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Arisa [49]
3 years ago
7

A suspension bridge has twin towers that are 1300 feet apart. Each tower extends 180 feet above the road surface. The cables are

parabolic in shape and are suspended from the tops of the towers. The cables touch the road surface at the center of the bridge. Find the height of the cable at a point 200 feet from the center of the bridge.
Physics
1 answer:
zhannawk [14.2K]3 years ago
8 0

Answer:

y = 17.04 ft

Explanation:

Since the cable touches the road at the mid point of two towers

so here we have vertex at that mid point taken to be origin

now the maximum height on the either side is given as

y = 180 ft

horizontal distance of the tower from mid point is given as

x = \frac{1300}{2} = 650 ft

now from the equation of parabola we have

y = k x^2

180 = k(650^2)

k = 4.26 \times 10^{-4}

now we have

y = (4.26 \times 10^{-4})x^2

now we need to find the height at distance of 200 ft from center

so we have

y = (4.26 \times 10^{-4})(200^2)

y = 17.04 ft

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So if it reaches the ground in <em>t</em> = 1.54 s, then the height of the building <em>y</em> is

<em>y</em> = (12.0 m/s) (1.54 s) + 1/2 (9.80 m/s²) (1.54 s)² ≈ 30.1 m

(b) The stone's (downward) velocity <em>v</em> at time <em>t </em>is

<em>v</em> = 12.0 m/s + <em>g t</em>

so that after <em>t</em> = 1.54 s, its velocity is

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What is the resultant displacement? Explain please to
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Answer:

41.4496148484\ m/s

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\Delta x = 0.57-0.26

q = Charge = 6.9\times 10^{-9}\ C

m = Mass of object = 8.3\times 10^{-9}\ kg

Electric field due to a sheet is given by

E=\dfrac{\sigma}{2\epsilon_0}\\\Rightarrow E=\dfrac{5.9\times 10^{-8}}{2\times 8.85\times 10^{-12}}\\\Rightarrow E=3333.33\ V/m

Electric field is given by

E=\dfrac{V}{d}

Voltage is given by

V=E\Delta x

Kinetic energy is given by

K=qV

\dfrac{1}{2}mv^2=qE\Delta x\\\Rightarrow v=\sqrt{\dfrac{2qE\Delta x}{m}}\\\Rightarrow v=\sqrt{\dfrac{2\times 6.9\times 10^{-9}\times 3333.33\times (0.57-0.26)}{8.3\times 10^{-9}}}\\\Rightarrow v=41.4496148484\ m/s

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The orbiting body experience a centripetal force that is equal to the gravitational force of the second object towards the body.

The velocity of the orbit is given by the relation,

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