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)
You used density, because water/ice has a density of 1, and ice will sink in anything with a lesser density
Answer:
50 m
Explanation:
Acceleration= force/mass
3000/3000=1m/s^-2
Applying equation of motion:
V^2=U^2+2as; V is final velocity, u is initial velocity, a is acceleration and s is the distance covered.
0=10^2 -2*1s;
Solve for s
Answer:
it will be 1/√2 of its original period.
Explanation:
Answer:
6.75 seconds
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration = 16 m/s²
g = Acceleration due to gravity = 9.81 m/s²
Let y be the distance the rocket is accelerating
960-y is the distance traveled in free fall

In free fall

The distance the rocket will keep accelerating is 364.881828749 m
After which it will travel 960-364.881828749 = 595.118171251 m in free fall

The time the rocket is accelerating is 6.75 seconds