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Galina-37 [17]
3 years ago
13

In an experiment designed to measure the Earth's magnetic field using the Hall effect, a copper bar 0.410 cm thick is positioned

along an east–west direction. Assume n = 8.46 × 1028 electrons/m3 and the plane of the bar is rotated to be perpendicular to the direction of B. If a current of 8.00 A in the conductor results in a Hall voltage of 4.75 10-12 V, what is the magnitude of the Earth's magnetic field at this location?
Physics
1 answer:
ehidna [41]3 years ago
3 0

Answer:

B=3.2\times 10^{-5}\ T

Explanation:

Given that

t = 0.410 cm

n = 8.46 \times 10^{28}\ electrons/m^3

I= 8 a

V_h= 4.75\times 10^{-12}\ V

q=1.6\times 10^{-19}\ C

We know that magnetic field given as

B=\dfrac{nqtV_h}{I}

B=\dfrac{8.46 \times  10^{28}\times 1.6\times 10^{-19}\times 0.41\times 10^{-2}\times 4.75\times 10^{-12}}{8}

B=3.2\times 10^{-5}\ T

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Usimov [2.4K]

Answer:

a) v'=-0.227\ m.s^{-1}

b) v=1.36\ m.s^{-1}

Explanation:

Given:

mass of the lighter block, m'=0.02\kg

velocity of the lighter block, u'=0.68\ m.s^{-1}

mass of the heavier block, m=0.04\ kg

velocity of the heavier block, u=0\ m.s^{-1}

a)

Using conservation of linear momentum:

m'.u'+m.u=m'.v'+m.v

where:

v'= final velocity of the lighter block

v= final velocity of the heavier block

m'.u'=m'.v'+m.v

m'(u'-v')=m.v ........................(1)

Since kinetic energy is conserved in elastic collision:

\frac{1}{2}m'.u'^2=\frac{1}{2}m'.v'^2+\frac{1}{2}m.v^2

m'(u'^2-v'^2)=m.v^2

m'(u'-v')(u'+v')= m.v^2

divide the above equation by eq. (1)

v=u'+v' .............................(2)

now we substitute the value of v from eq. (2) in eq. (1)

m'(u'-v')=m(u'+v')

\frac{m'+m}{m'-m} =\frac{u'}{v'}

\frac{0.02+0.04}{0.02-0.04} =\frac{0.68}{v'}

v'=-0.227\ m.s^{-1} (negative sign denotes that the direction is towards left)

b)

now we substitute the value of v' from eq. (2) in eq. (1)

m'(u'-v+u')=m.v

2m'.u'=(m-m')v

2\times 0.02\times 0.68=(0.04-0.02)\times v

v=1.36\ m.s^{-1}

6 0
3 years ago
A capacitor is connected across an ac source. Suppose the frequency of the source is doubled. What happens to the capacitive rea
fgiga [73]

The capacitive reactance is reduced by a factor of 2.

<h3>Calculation:</h3>

We know the capacitive reactance is given as,

Xc = \frac{1}{2\pi fC}

where,

Xc\\ = capacitive reactance

f = frequency

C = capacitance

It is given that frequency is doubled, i.e.,

f' = 2f

To find,

Xc =?

Xc' = \frac{1}{2\pi f'C}

      = \frac{1}{2\pi 2f C}

      = \frac{1}{2} (\frac{1}{2\pi fC} )\\

Xc' = \frac{1}{2} Xc

Therefore, the capacitive reactance is reduced by a factor of 2.

I understand the question you are looking for is this:

A capacitor is connected across an AC source. Suppose the frequency of the source is doubled. What happens to the capacitive reactant of the inductor?

  1. The capacitive reactance is doubled.
  2. The capacitive reactance is traduced by a factor of 4.
  3. The capacitive reactance remains constant.
  4. The capacitive reactance is quadrupled.
  5. The capacitive reactance is reduced by a factor of 2.

Learn more about capacitive reactance here:

brainly.com/question/23427243

#SPJ4

3 0
1 year ago
The moment of water from ocean through the atmosphere and back
melisa1 [442]

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There is always water in the atmosphere. Clouds are, of course, the most visible manifestation of atmospheric water, but even clear air contains water—water in particles that are too small to be seen. One estimate of the volume of water in the atmosphere at any one time is about 3,100 cubic miles (mi3) or 12,900 cubic kilometers (km3). That may sound like a lot, but it is only about 0.001 percent of the total Earth's water volume of about 332,500,000 mi3 (1,385,000,000 km3), If all of the water in the atmosphere rained down at once, it would only cover the globe to a depth of 2.5 centimeters, about 1 inch.

6 0
3 years ago
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An object experiences an acceleration of -6.8 m/s​2.​ As a result, it accelerates from 54 m/s to a complete stop. How much dista
inessss [21]

Answer:

The distance traveled during its acceleration, d = 214.38 m

Explanation:

Given,

The object's acceleration, a = -6.8 m/s²

The initial speed of the object, u = 54 m/s

The final speed of the object, v = 0

The acceleration of the object is given by the formula,

                                      a = (v - u) / t   m/s²

       ∴                              t = (v - u) / a

                                         = (0 - 54) / (-6.8)

                                         = 7.94 s

The average velocity of the object,

                                       V = (54 + 0)/2

                                           = 27 m/s

The displacement of the object,

                                 d = V x t   meter

                                    = 27 x 7.94

                                    = 214.38 m

Hence, the distance the object traveled during that acceleration is, a = 214.38 m

3 0
3 years ago
Kristen has two identical sized cubes, one is lead (Pb) and one is copper (Cu). Kristen is determining the density of each cube.
natta225 [31]

Answer:

volume

Explanation:

Identical size means volume will be the same in each calculation of

density =   mass / <u>volume</u>

3 0
2 years ago
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