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Sliva [168]
3 years ago
9

In Fig.25-46,how much charge is stored on the parallel-plate capacitors by the 12.0 V battery? One is filled with air,and the ot

her is filled with a dielectric for which k ! 3.00;both capacitors have a plate area of 5.00 $ 10#3 m2 and a plate separation of 2.00 mm.
Physics
1 answer:
Fittoniya [83]3 years ago
6 0

Answer:

Q=7.9\times 10^{-10}\ C

Explanation:

Given that

V= 12 V

K=3

d= 2 mm

Area=5.00 $ 10#3 m2

Assume that

$ = Multiple sign

# = Negative sign

A=5\times 10^{-3}\ m^2

We Capacitance given as

For air

C_1=\dfrac{\varepsilon _oA}{d}

C_1=\dfrac{\varepsilon _oA}{d}

C_1=\dfrac{8.85\times 10^{-12}\times 5\times 10^{-3}}{2\times 10^{-3}}

C_1=2.2\times 10^{-11}\ F

C_2=\dfrac{K\varepsilon _oA}{d}

C_2=\dfrac{3\times 8.85\times 10^{-12}\times 5\times 10^{-3}}{2\times 10^{-3}}\ F

C_2=6.6\times 10^{-11}\ F

Net capacitance

C=C₁+C₂

C=8.8\times 10^{-11}\ F

We know that charge Q given as

Q= C V

Q=12\times 6.6\times 10^{-11}\ C

Q=7.9\times 10^{-10}\ C

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3 0
4 years ago
Mary is an avid game show fan and one of the contestants on a popular game show. She spins the wheel, and after 5.5 revolutions,
spin [16.1K]

Answer:

The angle is 23.2 radians, equivalent to 3.69 revolutions.

Explanation:

First, we need to find the angular acceleration of the wheel. This can be done using one of the kinematic formulas:

\omega^{2}=\omega_0^{2}+2\alpha\theta\\\\\implies \alpha=\frac{\omega^{2}-\omega_0^{2}}{2\theta}

Since the final angular velocity is zero after 5.5 revolutions (equivalent to 11π radians) we have that:

\alpha=\frac{-(3.15rad/s)^{2}}{2(11\pi rad)}\\\\\alpha=-0.144rad/s^{2}

Now, using the same equation, we can solve for the requested angle:

\theta=\frac{\omega^{2}-\omega_0^{2}}{2\alpha}\\\\\theta=\frac{(1.80rad/s)^{2}-(3.15rad/s)^{2}}{2(-0.144rad/s^{2})}\\\\\theta=23.2rad

Finally, it means that the angle through which the wheel has turned when the angular speed reaches 1.80 rad/s is 23.2 radians, equivalent to 3.69 revolutions.

8 0
3 years ago
Which branches of natural science include the study of an organism that lived 10 million years ago?
Verdich [7]

Analyzing the components and decay products of school lunches.

7 0
3 years ago
Read 2 more answers
1. What is the momentum of a 0.15 kg arrow that is traveling at 120 m/s?
Korvikt [17]

For this case we have that by definition, the momentum equation is given by:

p = m * v

Where:

m: It is the mass

v: It is the velocity

According to the data we have:

m = 0.15 \ kg\\v = 120 \frac {m} {s}

Substituting:

p = 0.15 * 120\\p = 18 \frac {kg * m} {s}

On the other hand, if we clear the variable "mass" we have:

m = \frac {p} {v}

According to the data we have:

p = 250 \frac {kg * m} {s}\\v = 5 \frac {m} {s}\\m = \frac {250} {5}\\m = 50 \ kg

Thus, the mass is 50 \ kg

Answer:

p = 18 \frac {kg * m} {s}\\m = 50 \ kg

3 0
3 years ago
Review. (c) Assume the dipole is a compass needle-a light bar magnet-with a magnetic moment of magnitude μ . It has moment of in
hammer [34]

The frequency of oscillation is \frac{1}{2\pi }  \sqrt{\frac{\mu B}{I} }.

<h3>What is a magnetic moment?</h3>

The magnetic moment is the magnetism of a magnet or other item that creates a magnetic field, as well as its orientation and strength. Electromagnets, permanent magnets, elementary particles like electrons, different compounds, and a variety of celestial objects are examples of things that have magnetic moments (such as many planets, some moons, stars, etc). The phrase "magnetic moment" typically refers to a system's magnetic dipole moment, which can be represented by an analogous magnetic dipole, which has a magnetic north and south pole that are barely separated from one another. For sufficiently small magnets or at sufficiently wide distances, the magnetic dipole component is adequate. For extended objects, additional terms may be required in addition to the dipole moment, such as the magnetic quadrupole moment.

T = \frac{ Id^{2}\theta }{dt^{2} }

-\mu B\theta=\frac{ Id^{2}\theta }{dt^{2} }

\frac{d^{2}\theta }{dt^{2} } = -(\frac{\mu BI}{\theta} )

By Comparing the above equation with the SHM equation

\frac{d^2 \theta} {dt^{2} } = -\omega^{2} \theta

\omega^{2} =\frac{ \mu B}{I}

Frequency = \frac{\mu}{2\pi }

=\frac{\sqrt{\frac{\mu B}{I} } }{2\pi}

=\frac{1}{2\pi }  \sqrt{\frac{\mu B}{I} }

To learn more about a magnetic moment, visit:

brainly.com/question/17000031

#SPJ4

8 0
1 year ago
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