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3 years ago

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A very large sheet of a conductor carries a uniform charge density of on its surfaces. What is the electric field strength 3.00

mm outside the surface of the conductor?

1 answer:

olga55 [171]3 years ago

5 0

Complete Question

A very large sheet of a conductor carries a uniform charge density of $4.00\ pC/mm^2$ on its surfaces. What is the electric field strength 3.00 mm outside the surface of the conductor?

Answer:

The electric field is $E = 4.5198 *10^{5} \ N/C$

Explanation:

From the question we are told that

The charge density is $\sigma = 4.00pC /mm^2 = 4.00 * 10^{-12 } * 10^{6} = 4.00 *10^{-6}C/m$

The position outside the surface is $a = 3.00 \ mm = 0.003 \ m$

Generally the electric field is mathematically represented as

$E = \frac{\sigma}{\epsilon _o }$

Where $\epsilon_o$ is the permitivity of free space with values $\epsilon _o = 8.85 *10^{-12} F/m$

substituting values

$E = \frac{4.0*10^{-6}}{8.85 *10^{-12} }$

$E = 4.5198 *10^{5} \ N/C$

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Radio wave radiation falls in the wavelength region of 10.0 to 1000 meters. What is the energy of radio wave radiation that has

dmitriy555 [2]

Answer:

Explanation:

Given

Wavelength of radiation $\lambda =784\ m$

We know Energy of wave with wavelength $\lambda$ is given by

$E=\frac{hc}{\lambda }$

where h=Planck's constant

c=velocity of light

$\lambda$ =wavelength of wave

$E=\frac{6.626\times 10^{-34}\times 3\times 10^8}{784}$

$E=2.53\times 10^{-28}\ J$

Hence the energy of the wave with wavelength 784 m is $2.53\times 10^{-28}\ J$

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3 years ago

Here is something to blow your mind. In one slice of "everything" pizza there are 650 Calories (1000 chemistry calories). That e

amid [387]

Answer:

Number of slices of pizza is 993.

Explanation:

It is given that, in one slice of "everything" pizza there are 650 Calories. The conversion factor from calories to joules is :

1 calorie = 4.184 joules

650 calories = 2719.6 joules

Total energy in the pizza, E = 2700000 J

Let there are n number of slices of pizza. It is given by :

$n=\dfrac{2700000}{2719.6}$

n = 992.79

or

n = 993

So, there are 993 slices of pizza. Hence, this is the required solution.

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3 years ago

A very thin circular disk of radius R = 20.00 cm has charge Q = 30.00 mC uniformly distributed on its surface. The disk rotates

lutik1710 [3]

Answer:

$B= 7.5*10^{-15}T$

Explanation:

The magnetic field strenght on the z-axis at a distance d from the center is,

$B= \frac{\mu_0 Q\omega R^2}{8\pi d^3}$

Our values are:

$R=20cm\\Q=30mc\\w=5rad/s\\d=2*10^3cm=20m$

Replacing,

$B= \frac{(4\pi*10^{}-7)(30*10^{-3})(5)(0.2)^2}{8\pi(20)}$

$B= 7.5*10^{-15}T$

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3 years ago

Which of the following describes the relationship between the weight of fluid

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The answer is archimedes principle

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3 years ago

A student sits on a rotating stool holding two 3.09-kg masses. When his arms are extended horizontally, the masses are 1.08 m fr

schepotkina [342]

Answer:

a

The New angular speed is $w_f = 2.034 rad/s$

b

The Kinetic energy before the masses are pulled in is $KE_i = 3.101 \ J$

c

The Kinetic energy after the masses are pulled in is $KE_f = 8.192 \ J$

Explanation:

From the we are told that masses are 1.08 m from the axis of rotation, this means that

The radius $r =1.08m$

The mass is $m = 3.09\ kg$

The angular speed $w = 0.770 \ rad/sec$

The moment of inertia of the system excluding the two mass $I = 3.25 \ kg \cdot m^2$

New radius $r_{new} = 0.34m$

Generally the conservation of angular momentum can be mathematical represented as

$w_f = [\frac{I_i}{I_f} ]w_i .....(1)$

Where $w_f$ is the final angular speed

$w_i$ is the initial angular speed

$I_i$ is the initial moment of inertia

$I_f$ is the final moment of inertia

Moment of inertia is mathematically represented as

$I = m r^2$

Where I is the moment of inertia

m is the mass

r is the radius

So the Initial moment of inertia is given as

$I_i = moment \ of \ inertia \ of\ the \ two \ mass \ + 3.25 \ kg \cdot m^2$

$I_i = 2m r^2 + 3.25$

The multiplication by is because we are considering two masses

$I_i = 2 [(3.09)(1.08)^2] +3.25 = 10.46 kg \cdot m^2$

So the final moment of inertia is given as

$I_f = 2[(3.09)(0.34)^2] +3.25 = 3.96 \ kg \cdot m^2$

Substituting these values into equation 1

$w_f = [\frac{10.46}{3.96} ] * 0.77 = 2.034 \ rad/sec$

Generally Kinetic energy is mathematically represented in term of moment of inertia as

$KE = \frac{1}{2} * I * w^2$

Now considering the kinetic energy before the masses are pulled in,

$KE_i = \frac{1}{2} * I_i * w^2_i$

The Moment of inertia would be $I_i = 10.46 \ Kg \cdot m^2$

The Angular speed would be $w_i = 0.77 \ rad/s$

Now substituting these value into the equation above

$KE_i = \frac{1}{2} * (10.46) * (0.770)^2 = 3.101 J$

Now considering the kinetic energy after the masses are pulled in,

$KE_f = \frac{1}{2} * I_f * w^2_f$

The Moment of inertia would be $I_f = 3.96 \ Kg \cdot m^2$

The Angular speed would be $w_f = 2.034 \ rad/s$

Now substituting these value into the equation above

$KE_f= \frac{1}{2} *(3.96)(2.034)^2$

$= 8.192J$

8 0

3 years ago