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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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3 years ago
5.0 kg, with a bullet of mass 0.1 kg. The target was mounted on
kari74 [83]

Answer:

306 m/s

Explanation:

Law of conservation of momentum

m1v1 + m2v2 = (m1+m2)vf

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v1 is what we're trying to solve

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plugging in, we get

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8 0
3 years ago
During the latter part of your European vacation, you are hanging out at the beach at the gold coast of Spain. As you are laying
Jlenok [28]

Well, I guess you can come close, but you can't tell exactly.

It must be presumed that the seagull was flying through the air
when it "let fly" so to speak, so the jettisoned load of ballast
of which the bird unburdened itself had some initial horizontal
velocity.

That impact velocity of 98.5 m/s is actually the resultant of
the horizontal component ... unchanged since the package
was dispatched ... and the vertical component, which grew
all the way down in accordance with the behavior of gravity.

  98.5 m/s  =  √ [ (horizontal component)² + (vertical component)² ].

The vertical component is easy; that's (9.8 m/s²) x (drop time).
Since we're looking for the altitude of launch, we can use the
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             Height = (1/2) (acceleration) (time²) .

If the impact velocity were comprised solely of its vertical
component, then the solution to the problem would be a
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                  Time = (98.5 m/s) / (9.81 m/s²) = 10.04 seconds
whence
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As noted, this solution applies only if the gull were hovering with
no horizontal velocity, taking careful aim, and with malice in its
primitive brain, launching a remote attack on the rich American.

If the gull was flying at the time ... a reasonable assumption ... then
some part of the impact velocity was a horizontal component.  That
implies that the vertical component is something less than 98.5 m/s,
and that the attack was launched from an altitude less than 494 m.   

8 0
3 years ago
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Answer:

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Explanation:

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