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coldgirl
17 days ago
8

A sphere of radius 0.03m has a point charge of q= 7.6 micro C located at it’s centre. Find the electric flux through it?

Physics
1 answer:
Sav [1.1K]17 days ago
7 0

Answer:

The electric flux going through the sphere is 8.58 *10^{5} \frac{Nm^2}{C}

Explanation:

Given

Radius,\ r = 0.03m\\Charge,\ q =7.6\µC

Required

Calculate the electric flux

Electric flux can be computed using the formula;

Ф = q/ε

Where ε stands for the electric constant permittivity

ε = 8.8542 * 10^{-12}

Substituting ε = 8.8542 * 10^{-12} and q =7.6\µC; the formula simplifies to

Ф = \frac{7.6\µC}{8.8542 * 10^{-12}}

Ф = \frac{7.6 * 10^{-6}}{8.8542 * 10^{-12}}

Ф = \frac{7.6}{8.8542} *\frac{10^{-6}}{10^{-12}}

Ф = \frac{7.6}{8.8542} *10^{12-6}}

Ф = 0.85834970974 *10^{12-6}}

Ф = 0.85834970974 *10^{6}}

Ф = 8.5834970974 *10^{5}}

Ф = 8.58 *10^{5} \frac{Nm^2}{C}

Thus, the electric flux through the sphere is 8.58 *10^{5} \frac{Nm^2}{C}

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Answer:

If the starting and ending points are identical, the overall work equals zero.

Explanation:

Option (D) is correct.

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Consult Conceptual Example 9 in preparation for this problem. Interactive LearningWare 6.3 also provides useful background. The
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Answer:

11.56066 m/s

Explanation:

m = Mass of individual

v = Velocity of individual = 13.4 m/s

g = Gravitational acceleration = 9.81 m/s²

v' = Velocity of the individual after dropping

At the surface, kinetic and potential energy will equalize

\dfrac{1}{2}mv^2=mgh\\\Rightarrow h=\dfrac{v^2}{2g}\\\Rightarrow h=\dfrac{13.4^2}{2\times 9.81}\\\Rightarrow h=9.15188\ m

The cliff's height is 9.15188 m

Define fall height as h' = 2.34 m

\dfrac{1}{2}mv'^2+mgh'=mgh\\\Rightarrow v'=\sqrt{2g(h-h')}\\\Rightarrow v'=\sqrt{2\times 9.81(9.15188-2.34)}\\\Rightarrow v'=11.56066\ m/s

The person's speed is 11.56066 m/s

3 0
6 days ago
A bowling pin is thrown vertically upward such that it rotates as it moves through the air, as shown in the figure. Initially, t
Ostrovityanka [942]

Answer:

Explanation:

The equation used to determine the maximum height of the bowling pin during its trajectory is given by;

H = u²/2g

where u, the initial speed/velocity, equals 10m/s

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Substituting in the values gives us

H = 10²/2(9.81)

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3 0
13 days ago
A charge of 4.9 x 10-11 C is to be stored on each plate of a parallel-plate capacitor having an area of 150 mm2 and a plate sepa
Keith_Richards [1034]

Answer:2.53*10^-10F

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C=£o£r*A/d

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£o= 8.85*10^-12f/m

£r=6.3

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d=3.3mm= 0.0033m

C=8.85*10^-12*6.3*0.015/0.0033

C=8.85*6.3*10^-12*0.015/0.0033

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7 0
1 day ago
1)After catching the ball, Sarah throws it back to Julie. However, Sarah throws it too hard so it is over Julie's head when it r
Softa [913]

Answer:

1)

v_{oy}=11.29\ m/s

2)

y=7.39\ m

Explanation:

Projectile Motion

When an object is projected near the surface of the Earth at an angle \theta to the horizontal, it follows a trajectory known as a parabola. The only force acting on it (ignoring wind resistance) is gravity, affecting the vertical axis.

The height of a projectile can be calculated using

\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}

where y_o represents the initial height from ground level, v_{oy} is the vertical component of the initial velocity, and t is the elapsed time.

The vertical speed component is expressed as

v_y=v_{oy}-gt

1) To proceed, we will determine the initial vertical velocity component since we lack sufficient data to calculate the absolute value of v_o.

The peak height is attained when v_y=0, which allows us to compute the time to reach that height.

v_{oy}-gt_m=0

Solving for t_m

\displaystyle t_m=\frac{v_{oy}}{g}

Thus, the maximum height reached is

\displaystyle y_m=y_o+\frac{v_{oy}^2}{2g}

We know this value is equal to 8 meters

\displaystyle y_o+\frac{v_{oy}^2}{2g}=8

Continuing with the calculations for v_{oy}

\displaystyle v_{oy}=\sqrt{2g(8-y_o)}

Substituting known values yields

\displaystyle v_{oy}=\sqrt{2(9.8)(8-1.5)}

\displaystyle v_{oy}=11.29\ m/s

2) At t=1.505 seconds, the ball is positioned above Julie’s head; we can calculate

\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}

\displaystyle y=1.5+(11.29)(1.505)-\frac{9.8(1.505)^2}{2}

\displaystyle y=1.5\ m+16,991\ m-11.098\ m

y=7.39\ m

5 0
12 days ago
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