period of Orbit-C .

Orbital period ≈ (proportional to) (the orbital distance) ^ 3/2 power.

This was empirically demonstrated about 350 years ago by Johannes
and his brilliant Kepple, and derived about 100 years later by Newton
from his formula for the forces of gravity.

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What was the greatest contribution of the monasteries? A. Illuminated manuscripts. B. Cloisters. C. Sculpture.d. Transepts

REY [17]

I am not entirely sure but i believe the answer is C scupture

3 0

3 years ago

Read 2 more answers

If you were to travel from the equator to the higher latitudes (near either pole) on Saturn, the differential rotation would cau

Dmitrij [34]

Answer:

Saturn's differential rotation will cause the length of a day measures to be longer by 0.4 hours

Explanation:

Differential rotation occurs due to the difference in angular velocities of an object as we move along the latitude of the or as we move into different depth of the object, indicating the observed object is in a fluid form

Saturn made almost completely of gas and has differential motion given as follows

Rotation at the equator = 10 hours 14 minutes

Rotation at high altitude = 10 hours 38 minutes

Therefore;

The differential rotation = 10 hours 38 minutes - 10 hours 14 minutes

The differential rotation = 24 minutes = 24 minutes × 1 hour/(60 minutes) = 0.4 hours

The differential rotation = 0.4 hours

Therefore, the measured day at the higher altitude will be 0.4 longer than at the equator.

8 0

3 years ago

Acar accelerates from 4 meters/second to 16 meters/second in 4 seconds. The car's acceleration is

s2008m [1.1K]

To understand this question, you need to understand the concept of acceleration first. Have you ever been in a car and noticed that it was getting faster and faster? That "speeding up" of the car is known as acceleration! Acceleration is essentially the rate at which you speed up.

Okay, so we now know what acceleration is. What are its units? The unit of acceleration is the change in velocity over a period of time: $\frac{∆v}{t}$

If you haven't learned about velocity yet, just think about it as speed for now. The funny-looking triangle, ∆, is a symbol for "the change of." For example, if I started walking at 3 $\frac{feet}{second}$ then sped up to 5 $\frac{feet}{second}$ , then the change in my speed would be 2 $\frac{feet}{second}$ , because I started walking 2 $\frac{feet}{second}$ faster!

Okay, enough with all the explanations. Hopefully, you understand the units now. Let's take a look at the question. A car accelerates from 4 $\frac{meters}{second}$ to 16 $\frac{meters}{second}$ in 4 seconds. What would the acceleration be? Let's set up an equation:

a = $\frac{∆v}{t}$

a is the acceleration, ∆v is the change in velocity, and t is the time elapsed.

Now, let's plug in our values! ∆v is the change in velocity, and to find that we simply have to subtract 16 $\frac{meters}{second}$ by 4 $\frac{meters}{second}$ . That makes sense, right? Back to the equation.

a = $\frac{∆v}{t}$
a = $\frac{16-4}{4}$

(16 - 4 is the change in velocity, and 4 is the number of seconds the car was accelerating)

a = $\frac{12}{4}$

a = 3 ( $\frac{meters}{second^{2}}$ )

We have our answer! The car's acceleration is 3 meters per second^{2}.

(You might be thinking: Wait. Meters per second squared? The reason for that is because acceleration is the rate at which the speed increases! That makes the unit $\frac{\frac{meters}{second}}{second}$ , which can be simplified down to $\frac{meters}{second^{2} }$ )

Let me know if you need clarification on anything I explained here!
- breezyツ

6 0

2 years ago

what is the approximate weight of a 20-kg cannonball on the moon if the acceleration due to gravity is 1.6m/s^2

monitta

On Earth, a cannonball with a mass of 20 kg would weigh 196 Newtons.
With the formula F=mg, where F is the weight in Newtons, m is the mass, and g is the acceleration due to gravity on the Earth which is 9.8m/s^2.
F=20kg x 9.8m/s^2= 196 Newtons

BUT on the moon, acceleration due to gravity is 1.6 m/s^2,
so F=mg=20kgx1.6m/s^2= 32 N

5 0

3 years ago

What two factors determine how bright a star appears to be in the sky?

Delicious77 [7]

Answer:

1. Luminosity

2.Apparent brightness

Explanation:

There are two factors on which brightness of star appear to be in the sky

The two factors are

1. Luminosity

2.Apparent brightness

1.Luminosity :It is defined as the total energy emitted by the object in a given time.Luminosity vary with the distance of observer from the star.Luminosity is a intrinsic property which depends on the fundamental chemical composition and structure of the material.Luminosity is depends on the size of star.Lager the star luminosity will be more.

2.Apparent brightness: It is defined as how bright a star appears from an observer on the earth and the amount of starlight reaching the earth.if the distance is large then the brightness decreases.When the distance of star from us small then the brightness of star increases.Distance is inversely proportional to brightness of the star.

7 0

3 years ago