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DanielleElmas [232]
3 years ago
6

What speed would a proton need to achieve in order to circle Earth 1790 km above the magnetic equator, where the Earth's mag- ne

tic field is directed on a line between mag- netic north and south and has an intensity of 4 x 108 T? The mass of a proton is 1.673 Ã 10-21 kg
Physics
1 answer:
irinina [24]3 years ago
6 0

Answer:

The velocity is 31.25 m/s and direction is toward west.

Explanation:

Given that,

Distance h= 1790 km = 1.790\times10^{6}\ m

Magnetic field B=4\times10^{-8}\ T

Mass of proton m=1.673\times10^{-21}\ Kg

Radius of earth R =6.38\times10^{6}\ m

Radius of orbit r=R+h

r=6.38\times10^{6}+1.790\times10^{6}

r=8170000\ m

We need to calculate the speed

Using formula of magnetic field

Bvq=\dfrac{mv^2}{r}

v=\dfrac{Bqr}{m}

Put the value into the formula

v=\dfrac{4\times10^{-8}\times1.6\times10^{-19}\times8170000}{1.673\times10^{-21}}

v=31.25\ m/s

Hence, The velocity is 31.25 m/s and direction is toward west.

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Two solenoids A and B, spaced close to each other and sharing the same cylindrical axis, have 430 and 610 turns, respectively. A
KIM [24]

Answer

given,

Two solenoids A and B

Number of turn

Na = 430 turns          Nb = 610 turns

Current = 2.80 A

Average flux through  A  = 300 μWb

Average of flux through B = 90.0  μ Wb

a) L = \dfrac{N \phi}{I}

   L = \dfrac{610\times 90 \times 10^{-6}}{2.80}

   L =19.6 mH

b) inductance of A

   L = \dfrac{N_A \phi_A}{I_A}

   L = \dfrac{430\times 300 \times 10^{-6}}{2.80}

   L =46 mH

c) magnitude of the emf

    \epsilon_B = -L_B\dfrac{dI}{dT}

    \epsilon_B = -(19.6\times 10^{-3})(0.5)

    \epsilon_B = -9.8\times 10^{-3}\ V

    \epsilon_B = -9.8\ mV

7 0
3 years ago
One problem for humans living in outer space is that they are apparently weightless. One way around this problem is to design a
Maurinko [17]

This question is not complete.

The complete question is as follows:

One problem for humans living in outer space is that they are apparently weightless. One way around this problem is to design a space station that spins about its center at a constant rate. This creates “artificial gravity” at the outside rim of the station. (a) If the diameter of the space station is 800 m, how many revolutions per minute are needed for the “artificial gravity” acceleration to be 9.80m/s2?

Explanation:

a. Using the expression;

T = 2π√R/g

where R = radius of the space = diameter/2

R = 800/2 = 400m

g= acceleration due to gravity = 9.8m/s^2

1/T = number of revolutions per second

T = 2π√R/g

T = 2 x 3.14 x √400/9.8

T = 6.28 x 6.39 = 40.13

1/T = 1/40.13 = 0.025 x 60 = 1.5 revolution/minute

6 0
3 years ago
A light bulb, a speaker, and a toaster ovens are all examples of what in an electric circuit.
julia-pushkina [17]
Conductor because it creates heat.
6 0
3 years ago
A car of mass 500 kg is moving at a speed of 1.2 m/s. A man pushes the car,
rusak2 [61]
The answer is D I’m not really sure yet
6 0
3 years ago
A loop of wire is in a magnetic field such that its axis is parallel with the field direction. Which of the following would resu
ahrayia [7]

Answer:

All the given options will result in an induced emf in the loop.

Explanation:

The induced emf in a conductor is directly proportional to the rate of change of flux.

emf = -\frac{d \phi}{dt} \\\\where;\\\\\phi \ is \ magnetic \ flux\\\\\phi = BA\ cos \theta

where;

A is the area of the loop

B is the strength of the magnetic field

θ is the angle between the loop and the magnetic field

<em>Considering option </em><em>A</em>, moving the loop outside the magnetic field will change the strength of the magnetic field and consequently result in an induced emf.

<em>Considering option </em><em>B</em>, a change in diameter of the loop, will cause a change in the magnetic flux and in turn result in an induced emf.

Option C has a similar effect with option A, thus both will result in an induced emf.

Finally, <em>considering option</em> D, spinning the loop such that its axis does not consistently line up with the magnetic field direction will<em> </em>change the angle<em> </em>between the loop and the magnetic field. This effect will also result in an induced emf.

Therefore, all the given options will result in an induced emf in the loop.

4 0
3 years ago
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