3 years ago

What is the orbital period in years of a planet with a semi major axis of 35 au

Physics

1 answer:

mezya [45]3 years ago

6 0

Answer:

Orbital period of the planet will be 207.06 year

Explanation:

We have given the planet have the semi major axis as 35 au

We have to find the orbital period of the planet

From Keplar's third law there is relation between the orbital period and semi major axis which is t $T^2=R^3$

So $T^2=35^3$

$T^2=42875$

$T=207.06year$

So orbital period of the planet will be 207.06 year

You might be interested in

What do refrigerators and air conditioners use to move heat?

Aleksandr [31]

your answer is Freon

3 0

3 years ago

Read 2 more answers

⦁ Determining the magnetic flux, A rectangular piece of stiff paper measures 10 cm x 5 cm. You hold the piece of paper in a unif

MArishka [77]

Answer:

jguewjdofe

Explanation:

4 0

2 years ago

Main ingredient of all cells

Sati [7]

Chloroplast...................

7 0

3 years ago

If 2 electrically charged objects repel one another , which of the statements could be true, one object has more electrons, the

fiasKO [112]

Both objects have the same electrical charge. Opposite charges attract. And if they were neutral they would not do anything.

6 0

3 years ago

Read 2 more answers

Your 64-cm-diameter car tire is rotating at 3.3 rev/swhen suddenly you press down hard on the accelerator. After traveling 250 m

umka21 [38]

Answer:

0.76 rad/s^2

Explanation:

First, we convert the original and final velocity from rev/s to rad/s:

$v_o = 3.3\frac{rev}{s} * \frac{2\pi rad}{1rev} =20.73 rad/s$

$v_f = 6.4\frac{rev}{s} * \frac{2\pi rad}{1rev}=40.21 rad/s$

Now, we need to find the number of rads that the tire rotates in the 250m path. We use the arc length formula:

$D = x*r \\x = \frac{D}{r} = \frac{250m}{0.64m/2} = 781.25 rads$

Now, we just use the formula:

$w_f^2-w_o^2=2\alpha*x$

$\alpha =\frac{w_f^2-w_o^2}{2x} = \frac{(40.21rad/s)^2-(20.73rad/s)^2}{2*781.25rad} = 0.76 rad/s^2$

6 0

2 years ago

Read 2 more answers