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ohaa [14]
3 years ago
10

A solid nonconducting sphere of radius R has a charge Q uniformly distributed throughout its volume. A Gaussian surface of radiu

s r with r < R is used to calculate the magnitude of the electric field E at a distance r from the
center of the sphere. Which of the following equations results
from a correct application of Gauss’s law for this situation? And Why?

1. E (4 p r^2) = (Q r^3)/( epsilon sub0 * R3)
2. E(4pr^2)= (Q 3r^3)/ (epsilon sub0 * 4 p R)
3. E (4 p r^2) = (Q) / (epsilon sub0)
Physics
1 answer:
valentina_108 [34]3 years ago
6 0

Answer:

E(4 \pi r^{2})={\frac{ Q r^{3}}{R^{3}\epsilon _{0}}}

or

E(4 \pi)={\frac{ Q r}{R^{3}\epsilon _{0}}}

Explanation:

We know that Gauss's law states that the Flux enclosed by a Gaussian surface is given by

E.S=\frac{q}{\epsilon_{0}}

Here , E is electric field and S is surface are and q is charge enclosed by the surface and e is electrical permeability of the medium.

Here the Gaussian is of radius r<R so area of surface is

S=4 \pi r^{2}

Also, charge enclosed by the surface is

Charge =\frac{Total \: Charge }{Total \:Volume} \times Volume \: of \: Gaussian \: surface

therefore,

q=\frac{Q}{\frac{4}{3} \pi R^{3} }\frac{4}{3} \pi r^{3} =\frac{ Q r^{3}}{R^{3}}

Here Q is total charge,

Insert values in Gauss's law

E(4 \pi r^{2})=\frac{\frac{ Q r^{3}}{R^{3}}}{\epsilon _{0}}

Rearrange them

E(4 \pi r^{2})={\frac{ Q r^{3}}{R^{3}\epsilon _{0}}}

on further solving

E(4 \pi)={\frac{ Q r}{R^{3}\epsilon _{0}}}

This is the required form.

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A ball is thrown vertically upward with a speed of 25.0 m/s from a height of 2.0 m.
Marrrta [24]

Answer:

The answer to your question is

a) t = 2.55 s

b) t = 5.5 s

Explanation:

Data

vo = 25 m/s

h = 2 m

g = 9.81 m/s

Formula

t = -(vo)/g

a)

        t = -(25)/9.81

        t = 2.55 s

b)

        Tt = 2t

         Tt = 2(2.55)

         Tt = 5.1 s

Time in the last 10 m

         10 = 25t + (1/2)(9.81)t²

Simplify

          10 = 25t 4.91t²

          4.9t² + 25t - 10 = 0

Solve the equation using an online calculator

    t₁ = 0.37 s               t₂ = -5.47 s

The correct answer is t₁, t₂ is incorrect because there are no negative answers.

Total time = 5.1 + 0.37

                 = 5.5 s

4 0
3 years ago
During a testing process, a worker in a factory mounts a bicycle wheel on a stationary stand and applies a tangential resistive
olganol [36]

Answer:

The force is     F_c  =  789.03 \  N    

Explanation:

From the question we are told that

   The tangential  resistive force is F_t  =   115 \ N

   The mass of the wheel is  m  = 1.80 kg

  The diameter of the wheel is  d =  50.0 cm  = 0.5 \ m

   The diameter of the sprocket is  d_c  =  8.50 \ cm =0.085 \ m

  The angular acceleration considered is  \alpha  =  4.30\ rad/s^2

Generally the radius of the wheel is

       r = \frac{d}{2}

=>     r = \frac{0.5}{2}

=>     r = 0.25 \ m

Generally the radius of the sprocket is

       r_c = \frac{d_c}{2}

=>     r_c = \frac{0.085}{2}

=>     r_c = 0.0425 \ m

Generally the moment of inertia of the wheel is mathematically represented as

      I  =  m  *  r^2

=>    I  =  1.80  *  0.25^2

=>    I  = 1.1125 \ kg \cdot m^2

Generally the torque experienced by the wheel due to the forces acting on it  is mathematically represented as

      \tau =  F_c *  r_c  -  F_t  * r

Here  F_c is the force acting on the sprocket

So  

      \tau =  F_c *  0.0425 - 115  * 0.25

       \tau = 0.0425F_c  -  28.75

Generally the torques that will cause the wheel to move with \alpha  =  4.30\ rad/s^2 is mathematically represented as

       \tau  =  I  * \alpha

So

        0.0425F_c  -  28.75  =   I  * \alpha

        0.0425F_c  -  28.75  =   1.1125  *4.30    

       0.0425F_c  -  28.75  =   1.1125  *4.30    

        F_c  =  789.03 \  N    

5 0
3 years ago
A 65-cm segment of conducting wire carries a current of 0.35
algol13
The intensity of the magnetic force exerted on the wire due to the presence of the magnetic field is given by
F=ILB \sin \theta
where
I is the current in the wire
L is the length of the wire
B is the magnetic field intensity
\theta is the angle between the direction of the wire and the magnetic field

In our problem, L=65 cm=0.65 m, I=0.35 A and B=1.24 T. The force on the wire is F=0.26 N, therefore we can rearrange the equation to find the sine of the angle:
\sin \theta= \frac{F}{ILB}= \frac{0.26 N}{(0.35 A)(0.65 m)(1.24 T)}=0.922

and so, the angle is
\theta=\arcsin(0.921)=67.1^{\circ}
6 0
3 years ago
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DedPeter [7]

Answer:

d' = 75.1 cm

Explanation:

It is given that,

The actual depth of a shallow pool is, d = 1 m

We need to find the apparent depth of the water in the pool. Let it is equal to d'.

We know that the refractive index is also defined as the ratio of real depth to the apparent depth. Let the refractive index of water is 1.33. So,

n=\dfrac{d}{d'}\\\\d'=\dfrac{d}{n}\\\\d'=\dfrac{1\ m}{1.33}\\\\d'=0.751\ m

or

d' = 75.1 cm

So, the apparent depth is 75.1 cm.

4 0
3 years ago
Looking at the wave diagram which best describes the wave
Ad libitum [116K]

Answer:

B describes the best ...

5 0
2 years ago
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