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3 years ago

13

A coil of resistance 100ohm is placed in a magnetic field of 1mWb the coil has 100 turns and a galvanometer of 400ohm resistance

is connected in series with it. find the average emf and the current if the coil is moved in one tenth of a second from the given field to a field of 2.0mWb

1 answer:

Lina20 [59]3 years ago

6 0

Answer:

C

Explanation:

C

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1. If I dig a 6FT hole how deep is that hole?

iren [92.7K]

Answer:

For the first one its about 25 feet

Explanation:

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3 years ago

Read 2 more answers

enot [183]

Answer:

A. $U_0 = \dfrac{\epsilon_0 A V^2}{2d}$

B. $U_1 = \dfrac{\epsilon_0 A V^2}{6d}$

C. $U_2 = \dfrac{K\epsilon_0 A V^2}{2d}$

Explanation:

The capacitance of a capacitor is its ability to store charges. For parallel-plate capacitors, this ability depends the material between the plates, the common plate area and the plate separation. The relationship is

$C=\dfrac{\epsilon A}{d}$

$C$ is the capacitance, $A$ is the common plate area, $d$ is the plate separation and $\epsilon$ is the permittivity of the material between the plates.

For air or free space, $\epsilon$ is $\epsilon_0$ called the permittivity of free space. In general, $\epsilon=\epsilon_r \epsilon_0$ where $\epsilon_r$ is the relative permittivity or dielectric constant of the material between the plates. It is a factor that determines the strength of the material compared to air. In fact, for air or vacuum, $\epsilon_r=1$ .

The energy stored in a capacitor is the average of the product of its charge and voltage.

$U = \dfrac{QV}{2}$

Its charge, $Q$ , is related to its capacitance by $Q=CV$ (this is the electrical definition of capacitance, a ratio of the charge to its voltage; the previous formula is the geometric definition). Substituting this in the formula for $U$ ,

$U = \dfrac{CV^2}{2}$

A. Substituting for $C$ in $U$ ,

$U_0 = \dfrac{\epsilon_0 A V^2}{2d}$

B. When the distance is $3d$ ,

$U_1 = \dfrac{\epsilon_0 A V^2}{2\times3d}$

$U_1 = \dfrac{\epsilon_0 A V^2}{6d}$

C. When the distance is restored but with a dielectric material of dielectric constant, $K$ , inserted, we have

$U_2 = \dfrac{K\epsilon_0 A V^2}{2d}$

6 0

3 years ago

A rocket is launched from the surface of Earth with a speed v0 that will allow the rocket to escape the gravitational field of E

weeeeeb [17]

Answer:

option ( a ) is correct .

Explanation:

Escape velocity on the earth = √ ( 2 GM / R )

where G is universal gravitational constant , M is mass of the earth and R is radius .

V₀ = √ ( 2 GM / R )

escape velocity on the planet where mass is equal is earth's mass and radius is 4 times that of the earth

Radius of the planet = 4 R

escape velocity of planet = √ ( 2 GM / 4R )

= .5 x √ ( 2 GM / R )

= .5 V₀

option ( a ) is correct .

8 0

3 years ago

The earth has a vertical electric field at the surface,pointing down, that averages 102 N/C. This field is maintained by various

Schach [20]

Answer:

$q = -461532.5 \ C$

Explanation:

From the question we are told that

The electric filed is $E = 102 \ N/C$

Generally according to Gauss law

=> $E A = \frac{q}{\epsilon_o }$

Given that the electric field is pointing downward , the equation become

$- E A = \frac{q}{\epsilon_o }$

Here $q$ is the excess charge on the surface of the earth

$A$ is the surface area of the of the earth which is mathematically represented as

$A = 4\pi r^2$

Where r is the radius of the earth which has a value $r = 6.3781*10^6 m$

substituting values

$A = 4 * 3.142 * (6.3781*10^6 \ m)^2$

$A =5.1128 *10^{14} \ m^2$

So

$q = -E * A * \epsilon _o$

Here $\epsilon_o$ s the permitivity of free space with value

$\epsilon_o = 8.85*10^{-12} \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2$

substituting values

$q = -102 * 5.1128 *10^{14} * 8.85 *10^{-12}$

$q = -461532.5 \ C$

6 0

4 years ago

We decided to make an iced latte by adding ice to a 200 mL hot latte at 45 °C. The ice starts out at 0 C. How much ice do we nee

tankabanditka [31]

Answer:

m = 77.75 g

Explanation:

Here we know that at equilibrium the temperature of the system will be 10 degree C

so heat given by hot latte = heat absorbed by the ice

now we have

heat given by latte = $m s\Delta T$

$Q_1 = (200)(4.186)(45 - 10)$

$Q_1 = 29302 J$

now heat absorbed by ice is given as

$Q_2 = mL + ms\Delta T$

$Q_2 = m(335 + 4.186(10 - 0))$

$Q_2 = m(376.86)$

now by heat balance we have

$Q_1 = Q_2$

$29302 = m(376.86)$

$m = 77.75 g$

6 0

3 years ago