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Aleksandr [31]
2 years ago
14

Question 3

Physics
1 answer:
pickupchik [31]2 years ago
4 0
When a police officer is trying to decide if a driver is speeding, what is his point of reference. The speed limit
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I need help plzzz!!!!!
Masteriza [31]

Answer:

I truly don't know

Explanation:

5 0
3 years ago
Motivation is best described as the _____. ability to be successful with your exercise way you see your exercise desire to conti
Olin [163]

Answer:

The desire to continue with your exercise would be the correct answer! :D

6 0
3 years ago
Read 2 more answers
The more you heat an object, the
zhannawk [14.2K]
The answer is D.  The temperature obviously doesnt rise slower or faster, and if you are heating an object, it would make no sense to say that less heat is being transferred.
8 0
3 years ago
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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
How many nanoseconds are there in 1.90 yr ?<br> Express your answer using three significant figures.
shepuryov [24]

       (1.9 yr) x (365.24 day/yr) x (86,400 sec/day) x (10⁹ nsec/sec)

   =  (1.9 x 365.24 x 86,400 x 10⁹) nanosec

   =  6.00 x 10¹⁶ nanoseconds

5 0
3 years ago
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