A cube of side 3.56 cm has a charge of 9.11 μ C placed at its center. Calculate the electric flux through one side of the cube

Answers

Answer 1

Answer:

Answer:

$\phi_(surface) = 1.72 * 10^5 N/Cm^2$

Explanation:

As we know by the theory of flux that if charge "q" is enclosed in a closed surface then total flux through the surface is given as

$\phi = (q)/(\epsilon_0)$

so here total flux passing through the cube is given as

$\phi_(total) = (9.11\muC)/(8.85 * 10^(-12))$

now we will have

$\phi_(total) = 1.03* 10^6$

now as we know that charge is placed at the center of the cube

so the total flux is uniformly passing through all faces of the cube

so flux passing one side of the cube is given as

$\phi_(surface) = (1.03* 10^6)/(6)$

$\phi_(surface) = 1.72 * 10^5 N/Cm^2$

Answer 2

Answer: Flux=q/e0
e0=8.85*10^-12
!!! Cube has 6 sides, so flux through one side is equal to total flux/6
Ans=9.11*10^(-6)/((8.85*10(-12))*6)=(approximately)1.72*10^5Nm^2/C

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Charge q is 1 unit of distance away from the source charge S. Charge p is two times further away. The force exerted between S and q is _____ the force exerted between S and p. a. 1/2
b. 2 times
c. 1/4
d. 4 times

Answers

Answer : The correct option is, (d) 4 times

Solution :

According to the Coulomb's law, the electrostatic force of attraction or repulsion between two charges is directly proportional to the product of the charges and is inversely proportional to the square of the distance between the the charges.

Formula used :

$F=k_e(q_1q_2)/(r^2)$

where,

F = electrostatic force of attraction or repulsion

$k_e$ = Coulomb's constant

$q_1$ and $q_2$ are the charges

r = distance between two charges

First we have to calculate the force exerted between S and q when the distance between the charge is 1 unit and let us assumed that the charge be 'q'

$F_(sq)=k_e(qq)/(1^2)=k_e* q^2$ ..........(1)

Now we have to calculate the force exerted between S and p when the distance between the charge is 2 unit at the same charge.

$F_(sp)=k_e(qq)/(2^2)=k_e(q^2)/(4)$ ...........(2)

Equation equation 1 and 2, we get

$(F_(sq))/(F_(sp))=(1)/(4)$

$F_(sq)=4* F_(sp)$

Therefore, the force exerted between S and q is 4 times the force exerted between S and p.

I believe the answer is "4 times".

Give two examples of workplace environments where considerations must be made with respect to the possibility of electric discharges. Explain why this is a necessary concern

Answers

Answer:

1.) Oil and gas environment

2.) Mechanical engineering and electrical environment.

Explanation:

1.) Oil and gas environment.

Electricity and electric discharge are recognized as serious workplace hazard in oil and gas companies, exposing employees to burns, fires, and explosions.

2.) Mechanical engineering and electrical environment.

Electricity or electric discharge has long been recognized as a serious workplace hazard, exposing employees to electric shock, electrocution, burns, fires, and explosions. Workers could die from electrocutions at work. This is

accounting for on-the-job fatalities and what makes these more tragic is that most of these fatalities could have been easily avoided.

If the speed of an object is tripled, its kinetic energy will be

Answers

If the speed of an object is tripled, its kinetic energy will be 9 times the initial value.

What is kinetic energy?

The kinetic energy of an object is the energy possessed by the object due to its motion.

K.E = ¹/₂mv²

where;

m is mass of the object

v is speed of the object

When the speed is tripled

K.E = ¹/₂m(3v)²

K.E = 9(¹/₂mv²)

Thus, if the speed of an object is tripled, its kinetic energy will be 9 times the initial value.

Learn more about kinetic energy here: brainly.com/question/25959744

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A puck moves 2.35 m/s in a -22° direction. A hockey stick pushes it for 0.215 s, changing its velocity to 6.42 m/s in a 50.0° direction. What was the direction of the acceleration?

Answers

The puck starts with velocity vector

$\vec v_0=\left(2.35(\rm m)/(\rm s)\right)(\cos(-22^\circ)\,\vec\imath+\sin(-22^\circ)\,\vec\jmath)=(2.18\,\vec\imath-0.880\,\vec\jmath)(\rm m)/(\rm s)$

Its velocity at time $t$ is

$\vec v=\vec v_0+\vec at$

Over the 0.215 s interval, the velocity changes to

$\vec v=\left(6.42(\rm m)/(\rm s)\right)(\cos50.0^\circ\,\vec\imath+\sin50.0^\circ\,\vec\jmath)=(4.13\,\vec\imath+4.92\,\vec\jmath)(\rm m)/(\rm s)$

Then the acceleration must have been

$\vec v=\vec v_0+(0.215\,\mathrm s)\vec a\implies\vec a=(\vec v-\vec v_0)/(0.215\,\rm s)=(9.06\,\vec\imath+27.0\,\vec\jmath)(\rm m)/(\mathrm s^2)$

which has a direction of about $71.4^\circ$ .

Final answer:

The direction of the acceleration is determined by the direction of the change in velocity. This would be calculated by subtracting the initial velocity vector from the final velocity vector. However, the calculation would involve complex trigonometric functions.

Explanation:

In order to find the direction of the acceleration, we need to calculate the direction of the change in velocity and that direction will be the direction of the acceleration.

To calculate the change in velocity, we subtract the initial velocity from the final velocity: (6.42 m/s, 50.0°) - (2.35 m/s, -22°). We then calculate the angle of this vector which represents the change in velocity, and hence the direction of acceleration.

However, this calculation is not straightforward because it involves vector operations and would require the use of trigonometric functions to solve. This is due to the fact that velocity is a vector, meaning it has both a magnitude and a direction. Consequently, this becomes a multi-step process involving trigonometry and physics.

Learn more about Direction of Acceleration here:

brainly.com/question/33720661

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What is single kind of organism that can reproduce on its own?

Answers

A cell will reproduce on its own

A single cell organism can reproduce on its own, and a cell can reproduce on its own

How do you do a proper push up and sit-up

Answers

How to do PUSH UPS:

You put your hands on the floor. Lift your arms repeatedly without using your legs.

How to do SIT UPS:

Sit on the floor. Lay down with your knees up. Get up and down repeatedly without using your waist.

(Hope this helps!! LIKE A BOSS!!!!!!)

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