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2 years ago

Similarities between the flow of water and an electric circuit.

Physics

1 answer:

krek1111 [17]2 years ago

3 0

They both flow in currents. Water has a pump that works like a battery and pipes that work like a circuit.

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A solid conducting sphere with radius R that carries positive charge Q is concentric with a very thin insulating shell of radius

chubhunter [2.5K]

Answer:

a) 0 < r < R: E = 0, R < r < 2R: E = KQ/r^2, r > 2R: E = 2KQ/r^2

b) See the picture

Explanation:

We can use Gauss's law to find the electric field in all the regions:

EA = qen/e0 where qen is the enclosed charge

Remember that the electric field everywhere outside a sphere is:

E(r) = q/(4*pi*eo*r^2) = Kq/r^2

For 0 < r < R: There is not enclosed charge because all of it remains on the outer layer of the conducting sphere, therefore E = 0 EA = 0/e0 = 0 E = 0
For R < r < 2R: Here the enclosed charge is equal Q E = Q/(4*pi*eo*r^2) = KQ/r^2
For r > 2R: Here the enclosed charge is equal 2Q E = Q/(4*pi*eo*r^2) + Q/(4*pi*eo*r^2) = 2Q/(4*pi*eo*r^2) = 2KQ/r^2

b) At the beginning there is no electric field this is why you see a line in zero, In R the electric field is maximum and then it starts to decrease exponentially with the distance and finally in 2R the field increase a little due to the second sphere to then continue decreasing exponentially with the distance

7 0

2 years ago

A car with an initial speed of 6.5 m/s accelerates at a uniform rate of 0.92 m/s2 for 3.6 s. Find the final speed of the car dur

geniusboy [140]

U = 6.5 m/s, initial speed
t = 3.6 s, time
a = 0.92 m/s², acceleration

Let v = the final velocity.
Then
v = u +at
v = (6.5 m/s) + (0.92 m/s²)*(3.6 s) = 9.812 m/s

Answer: 9.81 m/s

8 0

2 years ago

contrast the location of genetic material in bacteria cells to its location in plant and animal cells.

Agata [3.3K]

First bacterial cells are usually much smaller than plant or animal cells. a human skin cells , for example is about times as larger as an average bacterial cell.

6 0

3 years ago

The Sun orbits the center of the Milky Way galaxy once each 2.60 × 108 years, with a roughly circular orbit averaging 3.00 × 104

Mamont248 [21]

To solve this problem it is necessary to apply the kinematic equations of linear and angular motion, as well as the given definitions of the period.

Centripetal acceleration can be found through the relationship

$a_c = \frac{v^2}{R}$

Where

v = Tangential Velocity

R = Radius

At the same time linear velocity can be expressed in terms of angular velocity as

$v = R\omega$

Where,

R = Radius

$\omega =$ Angular Velocity

PART A) From this point on, we can use the values used for the period given in the exercise because the angular velocity by definition is described as

$\omega = \frac{2\pi}{T}$

T = Period

So replacing we have to

$\omega = \frac{2\pi}{2.6*10^8years}\\\omega = 2.4166*10^{-8}rad/years\\\omega = 2.4166*10^{-8}rad/years(\frac{1years}{365days})(\frac{1day}{86400s})\\\omega = 7.663*10^{-16}rad/s$

Since $1 Light year = 9.48*10^{15}m$

Then the radius in meters would be

$R = (3*10^4ly)(\frac{9.48*10^{15}m}{1ly})$

$R = 2.844*10^{20}m$

Then the centripetal acceleration would be

$a_c = \frac{v^2}{R}\\a_c = \frac{(R\omega)^2}{R}\\a_c = R\omega^2 \\a_c = 2.844*10^{20}(7.663*10^{-16})^2\\a_c = 1.67*10^{-10}m/s^2$

From the result obtained, considering that it is an unimaginably low value of an order of less than $10^{-10}$ it is possible to conclude that it supports the assertion on the inertial reference frame.

8 0

2 years ago

Its volume is 20 cm3, and its mass is 100 grams. What is the sample’s density?

steposvetlana [31]

Well the density is mass/volume
100/20 = 5
The density is 5

3 0

2 years ago