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AlexFokin [52]
3 years ago
13

Calculate the average induced voltage between the tips of the wings of an airplane flying above East Lansing at a speed of 885 k

m/h. The wingspan of the airplane is 50.4 m, the vertical component of the Earth's magnetic field is 6.90×10-5 T.
Physics
1 answer:
Yakvenalex [24]3 years ago
6 0

Answer:

=0.855V

Explanation:

The induced voltage can be calculated using below expression

E =B x dA/dt

Where dA/dt = area

B= magnetic field = 6.90×10-5 T.

We were given speed of 885 km/h but we will need to convert to m/s for consistency of unit

speed = 885 km/h

speed = 885 x 10^3 m/hr

speed = 885 x 10^3/60 x60 m/s

speed = 245.8 m/s

If The aircraft wing sweep out" an area

at t= 50.4seconds then we have;

dA/dt = 50.4 x 245.8

= 123388.32m^2/s

Then from the expression above

E =B x dA/dt substitute the values of each parameters, we have

E = 6.90 x 10^-5 x 12388.32 V

E =0.855V

Hence, the average induced voltage between the tips of the wings is =0.855V

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kupik [55]

Recall that average velocity is equal to change in position over a given time interval,

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so that the <em>x</em>-component of \vec v_{\rm ave} is

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and its <em>y</em>-component is

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Solve for x_2 and y_2, which are the <em>x</em>- and <em>y</em>-components of the copter's position vector after <em>t</em> = 1.60 s.

x_2 = -2.25\,\mathrm m + \left(2.70\dfrac{\rm m}{\rm s}\right)(1.60\,\mathrm s) \implies \boxed{x_2 = 2.07\,\mathrm m}

y_2 = 5.70\,\mathrm m + \left(-2.50\dfrac{\rm m}{\rm s}\right)(1.60\,\mathrm s) \implies \boxed{y_2 = 1.70\,\mathrm m}

Note that I'm reading the given details as

x_1 = -2.25\,\mathrm m \\\\ y_1 = -5.70\,\mathrm m \\\\ v_x = 2.70\dfrac{\rm m}{\rm s}\\\\ v_y=-2.50\dfrac{\rm m}{\rm s}

so if any of these are incorrect, you should make the appropriate adjustments to the work above.

8 0
3 years ago
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Assume that the driver begins to brake the car when the distance to the wall is d=107m, and take the car's mass as m-1400kg, its
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Answer:

Explanation:

a ) Let let the frictional force needed be F

Work done by frictional force = kinetic energy of car

F x 107 = 1/2 x 1400 x 35²

F = 8014 N

b )

maximum possible static friction

= μ mg

where μ is coefficient of static friction

= .5 x 1400 x 9.8

= 6860 N

c )

work done by friction for μ = .4

= .4 x 1400 x 9.8 x 107

= 587216 J

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= .5 x 1400 x 35 x 35

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Kinetic energy at the at of collision

= 857500 - 587216

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So , if v be the velocity at the time of collision

1/2 mv² = 270284

v = 19.65 m /s

d ) centripetal force required

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4 0
3 years ago
PLEASE HELP
Gennadij [26K]

Answer:

3.675 m

Explanation:

a_{x} =0 v_{xo}=100 a_{y} =-g  v_{yo}=0

X-direction     | Y-direction

R=x_{o}+ v_{xo} t  | y=y_{o}+v_{yo}t+\frac{1}{2}a_{y}t^2

75=100t         |y=0+0+\frac{1}{2} (9.8)(0.75)

\frac{75}{100} =t             | y=3.675 m

0.75s=t              

Hope it helps

3 0
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Answer:

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F = (m∆v)/t

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the magnitude of the average force exerted on the wall by the ball is 800N

4 0
3 years ago
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