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dexar [7]
3 years ago
13

A kite 100 ft above the ground moves horizontally at a speed of 11 ft/s. At what rate is the angle (in radians) between the stri

ng and the horizontal decreasing when 200 ft of string have been let out?
Physics
2 answers:
frutty [35]3 years ago
8 0

Answer:

-2.26×10^-4 radians

Explanation:

The solution involves a right angle triangle

Length is z while the horizontal is the height x

X^2+ 100^2=z^2

Taking the derivatives

2x(dx/dt)=Z^2(dz/dt)

Specific moments = Z= 200 ,X= 100sqrt3 and dx/dt= 11

dz/dt= 1100sqrt3/200 = 9.53

Sin a= 100/a

Taking derivatives in terms of t

Cos a(da/dt)=100/z^2 dz/dt

a= 30°

Cos (30°)da/dt= (-100/40000×9.5)

a= -2.26×10^-4radians

Andrew [12]3 years ago
8 0

Answer:

0.11 rad/s

Explanation:

Given  that,

kite is 100 ft high

and moves horizontally at 7 ft/s

Total string let out =200 ft

String length(l), vertical(y) & Horizontal(x) distance of kite will form a right angle triangle

L^2 = y^2 + x^2

differentiate both side

= 2y\frac{dy}{dt}  + 2x\frac{dx}{dt} \\y\frac{dy}{dt} = -x\frac{dx}{dt} \\100\frac{dy}{dt}  = \sqrt{3} * 100 * \frac{dx}{dt} \\\frac{dy}{dt} = 11\sqrt{3}

Now Lcosθ = x

diferentiate

Lsin\theta * \frac{d\theta }{dt} = \frac{dx}{dt} \\200 * \frac{100}{200} * \frac{d\theta}{dt}  = 11\\\frac{d\theta }{dt }  = 0.11 rad/s

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If a force of 1.0 N pulled up and parallel to the surface of the incline is required to raise the mass back to the top of the in
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The question is incomplete! The complete question along with answer and explanation is provided below.

Question:

A 0.5 kg mass moves 40 centimeters up the incline shown in the figure below. The vertical height of the incline is 7 centimeters.

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Given Information:  

Mass = m = 0.5 kg

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Work done = 0.39 N.m

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The potential energy is given by

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PE = 0.343 Joules

As you can see in the attached image

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θ = sin⁻¹(0.07/0.4)

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The horizontal component of the normal force is given by

Fx = Fncos(θ)

Fx = 1*cos(10.078)

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Work done is given by

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W = 0.984*0.4

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