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3 years ago

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To support a tree damaged in a storm, a 12-foot wire is secured from the ground to the tree at a point 10 feet off the ground. T

he tree meets the ground at a right angle. At approximately what angle does the wire meet the ground?
A)33.6°
B)39.8°
C)50.2°
D)56.4°

1 answer:

podryga [215]3 years ago

6 0

Answer:

The wire meet the ground at an angle of 56.4 degrees

Explanation:

It is given that,

To support a tree damaged in a storm, a 12-foot wire is secured from the ground to the tree at a point 10 feet off the ground.

The hypotenuse is, H = 12 foot

The perpendicular distance is, P = 10 feet

The angle between the tree and the ground is 90 degrees

Using Pythagoras theorem as :

$sin\theta=\dfrac{P}{H}$

$sin\theta=\dfrac{10}{12}$

$\theta=56.4^{\circ}$

So, the wire meet the ground at an angle of 56.4 degrees. Hence, the correct option is (d).

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The trains will collide at a distance 1660 m from the station

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Let the train traveling due north with a constant speed of 100 m/s be Train A.

Let the train traveling due south with a constant speed of 136 m/s be Train B.

From the question, Train B leaves a station 2,881 m away (that is 2,881 m away from Train A position).

Hence, the two trains would have traveled a total distance of 2,881 m by the time they collide.

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At the position where the trains will collide, the two trains must have traveled for equal time, t.

That is, At the point of collision,

$t_{A} = t_{B}$

$t_{A}$ is the time spent by train A

$t_{B}$ is the time spent by train B

From,

$Velocity = \frac{Distance }{Time }\\$

$Time = \frac{Distance}{Velocity}$

Since the time spent by the two trains is equal,

Then,

$\frac{Distance_{A} }{Velocity_{A} } = \frac{Distance_{B} }{Velocity_{B} }$

${Distance_{A} = x$ m

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${Velocity_{A} = 100$ m/s

${Velocity_{B} = 136$ m/s

Hence,

$\frac{x}{100} = \frac{2881 - x}{136}$

$136(x) = 100(2881 - x)\\136x = 288100 - 100x\\136x + 100x = 288100\\236x = 288100\\x = \frac{288100}{236} \\x = 1220.76m\\$

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This is the distance covered by train A by the time of collision.

Hence, Train B would have covered (2881 - 1221)m = 1660 m

Train B would have covered 1660 m by the time of collision

Since it is train B that leaves a station,

∴ The trains will collide at a distance 1660 m from the station.

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