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nordsb [41]
3 years ago
5

The top of a ladder slides down a vertical wall at a rate of 0.125 m/s. at the moment when the bottom of the ladder is 5 m from

the wall, it slides away from the wall at a rate of 0.3 m/s. how long is the ladder?
Physics
1 answer:
Contact [7]3 years ago
5 0
Vertically (y-axis), horizontally(x-axis)

dy/dt = -0.125 m/s (-ve since y decreasing )

dx/dt = +0.3 m/s (+ve, x increases)

and, x = 5 m

length of the ladder = k (a constant)

k^2 = x^2 + y^2

differentiating it wrt t,

0 = 2x dx/dt + 2y dy/dt

0 = 2(5)(0.3) + 2(y)(-0.125)

y = 12

which means, when the bottom of the ladder is 5m from the wall, the top of the ladder is 12m from the bottom.

thus, k^2 = 5^2 + 12^2

length of the ladder, k = 13 m
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If an object on Earth weighs 100N what is its weight in pounds?
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2 years ago
Find the value of F1 + F2 + F3.<br>​
Dovator [93]

Answer:

F = 0.78[N]

Explanation:

The given values correspond to forces, we must remember or take into account that the forces are vector quantities, that is, they have magnitude and direction. Since we have two X-Y coordinate axes (two-dimensional), we are going to decompose each of the forces into the X & y components.

<u>For F₁</u>

<u />F_{y}=2[N]<u />

<u>For F₂</u>

F_{x}=2*cos(60)\\F_{x}=1[N]\\F_{y}=-2*sin(60)\\F_{y}=-1.73[N]

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<u />F_{x}=-1*sin(60)\\F_{x}=-0.866[N]\\F_{y}=1*cos(60)\\F_{y}=0.5 [N]<u />

Now we can sum each one of the forces in the given axes:

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8 0
3 years ago
Two airplanes leave an airport at the same time. The velocity of the first airplane is 750 m/h at a heading of 51.3°. The veloci
abruzzese [7]
Refer to the diagram shown below.

In 2.4 hours, the distance traveled by the first airplane heading a 51.3° at 750 mph is 
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The second airplane travels
b = 620*2.4 = 1488 mile

The angle between the two airplanes is
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Let c =  the distance between the two airplanes after 2.4 hours.
From the Law of Cosines, obtain
c² = a² + b² - 2ab cos(111.7°)
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c = 2335.41 miles

Answer: 2335.4 miles

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3 years ago
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