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Mars2501 [29]
3 years ago
5

An automobile has a vertical radio antenna 1.45 m long. The automobile travels at 70.0 km/h on a horizontal road where Earth's m

agnetic field is 50.0 µT, directed toward the north and downward at an angle of 65.0° below the horizontal. (a) Specify the direction the automobile should move so as to generate the maximum motional emf in the antenna, with the top of the antenna positive relative to the bottom. north east west south (b) Calculate the magnitude of this induced emf. mV
Physics
1 answer:
Agata [3.3K]3 years ago
8 0

Answer:

595.73 X 10⁻⁶ V.

Explanation:

The magnetic field is towards the north with dip angle equal to 65 so the horizontal component of earth's magnetic field will be

H = 50 X 10⁻⁶ Cos 65 = 21.13 x 10⁻⁶ T.

length of rod L = 1.45 M

Velocity of rod = 70 km / h = 19.444 m /s

Since the top of the antenna rod is to have positive polarity , current induced in it will be from down to up . Magnetic field is towards the north .

To cut magnetic flux perpendicularly , rod should move towards the east .

We can easily get this direction by applying Fleming's right  hand Rule.

Induced emf =  H L v

= 21.13 x 10⁻⁶ X 1.45 X 19.444

= 595.73 X 10⁻⁶ V.

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gravitational constant value means it was never change in any particular area of the Earth

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Three plant like organisms which cannot produce their own food are
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A silver bar 0.125 meter long is subjected to a temperature change from 200 C to 100 C . What will be the length of the bar afte
dimulka [17.4K]
\Delta L= \alpha L_0 (T_f-T_i)

= (18 x 10^-6 /°C)(0.125 m)(100° C - 200 °C)

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8 0
3 years ago
Read 2 more answers
by how many times occur in the force of attraction between two bodies change when the distance between then is reduced to one th
xenn [34]

Answer:

<em>The force is now 9 times the original force</em>

Explanation:

<u>Coulomb's Law </u>

The electrostatic force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

Coulomb's formula is:

\displaystyle F=k\frac{q_1q_2}{d^2}

Where:

k=9\cdot 10^9\ N.m^2/c^2

q1, q2 = the particles' charge

d= The distance between the particles

Suppose the distance is reduced to d'=d/3, the new force F' is:

\displaystyle F'=k\frac{q_1q_2}{\left(\frac{d}{3}\right)^2}

\displaystyle F'=k\frac{q_1q_2}{\frac{d^2}{9}}

\displaystyle F'=9k\frac{q_1q_2}{d^2}

\displaystyle F'=9F

The force is now 9 times the original force

8 0
3 years ago
A thin soap bubble of index of refraction 1.33 is viewed with light of wavelength 550.0nm and appears very bright. Predict a pos
faust18 [17]

Answer:

The possible thickness of the soap bubble = 1.034\times 10^{-7}\ m.

Explanation:

<u>Given:</u>

  • Refractive index of the soap bubble, \mu=1.33.
  • Wavelength of the light taken, \lambda = 550.0\ nm = 550.0\times 10^{-9}\ m.

Let the thickness of the soap bubble be t.

It is given that the soap bubble appears very bright, it means, there is a constructive interference takes place.

For the constructive interference of light through a thin film ( soap bubble), the condition of constructive interference is given as:

2\mu t=\left ( m+\dfrac 12 \right )\lambda.

where m is the order of constructive interference.

Since the soap bubble is appearing very bright, the order should be 0, as 0^{th} order interference has maximum intensity.

Thus,

2\mu t=\left (0+\dfrac 12\right )\lambda\\t=\dfrac{\lambda}{4\mu}\\\ \ = \dfrac{550\times 10^{-9}}{4\times 1.33}\\\ \ = 1.034\times 10^{-7}\ m.

It is the possible thickness of the soap bubble.

6 0
3 years ago
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