When spring times comes around you ever be like, "k." If you feel like that every year during spring then there you go
Answer:
0.12
Explanation:
The acceleration due to gravity of a planet with mass M and radius R is given as:
g = (G*M) / R²
Where G is gravitational constant.
The mass of the planet M = 3 times the mass of earth = 3 * 5.972 * 10^24 kg
The radius of the planet R = 5 times the radius of earth = 5 * 6.371 * 10^6 m
Therefore:
g(planet) = (6.67 * 10^(-11) * 3 * 5.972 * 10^24) / (5 * 6.371 * 10^6)²
g(planet) = 1.18 m/s²
Therefore ratio of acceleration due to gravity on the surface of the planet, g(planet) to acceleration due to gravity on the surface of the planet, g(earth) is:
g(planet)/g(earth) = 1.18/9.8 = 0.12
Answer:
x ≤ 3.6913 m
Explanation:
Given
Mrod = 44.0 kg
L = 4.90 m
Tmax = 1450 N
Mman = 69 kg
A: left end of the rod
B: right end of the rod
x = distance from the left end to the man
If we take torques around the left end as follows
∑τ = 0 ⇒ - Wrod*(L/2) - Wman*x + T*Sin 30º*L = 0
⇒ - (Mrod*g)*(L/2) - (Mman*g)*x + Tmax*Sin 30º*L = 0
⇒ - (44*9.8)*(4.9/2) - (69*9.8)*x + (1450)*(0.5)*(4.9) = 0
⇒ x ≤ 3.6913 m
Answer:
Explanation:
Using Kepler's third law, we can relate the orbital periods of the planets and their average distances from the Sun, as follows:
Where 
and 
are the orbital periods of Mercury and Earth respectively. We have 
and 
. Replacing this and solving for
