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Sloan
2 months ago
10

Experimental tests have shown that hammerhead sharks can detect magnetic fields. In one such test, 100 turns of wire were wrappe

d around a 7.0-m-diameter cylindrical shark tank. A magnetic field was created inside the tank when this coil of wire carried a current of 1.5 A. Sharks trained by getting a food reward when the field was present would later unambiguously respond when the field was turned on.Part A)What was the magnetic field strength in the center of the tank due to the current in the coil?Part B)Is the strength of the coil's field at the center of the tank larger or smaller than that of the earth?- larger- smaller
Physics
1 answer:
Ostrovityanka [3.2K]2 months ago
7 0

B = µo*N*I/2r  

Thus, B = 4πx10^-7*150*1.6/2*3.5 = 4.31x10^-5T

You might be interested in
The parasailing system shown uses a winch to pull the rider in towards the boat, which is traveling with a constant velocity. Du
ValentinkaMS [3465]

Answer:

The force magnitude is  F_{net}= 1.837 *10^4N

and it is directed at 57.98° from the horizontal in a counterclockwise manner.

Explanation:

The problem states that

At t = 0, \theta = 20^o

The angular rate of increase is w = 2 \ ^o/s

Converting to revolutions per second gives us  \theta ' = 2 \ ^o/s * \frac{\pi}{180} =0.0349\ rps

The rope length is defined by

                      r = 125- \frac{1}{3}t^{\frac{3}{2} }    

    At \theta =30^o, Tension T of the rope is 18 kN.

     The weight of the para-sailor is M_p = 75kg

In analyzing the question, we observe that the equation for length can be represented as a linear displacement equation.

 The derivative of displacement results in velocity.

Hence,

           r' = -\frac{1}{3} [\frac{3}{2} ] t^{\frac{1}{2} }

signifies the velocity, and further differentiation yields acceleration.

Therefore,

         r'' = -\frac{1}{4} t^{-\frac{1}{2} }

Now considering the moment when the rope forms a 30° angle with the water,

typically angular velocity is expressed as

                      w = \frac{\Delta \theta}{\Delta t}

where \theta represents the angular displacement.

Next, evaluating the interval from 20^o \ to \ 30^o gives us

                 2 = \frac{30 -20 }{t -0}

making t the focal point.

             t = \frac{10}{2}

               = 5s

At this time, the displacement measures

             r = 125- \frac{1}{3}(5)^{\frac{3}{2} }  

                = 121.273 m

The linear velocity computes to

             r' = -\frac{1}{3} [\frac{3}{2} ] (5)^{\frac{1}{2} }

                = -1.118 m/s

Whereas linear acceleration calculates as

          r'' = -\frac{1}{4} (5)^{-\frac{1}{2} }

              = -0.112m/s^2

Generally, radial acceleration is given by

         \alpha _R = r'' -r \theta'^2

              = -0.112 - (121.273)[0.0349]^2

              = 0.271 m/s^2

Simultaneously, angular acceleration can be represented as

                 \alpha_t = r \theta'' + 2 r' \theta '

Then \theta '' = \frac{d (0.0349)}{dt} = 0

Thus,

             \alpha _t = 121.273 * 0 + 2 * (-1.118)(0.0349)

                   = -0.07805 m/s^2

The resultant acceleration is mathematically denoted as

                a = \sqrt{\alpha_R^2 + \alpha_t^2 }

                  = \sqrt{(-0.07805)^2 +(-0.027)^2}

                  = 0.272 m/s^2

Now the acceleration's direction is mathematically expressed as

                  tan \theta_a = \frac{\alpha_R }{\alpha_t }

                       \theta_a = tan^{-1} \frac{-0.271}{-0.07805}

                           = 73.26^o

               

The y-axis force acting on the para-sailor is mathematically shown as

               F_y = mg + Tsin 30^o + ma sin(90- \theta )

                    = (75 * 9.8) + (18 *10^3) sin 30 + (75 * 0.272)sin(90-73.26)

                    = 9.74*10^3 N

The x-axis force acting on the para-sailor is represented as

              F_x = mg + Tcos 30^o + ma cos(90- \theta )    

             = (75 * 9.8) + (18 *10^3) cos 30 + (75 * 0.272)cos(90-73.26)

             = 1.557 *10^4 N

The overall force is calculated as

                      F_{net} = \sqrt{F_x^2 + F_y^2}

                             =\sqrt{(1.557 *10^4)^2 + (9.74*10^3)^2}

                            F_{net}= 1.837 *10^4N

The directional force is evaluated as

              tan \theta_f = \frac{F_y}{F_x}

                   \theta_f = tan^{-1} [\frac{1.557*10^4}{9.74*10^3} ]

                       = tan^{-1} (1.599)

                       = 57.98^o

     

                     

7 0
3 months ago
The wavelength of light is 5000 angstrom. Express it in nm and m.
serg [3582]

Answer:

1 angstrom equals 0.1 nm.

To convert 5000 angstroms: 5000 angstrom = 5000 / 1 × 0.1 nm.

= 500 nm

1 \:  angstrom = 1 \times  {10}^{ - 10} m

To express 5000 angstroms in meters: 5000 angstrom = 5000 × 1 × 10^-10.

= 5 × 10^-7 m

Hope this explanation is useful for you.

7 0
4 months ago
A trolley of mass 200kg is travelling north at 2m\s. It collides head on with a second trolley of mass 500kg which is travelling
Ostrovityanka [3204]
200*2N+500*5S=(700)V
<span>400N-2500N=700V </span>
<span>determine V. A negative N indicates S</span>
6 0
2 months ago
Planting of crops to purify -lucern, shrubs, trees(possible job creation)
kicyunya [3294]
These plants were utilized to eliminate harmful substances from the environment. In today's world, such crops have evolved into a significant asset and can be a source of profit when managed properly, creating multiple job opportunities along the way. Hope this information is useful.
8 0
2 months ago
Read 2 more answers
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