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3 years ago

Which statement is true about a polarized object?

1 answer:

3 0

The statement which is true about a polarized object would be "the number of the positive and negative charges can be the same." Polarized objects became polarized and it has its center of positive charges separated from negative charges.

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A 50-cm wire placed in an east-west direction is moved horizontally to the north with a speed of 2.0 m/s. the horizontal compone

Strike441 [17]

When a wire is moved inside uniform magnetic field then its free electrons will experience magnetic force on it due to which wire will have potential difference at its ends.

Now here we will have magnetic field due to earth and wire is moving in this constant field so induced emf is given by formula

$EMF = v.(B x L)$

given that

$B = 25\mu Tj - 50\mu Tk$

$v = 2 m/s j$

$L = 0.50 m (-i)$

now by using the above formula we will have

$EMF = 2(j) .(25\mu j - 50\mu k) x (-0.50 i)$

$EMF = 2(j) .(12.5\mu k + 25\mu j)$

$EMF = 50 \mu Volts$

5 0

3 years ago

Read 2 more answers

10. Which of these devices is also referred to as a'key'?

brilliants [131]

Answer:

Electric switch is commonly known as the key of an electric circuit.

3 0

3 years ago

Read 2 more answers

The electric flux through a square-shaped area of side 5 cm near a very large, thin, uniformly-charged sheet is found to be 3\ti

deff fn [24]

Answer:

Explanation:

Given

side of square shape $a=5\ cm$

Electric flux $\phi =3\times 10^{-5}\ N.m^2/C$

Permittivity of free space $\epsilon_0=8.85\times 10^{-12} \frac{C^2}{N.m^2}$

Flux is given by

$\phi =EA\cos \theta$

where E=electric field strength

A=area

$\theta$ =Angle between Electric field and area vector

$E=\frac{\phi }{A\cos (0)}$

$E=\frac{3\times 10^{-5}}{25\times 10^{-4}\times \cos(0)}$

$E=0.012\ N/C$

and Electric field by a uniformly charged sheet is given by

$E=\frac{\sigma }{2\epsilon_0}$

where $\sigma$ =charge density

$=\frac{\sigma }{\epsilon_0}$

$\sigma =0.012\times 8.85\times 10^{-12}$

$\sigma =2.12\times 10^{-13}\ C/m^2$

5 0

3 years ago

Scientists plan to release a space probe that will enter the atmosphere of a gaseous planet. The temperature of the gaseous plan

ZanzabumX [31]

From my research, the image supports the question. From the graph given, we can construct the equation of the line using the two-point formula. Using the given value of 601 K, we can solve for the missing value of altitude.

y - y1 = [(y2 - y1)/(x2 - x1)](x- x1)
y - 147.52 = [ (567 - 147.54)/(78.11 - 18.4) ](x - 18.4)

Substituting y = 601 to solve for x:
601 - 147.52 = [ (567 - 147.54)/(78.11 - 18.4) ](x - 18.4<span>)
</span>x = 83

Therefore, the probe's instruments will fail at 83 kilometers.