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3 years ago

What is a Geographic test

Physics

2 answers:

Genrish500 [490]3 years ago

8 0

<em>It's a test on Geography!
</em>

maksim [4K]3 years ago

3 0

It could be a test for which each question is on a separate piece of paper located in a different country.

But I think it's actually a test with questions about stuff you would read in a Geography textbook or learn in a Geography class.

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Please please help i have to turn in

larisa [96]

Hi,

I've found a link that should assist you or answer your question.
http://click.dji.com/ANbvbbP7bwUWtSACp6U_?pm=link&as=0004

Have a nice day!

6 0

3 years ago

Suppose a small planet is discovered that is 16 times as far from the Sun as the Earth's distance is from the Sun. Use Kepler's

mamaluj [8]

Answer:

23376 days

Explanation:

The problem can be solved using Kepler's third law of planetary motion which states that the square of the period T of a planet round the sun is directly proportional to the cube of its mean distance R from the sun.

$T^2\alpha R^3\\T^2=kR^3.......................(1)$

where k is a constant.

From equation (1) we can deduce that the ratio of the square of the period of a planet to the cube of its mean distance from the sun is a constant.

$\frac{T^2}{R^3}=k.......................(2)$

Let the orbital period of the earth be $T_e$ and its mean distance of from the sun be $R_e$ .

Also let the orbital period of the planet be $T_p$ and its mean distance from the sun be $R_p$ .

Equation (2) therefore implies the following;

$\frac{T_e^2}{R_e^3}=\frac{T_p^2}{R_p^3}....................(3)$

We make the period of the planet $T_p$ the subject of formula as follows;

$T_p^2=\frac{T_e^2R_p^3}{R_e^3}\\T_p=\sqrt{\frac{T_e^2R_p^3}{R_e^3}\\}................(4)$

But recall that from the problem stated, the mean distance of the planet from the sun is 16 times that of the earth, so therefore

$R_p=16R_e...............(5)$

Substituting equation (5) into (4), we obtain the following;

$T_p=\sqrt{\frac{T_e^2(16R_e)^3}{(R_e^3}\\}\\T_p=\sqrt{\frac{T_e^24096R_e^3}{R_e^3}\\}$

$R_e^3$ cancels out and we are left with the following;

$T_p=\sqrt{4096T_e^2}\\T_p=64T_e..............(6)$

Recall that the orbital period of the earth is about 365.25 days, hence;

$T_p=64*365.25\\T_p=23376days$

4 0

3 years ago

What does the "coefficient of friction" tell you?

Arte-miy333 [17]

a coefficient of fraction is a value that shows the relationship between two objects and the normal reaction between the objects that are involved.

8 0

3 years ago

At speeds over 30 mph, you should maintain a following distance of at least ________ behind the vehicle ahead of you.

ICE Princess25 [194]

At speeds over 30 mph, you should maintain a following distance of at least <u>three full seconds</u> behind the vehicle ahead of you.

As a general rule and common sense at a speed of 30 mph you can leave three full seconds so that you can achieve a prudent distance between the car you are driving and the car in front in order to be able to perform some kind of maneuver if an accident or unforeseen event occurs.

To count the full three seconds you can use the technique of counting the Mississippis as follows: Mississippi one, Mississippi two, Mississippi three.

<h3>What is an accident?</h3>

An accident is an unexpected event that generally causes damage, injury or negative consequences.

Learn more about accident at: brainly.com/question/28070413

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