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3 years ago

Which of these statements about a dipole are correct? Select all that are true.

Physics

1 answer:

krok68 [10]3 years ago

7 0

Answer:

• The electric field at any location in space, due to a dipole, is the vector sum of the electric field due to the positive charge and the electric field due to the negative charge.

• At a distance d from a dipole, where d >> 5 (the separation between the charges), the magnitude of the electric field due to the dipole is proportional to 1/d^3

• A dipole consists of two particles whose charges are equal in magnitude but opposite in sign

Explanation:

A dipole is a pair of magnetized, equal or oppositely charged poles the are being separated by a distance.

The statements about a dipole that are correct are:

• The electric field at any location in space, due to a dipole, is the vector sum of the electric field due to the positive charge and the electric field due to the negative charge.

• At a distance d from a dipole, where d >> 5 (the separation between the charges), the magnitude of the electric field due to the dipole is proportional to 1/d^3

• A dipole consists of two particles whose charges are equal in magnitude but opposite in sign

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An object weigh 40N in air ,weigh 20N when submerged in water,and 30N when submerged in a liquid of unknown liquid density.what

muminat

Answer:

The density is $\rho_u =500 kg /m^3$

Explanation:

From the question we are told that

The weight in air is $W_a = 40 \ N$

The weight in water is $W_w = 20 \ N$

The weight in a unknown liquid is $W_u = 30 \ N$

Now according to Archimedes principle the weight of the object in water is mathematically represented as

$W_w = W_a -m _w g$

Where $m_w$ is he mass of the water displaced

substituting value

$m_w g = 40 -20$

$m_w g = 20 \ N --- (1)$

Now according to Archimedes principle the weight of the object in unknown is mathematically represented as

$W_u = W_a -m _u g$

Where $m_u$ is he mass of the unknown liquid displaced

substituting value

$m_u g = 40 -30$

$m_u g = 10 \ N ---(2)$

dividing equation 2 by equation 1

$\frac{m_ug}{m_wg} = \frac{10}{20}$

$\frac{m_u}{m_w} = \frac{1}{2}$

=> $m_u = 0.5 m_w$

Now since the volume of water and liquid displaced are the same then

$\rho _u = 0.5 \rho_w$

This because

$density = \frac{mass}{volume}$

So if volume is constant

mass = constant * density

Where $\rho_u$ is the density of the liquid

and $\rho_ w$ is the density of water which is a constant with a value $\rho_w = 1000 kg/m^3$

$\rho_u = 1000*0.5$

$\rho_u =500 kg /m^3$

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A 0.75μF capacitor is charged to 70 V . It is then connected in series with a 55Ω resistor and a 140 Ω resistor and allowed to d

Ipatiy [6.2K]

Answer:

Explanation:

Capacitor of 0.75μF, charged to 70V and connect in series with 55Ω and 140 Ω to discharge.

Energy dissipates in 55Ω resistor is given by V²/R

Since the 55ohms and 140ohms l discharge the capacitor fully, the voltage will be zero volts and this voltage will be shared by the resistor in ratio.

So for 55ohms, using voltage divider rule

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V=55/(55+140) ×70

V=19.74Volts is across the 55ohms resistor.

Then, energy loss will be

E=V²/R

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