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bulgar [2K]
4 years ago
10

For the magnetic field at some random angle to the plane of the small coil, draw a picture showing only the small coil, a vector

giving the direction of the magnetic field, the area vector for the coil, and the angle between the magnetic field and the area vector. Write an equation for the magnetic flux through the small coil at an instant of time when the area vector is at some angle to the magnetic field. Write an expression that shows how the magnetic flux through the small coil changes as it turns with a constant angular speed.

Physics
1 answer:
bearhunter [10]4 years ago
3 0

Answer:

Solution attached below

Explanation:

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The altitude of the International Space Station ttt minutes after its perigee (closest point), in kilometers, is given by \qquad
Dmitriy789 [7]

Answer:

T = 92.8 min

Explanation:

Given:

The altitude of the International Space Station t minutes after its perigee (closest point), in kilometers, is given by:

                               A(t) = 415 - sin(\frac{2*\pi (t+23.2)}{92.8})

Find:

- How long does the International Space Station take to orbit the earth? Give an exact answer.

Solution:

- Using the the expression given we can extract the angular speed of the International Space Station orbit:

                                 A(t) = 415 - sin({\frac{2*\pi*t }{92.8} + \frac{23.2*2*\pi }{92.8} )

- Where the coefficient of t is angular speed of orbit w = 2*p / 92.8

- We know that the relation between angular speed w and time period T of an orbit is related by:

                                T = 2*p / w

                                T = 2*p / (2*p / 92.8)

Hence,                     T = 92.8 min

7 0
3 years ago
A double nozzle lying in a horizontal x-y plane discharges water into the atmosphere at a rate of 0.5 m3 /s. Assume the water sp
Kisachek [45]

Answer:

The force is  F= 46.25kN

Explanation:

The diagram for this question is shown on the first uploaded image  

At Equilibrium the summation of the of force on the vertical axis is zero

         i.e   \sum F_y =0

=>            F_y sin \ 60^o =\rho Q (v_2 -v_1 cos \ 30^o)

 v_2 is the is the speed of water at the nozzle which can be mathematically evaluated as

                      v_2 = \frac{R}{A_n}

substituting  0.5m^3/s for R and \frac{\pi}{4}(12*\frac{1m}{100} )^2 for A_n

                    v_2 = \frac{0.5}{\frac{\pi}{4} * (12*\frac{1}{100} )^2 }

                         = 44.23 m/s

 v_1 is the is the speed of water at the pipe which can be mathematically evaluated as

                       v_1 = \frac{R}{A_p}

substituting  0.5m^3/s for R and \frac{\pi}{4}(30*\frac{1m}{100} )^2 for A_p

                                v_1 = \frac{0.5}{\frac{\pi}{4} * (30*\frac{1}{100} )^2 }

                                    = 7.07 m/s

\rho is he density of water with value \rho =1000 kg /m^3

Substituting values into the equation above

                  F_ysin 60^o = 1000 (0.5) (44.23 -7.07 cos 30)

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=>            F_y sin \ 30^o =\rho Q (v_2 -v_1 sin \ 30^o)

               Since The speed at both A and B nozzle are the same then v_2 remains the same

 Substituting values

               F_x sin30^o =1000 (0.5) (44.23 - 7.07*sin30)

=>                        F_x = 40.69kN

   Hence the force acting on the flange bolts required to hold the nozzle in place is

                      F = \sqrt{F_x^2 + F_y^2}

                         = \sqrt{40.69 ^2 + 21.99^2}

                         F= 46.25kN

                 

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