Answer: 31.6ft
Explanation:
Check the attachment for the diagram.
According to the right angle triangle AEC, we will use Pythagoras theorem to get |AC|. Note that |AE| = |AB| - |CD|
that is 20ft - 10ft = 10ft
According to the theorem, the square of the sum of the adjacent side and the opposite side is equal to the square of the hypotenuse.
|AE|^2 + |EC|^2 = |AC|^2
10^2 + 30^2 = |AC|^2
100 + 900 = |AC|^2
|AC| = √1000
|AC| = 31.6ft
Therefore, the wire should be anchored 31.6ft to the ground to minimize the amount of wire needed.
Answer:
25
Explanation:
Because u don't know the gravity so u take that force and just put it in the gravity spot
Answer:
F_A = 8 F_B
Explanation:
The force exerted by the planet on each moon is given by the law of universal gravitation
F = 
where M is the mass of the planet, m the mass of the moon and r the distance between its centers
let's apply this equation to our case
Moon A
the distance between the planet and the moon A is r and the mass of the moon is 2m
F_A = G \frac{2m M}{r^{2} }
Moon B
F_B = G \frac{m M}{(2r)^{2} }
F_B = G \frac{m M}{4 r^{2} }
the relationship between these forces is
F_B / F_A = 
= 1/8
F_A = 8 F_B
Answer:
1) 6 seconds
2) 60 m/s
Explanation:
Given:
Δy = 180 m
v₀ = 0 m/s
a = 10 m/s²
1) Find t.
Δy = v₀ t + ½ at²
180 m = (0 m/s) t + ½ (10 m/s²) t²
t = 6 s
2) Find v.
v² = v₀² + 2aΔy
v² = (0 m/s)² + 2 (10 m/s²) (180 m)
v = 60 m/s
