Answer:

Explanation:
We can solve the problem by using Kepler's third law, which states that the ratio between the cube of the orbital radius and the square of the orbital period is constant for every object orbiting the Sun. So we can write

where
is the distance of the new object from the sun (orbital radius)
is the orbital period of the object
is the orbital radius of the Earth
is the orbital period the Earth
Solving the equation for
, we find
![r_o = \sqrt[3]{\frac{r_e^3}{T_e^2}T_o^2} =\sqrt[3]{\frac{(1.50\cdot 10^{11}m)^3}{(365 d)^2}(180 d)^2}=9.4\cdot 10^{10} m](https://tex.z-dn.net/?f=r_o%20%3D%20%5Csqrt%5B3%5D%7B%5Cfrac%7Br_e%5E3%7D%7BT_e%5E2%7DT_o%5E2%7D%20%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B%281.50%5Ccdot%2010%5E%7B11%7Dm%29%5E3%7D%7B%28365%20d%29%5E2%7D%28180%20d%29%5E2%7D%3D9.4%5Ccdot%2010%5E%7B10%7D%20m)
Oooooo there's a spongy bone? that's cool! Lol okay okay, I will research it and help you out.
Here's what I found:
Cancellous bone<span>, also known as </span>spongy<span> or </span>trabecular bone<span>, is one of the </span>two<span> types of </span>bone<span> tissue found in the human body. ... It is very porous and contains red </span>bone<span>marrow, where blood cells are made.</span>
Answer:
For number 4: A vector pointing to the right with a magnitude of 2.0
Explanation:
Very simple- just subtract 6-2
I am not sure how to do #2- sorry!
Answer:
-0.912 m/s
Explanation:
When the package is thrown out, momentum is conserved. The total momentum after is the same as the total momentum before, which is 0, since the boat was initially at rest.

where
are the mass of the child, the boat and the package, respectively.
are the velocity of the package and the boat after throwing.



Answer:
Final velocity (v) = 36 m/s
Distance traveled (s) = 2,160 m
Explanation:
Given:
Initial velocity (u) = 0
Acceleration (a) = 0.3 m/s
Time travel (t) = 2 minutes = 120 seconds
Find:
Final velocity (v) = ?
Distance traveled (s) = ?
Computation:
v = u + at
v = 0 + 0.3(120)
v = 0.3(120)
v = 36 m/s
Final velocity (v) = 36 m/s
Distance traveled (s) = ut + (1/2)at²
Distance traveled (s) = (0.5)(0.3 × 120 × 120)
Distance traveled (s) = 2,160 m