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BARSIC [14]
4 years ago
7

A cube with sides of area 48 cm^2 contains a 28.7 nanoCoulomb charge. Find the flux of the electric field through the surface o

f the cube in unis of Nm^2/C.
Please conceptually explain this question answer to me! Thanks!!
Physics
1 answer:
Anuta_ua [19.1K]4 years ago
5 0

Answer:

The flux of the electric field through the surface is 3.24\times10^{3}\ Nm^/C[/tex].

Explanation:

Given that,

Area of cube = 48 cm²

Charge = 28.7 nC

We need to calculate the flux of the electric field through the surface

Using formula Gauss's law

The electric flux through any closed surface,

\phi =\dfrac{q}{\epsilon_{0}}

Where, q = charge

Put the value into the formula

\phi=\dfrac{28.7\times10^{-9}}{8.85\times10^{-12}}

\phi =3.24\times10^{3}\ Nm^/C

Hence, The flux of the electric field through the surface is 3.24\times10^{3}\ Nm^/C[/tex].

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Answer:

The drop time ball 1 is less than the drop time of ball 2. A further explanation is provided below.

Explanation:

The net force acting on the ball will be:

⇒ F_{net}=mg-F_r

Here,

F = Force

m = mass

g = acceleration

Now,

According to the Newton's 2nd law of motion, we get

⇒ F_{net} = ma

To find the value of "a", we have to substitute "F_{net}=ma" in the above equation,

⇒ ma=mg-F_r

⇒    a=g-\frac{F_r}{m}

We can see that, the acceleration is greater for the greater mass of less for the lesser mass. Thus the above is the appropriate solution.

8 0
3 years ago
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Tres móviles, A, B y C, con las mismas características tienen los siguientes estados: A se encuentra inicialmente en reposo y ca
Morgarella [4.7K]

Answer:

La respuesta es sí, hay una fuerza que actúa sobre el móvil A y es la única fuerza ya que A cae libremente bajo la influencia de la fuerza.

Explanation:

Según la primera ley de movimiento de Newton, un cuerpo continuará en su estado de reposo o en un movimiento uniforme en línea recta a menos que actúen sobre él fuerzas impresas.

Dado que el móvil A cae libremente, desde su estado de reposo inicial, según la primera ley de movimiento de Newton, experimenta una fuerza que actúa sobre él para hacer que caiga y continúe en caída libre.

El móvil B se mueve con una velocidad constante, por lo tanto, de acuerdo con la primera ley de movimiento de Newton, no hay fuerzas impresas que actúen sobre él.

El móvil C está completamente en reposo en el suelo, por lo tanto, tampoco hay fuerzas que actúen sobre él.

La respuesta es sí, hay una fuerza actuando sobre el móvil A y es la única fuerza cuando A cae libremente bajo la influencia de la fuerza.

6 0
4 years ago
A professional baseball player can throw the ball around 45 m/s if the distance between the pitcher and the batter is 18.39 m. H
svetlana [45]

Answer:

The time taken for the ball to get to the batter is 0.41 s.

Explanation:

Given;

initial velocity of the baseball, u = 45 m/s

horizontal distance between the pitcher and the batter, X = 18.39 m

The horizontal distance or range of a projectile is given as;

X = ut

where;

t is the time of flight

u is the initial velocity

t = X / u

t = 18.39 / 45

t = 0.41 s

Therefore, the time taken for the ball to get to the batter is 0.41 s.

6 0
3 years ago
During a blackout, you are trapped in a tall building. you want to call rescuers on your cell phone, but you can't remember whic
Gennadij [26K]

Let the height where we are trapped is H

now to find the time to reach the key at the bottom is given as

y = v_i t + \frac{1}{2}at^2

now we have

H = \frac{1}{2}gt^2

t = \sqrt{\frac{2H}{g}}

now if the speed of sound is considered to be 340 m/s then time taken by the sound to reach at the top is given as

t = \frac{H}{340}

now the total time is given as

\sqrt{\frac{2H}{g}} + \frac{H}{340} = 3.27

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3 0
3 years ago
The electric flux through a square-shaped area of side 5 cm near a very large, thin, uniformly-charged sheet is found to be 3\ti
deff fn [24]

Answer:

Explanation:

Given

side of square shape a=5\ cm

Electric flux \phi =3\times 10^{-5}\ N.m^2/C

Permittivity of free space \epsilon_0=8.85\times 10^{-12} \frac{C^2}{N.m^2}

Flux is given by

\phi =EA\cos \theta

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A=area

\theta=Angle between Electric field and area vector

E=\frac{\phi }{A\cos (0)}

E=\frac{3\times 10^{-5}}{25\times 10^{-4}\times \cos(0)}

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and Electric field  by a uniformly charged sheet is given by

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\sigma =2.12\times 10^{-13}\ C/m^2    

5 0
3 years ago
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