An object 1 of mass m1 is separated by some distance d from an object 2 of mass 2m1 . An object 3 of mass m3 is to be placed bet
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Answer:
The anchor points are 78.37 ft and 111.99 ft
Explanation:
If you look at the attached (Fig 1) you will find that the union of antenna and its guy wires forms two right triangles. To solve problems that involve this kind of triangles, you can apply trigonometric functions (sine, cosine, etc) and Pythagoras Theorem. Trigonometric functions states the relation between angles, sides and hypotenuse of a right triangle. If you look Fig2, considering α angle, "b" is the opposite side, "a" the adjacent side and "c" the hypotenuse. Then
a) Sine (α) = b/c it means opposite side/hypotenuse
b) Cosine (α)= a/c, it means adjacent side/hypotenuse
c) Tangent (α) = b/a opposite side/adjacent side.
Pythagoras theorem states that if you called "a" and "b", the sides of the right triangle, and "c" the hypotenuse, then:
a² + b² = c²
As the problem states the lengths 150 ft and 170 ft represents the value of the hypotenuse of each triangle and 65° is one of the angles of the triangle with 150 ft hypotenuse. So you can solve this using sin (65°) to find the height of the antenna (h) and then the two distances (x and y,).
Sine (65°) = h/ 150 ft ⇒ Sine (65°) x 150ft = h ⇒ h = 127.9 ft.
To find x : Cosine (65°) = x/ 150 ft ⇒ Cosine (65°) x 150 ft = x
⇒ x = 78.37 ft.
And finally, to find y we can apply Pythagoras theorem
(170 ft)² = (127.9 ft)² + y² ⇒ y² = (170 ft)² - (127.9 ft)² ⇒ y = 111.99 ft
Summarizing, the anchor points are 78.37 ft and 111.99 ft
Answer:
The value is
Explanation:
From the question we are told that
The radius of the inner conductor is
The radius of the outer conductor is
The potential at the outer conductor is
Generally the capacitance per length of the capacitor like set up of the two conductors is
Here is the permitivity of free space with value
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Generally given that the potential of the outer conductor with respect to the inner conductor is positive it then mean that the outer conductor is positively charge
Generally the line charge density of the outer conductor is mathematically represented as
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Generally the surface charge density is mathematically represented as
here
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