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gayaneshka [121]
3 years ago
10

A point charge of 6.0 nC is placed at the center of a hollow spherical conductor (inner radius = 1.0 cm, outer radius = 2.0 cm)

which has a net charge of –4.0 nC. Determine the resulting charge density on the inner surface of the conducting sphere.
Physics
1 answer:
egoroff_w [7]3 years ago
5 0

Explanation:

The given data is as follows.

             q = 6.0 nC = 6 \times 10^{-9} C

         inner radius (r) = 1.0 cm = 0.01 m   (as 1 cm = 100 m)

So, there will be same charge on the inner surface as the charge enclosed with an opposite sign.

Formula to calculate the charge density is as follows.

            \sigma = \frac{q_{in}}{A} .......... (1)

Since, area of the sphere is as follows.

               A = 4 \pi r^{2} ........... (2)

Hence, substituting equation (2) in equation (1) as follows.

      \sigma = \frac{q_{in}}{4 \pi r^{2}}

                   = \frac{6 \times 10^{-9} C}{4 \times 3.14 \times (0.01)^{2}}            

                   = 0.477 \times 10^{-5}

or,               = 4.77 \mu C/m^{2}

Thus, we can conclude that the resulting charge density on the inner surface of the conducting sphere is 4.77 \mu C/m^{2}.

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6050 J is the kinetic energy at D

<u>Explanation:</u>

In physics, the object's kinetic energy (K.E) defined as the energy it possesses during movement. It can be defined as the required work to accelerate a certain body weight in order to rest at a certain speed. When the body receives this energy as it speeds up (accelerates), it retains this energy unless speed varies. The equation is given as,

           K . E=\frac{1}{2} \times m \times v^{2}

Where,

m - mass of an object

v - velocity of the object

Here,

Given data:

m  = 100 kg

v = 11 m/s

By substituting the given values in the above equation, we get

            K . E=\frac{1}{2} \times 100 \times(11)^{2}=\frac{1}{2} \times 100 \times 121=\frac{12100}{2}=6050\ \mathrm{J}

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Mengapa indra peraba tidak dapat digunakan untuk membandingkan suhu dengan tepat?
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If the force on a hammer is 24 N and its mass is 1.6 kg, then the
Gnesinka [82]

Answer:

15 m/s²

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6 0
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Suppose you first walk 12.0 m in a direction 20? west of north and then 20.0 m in a direction 40.0? south of west. how far are y
Gnesinka [82]
The representation of this problem is shown in Figure 1. So our goal is to find the vector \overrightarrow{R}. From the figure we know that:

\left | \overrightarrow{A} \right |=12m \\ \\ \left | \overrightarrow{B} \right |=20m \\ \\ \theta_{A}=20^{\circ} \\ \\ \theta_{B}=40^{\circ}

From geometry, we know that:

\overrightarrow{R}=\overrightarrow{A}+\overrightarrow{B}

Then using vector decomposition into components:

For \ A: \\ \\ A_x=-\left | \overrightarrow{A} \right |sin\theta_A=-12sin(20^{\circ})=-4.10 \\ \\ A_y=\left | \overrightarrow{A} \right |cos\theta_A=12cos(20^{\circ})=11.27 \\ \\ \\ For \ B: \\ \\ B_x=-\left | \overrightarrow{B} \right |cos\theta_B=-20cos(40^{\circ})=-15.32 \\ \\ B_y=-\left | \overrightarrow{B} \right |sin\theta_B=-20sin(40^{\circ})=-12.85

Therefore:

R_x=A_x+B_x=-4.10-15.32=-19.42m \\ \\ R_y=A_y+B_y=11.27-12.85=-1.58m

So if you want to find out <span>how far are you from your starting point you need to know the magnitude of the vector \overrightarrow{R}, that is:
</span>
\left | \overrightarrow{R} \right |=&#10;\sqrt{R_x^2+R_y^2}=\sqrt{(-19.42)^2+(-1.58)^2}=\boxed{19.48m}

Finally, let's find the <span>compass direction of a line connecting your starting point to your final position. What we are looking for here is an angle that is shown in Figure 2 which is an angle defined with respect to the positive x-axis. Therefore:

</span>\theta_R=180^{\circ}+tan^{-1}(\frac{\left | R_y \right |}{\left | R_x \right |}) \\ \\ \theta_R=180^{\circ}+tan^{-1}(\frac{1.58}{19.42}) \\ \\ \theta_R=180^{\circ}+4.65^{\circ}=185.85^{\circ}


6 0
3 years ago
D
Rudiy27
Okay so yeah u have to minus then subtract then decide it it’s a method i was taught to do
5 0
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