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3 years ago

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A point charge of 4 µC is located at the center of a sphere with a radius of 25 cm. Find the electric flux through the surface o

f the sphere. The Coulomb constant is 8.98755 × 109 N · m2 /C 2 and the acceleration due to gravity is 9.8 m/s 2 .Answer in units of N · m2 /C.

1 answer:

givi [52]3 years ago

5 0

Answer:

Ф = 4.5176x10⁵ N . m² / C

Explanation:

In this case, we need to use two expressions in order to calculate the electric flux of the sphere.

First the Electric flux is calculated using this expression:

Ф = E * A (1)

Where:

E: Electric field

A: Area of the sphere

To get the electric field E, we use this expression:

E = K * q / r² (2)

If we replace (2) into (1) we have the following:

Ф = K * q * A / r² (3)

Finally, we need to know the expression to get the area of a sphere which is the following:

A = 4πr² (4)

Replacing into (3):}

Ф = K * Q * 4πr² / r² discarting r²:

Ф = K * Q * 4π (5)

Now, all we need to do is replace the given values and solve for the electric flux of the sphere:

Ф = 8.98755x10⁹ * 4x10⁻⁶ * 4 * π

<h2>Ф = 4.5176x10⁵ N . m² / C</h2>

Hope this helps

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In this case we have a gravitation force $F_{1}=16units$ , given by the formula written at the beginning. Let’s rename the distance $r$ as $d$ :

$F_{1}=G\frac{m_{1}m_{2}}{d^2}$ (1)

And we are asked to find the gravitation force $F_{2}$ with a given distance of $\frac{d}{2}$ :

$F_{2}=G\frac{m_{1}m_{2}}{({\frac{d}{2})}^{2}}$

$F_{2}=G\frac{m_{1}m_{2}}{{\frac{d^{2}}{4}}}$ (2)

The gravity constant is the same for both equations, and we are assuming both masses are constants, as well. So, let’s isolate $G m_{1}m_{2}$ in both equations:

From (1):

$Gm_{1}m_{2}=F_{1}{d}^{2}$ (3)

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$Gm_{1}m_{2}=F_{2}\frac{{d}^{2}}{4}$ (4)

If (3)=(4):

$F_{1}{d}^{2}=F_{2}\frac{{d}^{2}}{4}$ (5)

Now we have to find $F_{2}$ :

$F_{2}=F_{1}{d}^{2}\frac{4}{{d}^{2}}$

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$F_{2}=64 units$ >>>>This is the new force of attraction

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How long did it take Sally riding her bike to slow down from 18 km/h to 6 km/h if the deceleration was at a rate of 4 km/h/s?

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2 years ago

A car is traveling at a speed of 45 km/h into town. It takes the car 2 hours to get there. How far has the car traveled?

hjlf

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Explanation:

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3 years ago

Read 2 more answers

Determine the number of atoms per unit cell in a (a) face-centered cubic, (b) body- centered cubic, and (c) diamond lattice.

fredd [130]

Answer:

a) 4

b) 2

c) 8

Explanation:

In a cubic lattice, each atom of the vertex is shared among other 8 unit cells, so the atoms on the vertex contribute to 1/8 to a given unit cell.

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Therefore, the total atoms are:

8*(1/8) + 6*(1/2) = 4

b) In the body centered cubic structure, the centered atom is not shared with another cell, therefore it contribute to 1 to the given cell:

The number of atoms per unit cel is:

8*(1/8) + 1 = 2

c) The diamond lattice is similar to the face-centered cubic lattice but it contains two identical atoms per lattice point.

Therefore it must contain twice atoms than the face-centered cubic lattice:

that is, it has 8 atoms per unit cell

3 0

4 years ago

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Answer:

<em>a. 10 years</em>

<em>b. 6 years</em>

<em>c. 12 years behind</em>

<em></em>

Explanation:

speed of light c = 3x $10^{8}$ m/s

distance of sirius from earth = 8 lightyears = 8 x c x 60 x 60 x 24 x 365 = 252288000c meters

speed of spaceship = 0.8c

time taken to reach sirius = distance covered divided by speed

t = d/s = 252288000c/0.8c = 315360000 sec

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315360000 sec = 315360000/31536000 = <em>10 yrs</em>

<em>the time as observed by Goslo will be shorter than the real time as measured within the ship due to relativistic effects. using the relativistic relation for time dilation we have</em>

t' = t $\sqrt{1 - \frac{v^{2} }{c^{2} } }$

since v = 0.8c

v/c = 0.8

where t' is the relativistic time as observed by Goslo,

t is the time within the ship

t' = 10 x $\sqrt{1 - 0.8^{2} }$ = <em>6 yrs</em>

<em></em>

If speedo turns around immediately after reaching sirius, his total distance will be 252288000c x 2 = 50457600c meter

time taken = 50457600c/0.8 = 630720000 sec = 20 yrs

time within the ship will be calculated from time dilation

t' = t $\sqrt{1 - \frac{v^{2} }{c^{2} } }$

t' = 20 x $\sqrt{1 - 0.8^{2} }$ = 378432000 sec = <em>12 years behind</em>

4 0

3 years ago