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)
A )
T = mB g + mB a
T + mA a - mA g sin 35° = (Mi) mA g cos 35°
------------------------------------------------------------
T = 2.7 · 9.81 + 2.7 a
T = 26.487 + 2.7 a
26.487 + 2.7 a + 2.7 a - 2.7 · 9.81 · 0.574 = 0.15 · 2.7 · 9.81 · 0.819
5.4 a + 26.487 - 15.2023 = 3.2539
5.4 a = 8.0296
a = 1.487 ≈ 1.5 m/s²
B )
T = 2,7 · 9.81 = 26.487
26.487 - 15.2035 = (Mi) · 2.7 · 9.81 · 0.819
11.2835 = (Mi) · 21.69
(Mi) = 11.2835 : 21.69 = 0.52
Answer:
A. The amount of mass changes only slightly during a chemical
reaction.
The reason why there is a difference between free-fall acceleration is a centrifugal force.
I attached a diagram that shows how this force aligns with the force of gravity.
From the diagram we can see that:

Where g' is the free-fall acceleration when there is no centrifugal force, r is the radius of the planet, and w is angular frequency of planet's rotation.

is the latitude.
We can calculate g' and wr^2 from the given conditions in the problem.

Our final equation is:

Colatitude is:

The answer is:
The moment of inertia of the flywheel is 2.63 kg-
It is given that,
The maximum energy stored on the flywheel is given as
E=3.7MJ= 3.7×
J
Angular velocity of the flywheel is 16000
= 1675.51
So to find the moment of inertia of the flywheel. The energy of a flywheel in rotational kinematics is given by :
E = 

By rearranging the equation:
I = 
I = 2.63 kg-
Thus the moment of inertia of the flywheel is 2.63 kg-
.
Learn more about moment of inertia here;
brainly.com/question/13449336
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