All categories

2 years ago

A piece of plastic is uniformly charged with surface charge density n1.

Physics

1 answer:

Free_Kalibri [48]2 years ago

4 0

Answer:

Explanation:

Since the definition of density is mass per unit volume, the surface density has to do with the area of the surface.

If a piece of plastic is uniformly charged with surface charge density n1, that means that the charge density will be charged per unit area.

The magnitude of the charge depends on the surface areas.

As the surface areas reduce, the magnitude of the charge will be increasing.

Ranking in order, from largest to smallest, the surface charge densities n1 to n3, the correct option will be option E which is n3>n2>n1

You might be interested in

A 1090 kg car has four 12.7 kg wheels. When the car is moving, what fraction of the total kinetic energy of the car is due to ro

max2010maxim [7]

Answer:

$\frac{KE_{Rotational}}{KE_{Total}} = 0.018$

Explanation:

To develop this exercise we proceed to use the kinetic energy equations,

In the end we replace

$KE_{Total}=KE_{Translational}+KE_{Rotational}$

$KE_{Total}=\frac{1}{2}m_{car}+4*\frac{1}{2}*I*(\frac{v}{r})^2$

Here

$I=\frac{1}{2}m_{wheels}*r^2$ meaning the 4 wheels,

So replacing

$KE_{Rotational}=4\frac{1}{2}*(\frac{1}{2}m_{wheels}*r^2)*(\frac{v}{r})^2=m*v^2$

So,

$\frac{KE_{Rotational}}{KE_{Total}} = \frac{m_{wheels}*v^2}{\frac{1}{2}m_{car}*v^2+m_{wheels}*v^2}$

$\frac{KE_{Rotational}}{KE_{Total}} = \frac{m_{wheels}}{\frac{1}{2}m_{car}+m_{wheels}}$

$\frac{KE_{Rotational}}{KE_{Total}} = \frac{10}{545+10}$

$\frac{KE_{Rotational}}{KE_{Total}} = 0.018$

3 0

3 years ago

A typical atomic polarizability is 1 × 10-40 (C·m)/(N/C). If the q in p = qs is equal to the proton charge e, what charge separa

Alenkasestr [34]

Answer:

s = 6.25 10⁻²² m

Explanation:

Polarizability is the separation of electric charges in a structure, in the case of the atom it is the result of the separation of positive charges in the nucleus and the electrons in their orbits, macroscopically it is approximated by

p = q s

s = p / q

let's calculate

s = 1 10⁻⁴⁰ / 1.6 10⁻¹⁹

s = 0.625 10⁻²¹ m

s = 6.25 10⁻²² m

We see that the result is much smaller than the size of the atom, therefore this simplistic model cannot be taken to an atomic scale.

6 0

3 years ago

330 g of water at 55°c are poured into an 855 g aluminum container with an initial temperature of 10°

Olenka [21]

The final temperature of the system is 32.5°
we know, H = mcT
where, H = Heat content of the body
m = Mass,
c = Specific heat
T = Change in temperature
According to to the Principle of Calorimetry

The net heat remains constant i.e.
⇒ the heat given by water = heat accepted by the aluminum container.
⇒ 330 x 1 x (45 - T) = 855 x $\frac{900}{4200}$ x (T - 10)

⇒ 14,850 - 330T = 183.21T - 1832

⇒ - 513.21 T = - 16682

or T = 32.5°

3 0

2 years ago

Someone help this is a test and i need to finish this quick i will give brainliest of it lets me

tekilochka [14]

Sorry, I don't know but I think the correct answer is the first option.

7 0

2 years ago

NIST 800-14's Principles for Securing Information Technology Systems, can be used to make sure the needed key elements of a succ

mariarad [96]

Answer:

True

Explanation:

-NIST 800-14's are generally accepted principles for securing information technology systems.

- The defined principles, if adhered to and continously improved, will will ensure sytem security over it's lifetime as desired.

5 0

3 years ago