Two poles are connected by a wire that is also connected to the ground. The first pole is 20ft tall and the second pole is 10ft
tall. There is a distance of 30ft between the two poles. Where should the wire be anchored to the ground to minimize the amount of wire needed
1 answer:
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.
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(1)
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: is the final velocity = 15.0 m/s
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: is the final time = 0.15 s
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Therefore, the average force on the squid during the propulsion is 60 N.
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Answer:
The distance can the water be projected is 4.51 m
Explanation:
The speed of the water in the hose is equal to:
v1 = R/A1
If we solve the continuity for v2:
v2 = R/A2 (eq. 1)
The equation for the vertical position is:
yf = yi + vy*t - (1/2)gt²
yi = 0
vy = 0
Clearing t:
(eq. 2)
The equation for position is:
xf = xi + vxt = 0 + v2t = v2t (eq. 3)
Replacing equation 1 and 2 in equation 3: