Explanation:
4.
Given: Δd = 4 + 2 + 1 = 7km, Δt = 32 + 22 + 16 = 70mins
Required: V
Equation: V = Δd/Δt
Solution: V = 7/70
= 0.1km/min
Vertical line from the centre of mass is inside the base of the tower.
Well a isn’t it so i’m gonna go with d the american social statesman
Answer:
a. The Sun is 110.24 cm.
b. The distance of the nearest star at 4.24 light years is 3.1586 × cm.
Explanation:
Given a scale :
12700 Km : 1 cm
a. How large is the Sun of diameter 1.4 × Km?
Thus using the given scale, we have to compare the diameter of the Sun to that of the Earth.
=
= 110.2362
The Sun is 110.24 cm.
b. how far away is the nearest star at a distance of 4.24 light-years on this scale?
A light year is a term that is used to express the total distance that light would travel in an empty space in a year.
One light year = 9.461 × Km
So that,
4.24 light years = 4.24 × 9.461 × Km
= 40.11464 × Km
Using the given scale,
= 3.1586 ×
The distance of the nearest star at 4.24 light years is 3.1586 × cm.
We can solve the problem by requiring the equilibrium of the forces and the equilibrium of torques.
1) Equilibrium of forces:
where
is the weight of the person
is the weight of the scaffold
Re-arranging, we can write the equation as
(1)
2) Equilibrium of torques:
where 3 m and 2 m are the distances of the forces from the center of mass of the scaffold.
Using
and replacing T1 with (1), we find
from which we find
And then, substituting T2 into (1), we find