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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Answer:
(a) 181.05 m/s²
(b) 13.2°
Explanation:
Given:
Radius of the circle (R) = 0.610 m
Angular acceleration (α) = 67.6 rad/s²
Angular speed (ω) = 17.0 rad/s
(a)
Radial acceleration of the ball is given as:
Plug in the given values and solve for . This gives,
Now, tangential acceleration is given by the formula:
Plug in the given values and solve for . This gives,
Now, the magnitude of total acceleration is given as the square root of the sum of the squares of tangential and centripetal accelerations. Therefore,
Plug in the given values and solve for total acceleration, . This gives,
Therefore, the magnitude of total acceleration is 181.05 m/s².
(b)
Angle of total acceleration relative to radial direction is given by the formula:
Therefore, the total acceleration makes an angle of 13.2° relative to radial direction.
Answer:
0.125 m
Explanation:
In this problem, we have:
v = 0.50 m/s is the average velocity of the wave
T = 0.25 s is the period of the wave
We can find the frequency of the wave, which is equal to the reciprocal of the period:
The problem is asking us to find the distance between two crests of the wave: this is equivalent to the wavelength. The wavelength is related to the average velocity and the frequency by the formula:
Substituting the numerical values, we find