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alina1380 [7]
2 years ago
9

A cubical gaussian surface surrounds a long, straight, charged filament that passes perpendicularly through two opposite faces.

No other charges are nearby. (ii) Through how many of the cube's faces is the electric flux zero? Choose from the same possibilities as in part (i).
Physics
1 answer:
laiz [17]2 years ago
7 0

Answer:

The electric flux is zero through four cube surfaces given that a cubical gaussian surface surrounds a long, straight, charged filament that passes perpendicularly through two opposite faces.

Explanation:

Assuming the charged filament is quite long and you are not near the edges, the two opposing sides that the filament travels through have no flux. If the charge filament is long, which you may assume is indefinitely long, then there is the equal amount of charge on the left and right of where you are, therefore the electric field has no preference for left or right. This implies that the electric field can only travel in or out of the filament. No field lines run through the two faces of the cube that the filament goes through if the electric field is not moving left or right. There are electric field lines on the four sides of the filament.

To learn more about cubical gaussian surface and electric flux. Click brainly.com/question/13003911

#SPJ4

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A 10 kg rock that has been dropped from a 60 meter high cliff experiences an average force of air resistance of 30 N. Calculate
lesya692 [45]

Answer:

4086 J

Explanation:

The potential energy is transformed to kinetic energy less the frictional energy. Potential energy= mgh where m represent mass, g is acceleration due to gravity and h is the height of cliff

Since we have force of air resistance, work done due to air resistance will be product of force and distance

mgh-Fh= 0.5mv^{2}= KE

Substituting 10 Kg for m, 9.81 for g and 60 m for F then the kinetic energy at the bottom will be

KE= 10*9.81*60- (30*60)=4086 J

8 0
3 years ago
Which of the following examples does not make use of total internal reflection?
liubo4ka [24]

Answer:

laser pointer does not make use of total internal reflection

7 0
3 years ago
It is correct to say that impulse is equal toA) momentum.B) the change in momentum.C) the force multiplied by the distance the f
Elena-2011 [213]

Answer:

B) the change in momentum.

Explanation:

The impulse is defined as the product between the force applied on an object (F) and the duration of the collision (\Delta t):

J=F \Delta t (1)

We can rewrite the force by using Newton's second law, as the product between mass (m) and acceleration (a):

F=ma

So, (1) becomes

J=ma \Delta t

Now we can also rewrite the acceleration as ratio between the change in velocity and change in time: a=\frac{\Delta v}{\Delta t}. If we substitute into the previous equation, we find

J=m\frac{\Delta v}{\Delta t}\Delta t=m\Delta v

And the quantity m\Delta v is equivalent to the change in momentum, \Delta p.

6 0
3 years ago
Select all of the following that will produce a magnetic field:________.
zysi [14]

Answer:

a ,b , d ,e

Explanation:

took the k12 quiz and got a 100

4 0
2 years ago
An object is thrown upward with some velocity. If the object rises 77.5 m above the point of release, (a) how fast was the objec
jolli1 [7]

Answer:

v_o=39\ m/s\\t_m=4\ s

Explanation:

<u>Vertical Launch Upwards</u>

In a vertical launch upwards, an object is launched vertically up from a height H without taking into consideration any kind of friction with the air.

If vo is the initial speed and g is the acceleration of gravity, the maximum height reached by the object is given by:

\displaystyle h_m=H+\frac{v_o^2}{2g}

The object referred to in the question is thrown from a height H=0 and the maximum height is hm=77.5 m.

(a)

To find the initial speed we solve for vo:

\displaystyle v_o=\sqrt{2gh_m}

v_o=\sqrt{2\cdot 9.8\cdot 77.5}

v_o=39\ m/s

(b)

The maximum time or the time taken by the object to reach its highest  point is calculated as follows:

\displaystyle t_m=\frac{v_o}{g}

\displaystyle t_m=\frac{39}{9.8}

t_m=4\ s

7 0
3 years ago
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