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Arisa [49]
2 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]2 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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Explanation:

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) Force F = − + ( 8.00 N i 6.00 N j ) ( ) acts on a particle with position vector r = + (3.00 m i 4.00 m j ) ( ) . What are (a)
natali 33 [55]

To develop this problem it is necessary to apply the concepts related to the Cross Product of two vectors as well as to obtain the angle through the magnitude of the angles.

The vector product between the Force and the radius allows us to obtain the torque, in this way,

\tau = \vec{F} \times \vec{r}

\tau = (8i+6j)\times(-3i+4j)

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Therefore the torque on the particle about the origen is 50k

PART B) To find the angle between two vectors we apply the definition of the dot product based on the vector quantities, that is,

cos\theta = \frac{r\cdot F}{|\vec{r}|*|\vec{F}|}

cos\theta = \frac{(8*-3)+(4*3)}{\sqrt{(-3)^2+4^2}*\sqrt{8^2+6^2}}

cos\theta = -0.24

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5 0
2 years ago
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5 0
3 years ago
Read 2 more answers
The valu
natulia [17]

Answer:

\% Error = 2.6\%

Explanation:

Given

x: 1.54, 1.53, 1.44, 1.54, 1.56, 1.45

Required

Determine the percentage error

First, we calculate the mean

\bar x = \frac{\sum x}{n}

This gives:

\bar x = \frac{1.54+ 1.53+ 1.44+ 1.54+ 1.56+ 1.45}{6}

\bar x = \frac{9.06}{6}

\bar x = 1.51

Next, calculate the mean absolute error (E)

|E| = \sqrt{\frac{1}{6}\sum(x - \bar x)^2}

This gives:

|E| = \sqrt{\frac{1}{6}*[(1.54 - 1.51)^2 +(1.53- 1.51)^2 +.... +(1.45- 1.51)^2]}

|E| = \sqrt{\frac{1}{6}*0.0132}

|E| = \sqrt{0.0022}

|E| = 0.04

Next, calculate the relative error (R)

R = \frac{|E|}{\bar x}

R = \frac{0.04}{1.51}

R = 0.026

Lastly, the percentage error is calculated as:

\% Error = R * 100\%

\% Error = 0.026 * 100\%

\% Error = 2.6\%

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