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4 years ago

Which statement correctly stated kepler's 3rd law of planetary motion?

2 answers:

8 0

When you square the "year" of each planet and divide it by the cube of its distance, or axis from the sun, the number would be the same for all the planets

svetlana [45]4 years ago

6 0

Answer:D

Explanation: the square of the orbital periods of the planets are directly proportional to the cubes of their average distances from the sun

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As you move toward a warning siren, the pitch _____.

sleet_krkn [62]

Answer:

As you move toward the siren the pitch should get louder

Explanation:

4 0

3 years ago

Read 2 more answers

When you throw a ball, the work you do to accelerate it equals the kinetic energy the ball gains. If you do twice as much work w

aniked [119]

Answer:

No. Twice as much work will give the ball twice as much kinetic energy. But since KE is proportional to the speed squared, the speed will be $sqrt{2}$ times larger.

Explanation:

The work done on the ball is equal to the kinetic energy gained by the ball:

$W=K$

So when the work done doubles, the kinetic energy doubles as well:

$2W = 2 K$

However, the kinetic energy is given by

$K=\frac{1}{2}mv^2$

where

m is the mass of the ball

v is its speed

We see that the kinetic energy is proportional to the square of the speed, $v^2$ . We can rewrite the last equation as

$v=\sqrt{\frac{2K}{m}}$

which also means

$v=\sqrt{\frac{2W}{m}}$

If the work is doubled,

$W'=2W$

So the new speed is

$v'=\sqrt{\frac{2(2W)}{m}}=\sqrt{2}\sqrt{\frac{2W}{m}}=\sqrt{2} v$

So, the speed is $\sqrt{2}$ times larger.

5 0

3 years ago

Which of the following is a correct step in the process of adding 8.0 x 10^ -2 and 6.0 x 10^ -3

dybincka [34]

Explanation:

In order to compute correctly the sum of the two terms, we have to rewrite one of them such that they have the same exponent.

The two terms are:

$8.0 \cdot 10^{-2}$

$6.0 \cdot 10^{-3}$

For instance, we can re-write the second term such as it also has a power $10^{-2}$ . In order to do that, we have to move the decimal point one place to the left, therefore: