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goldfiish [28.3K]
3 years ago
7

For a charged solid metal sphere with total charge Q and radius R centered on the origin: Select "True" or "False" for each stat

ement.The electric field near the metal surface on the outside is perpendicular to the surface. The net charge on the inside of the solid metal sphere is negative. If the solid sphere is an insulator (instead of metal) with net charge QQ, the electric field for r >> R would be the same as that of a conductor with the same shape and charge. If the solid sphere is an insulator (instead of metal) with net charge QQ, the net charge on the inside of the solid sphere is negative. The electric field for the metal sphere at r << R will be the same as the field of a point charge, Q, at the origin. The electric field inside the solid metal sphere is never zero
Physics
1 answer:
mario62 [17]3 years ago
7 0

Answer:

the answer is true

Explanation:

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consider an electric dipole lying along x-axis with mid-point O as the origin of coordinate system.find the electric potential V
velikii [3]

Answer:

axial   V = 0

equatorial  V = k q 2a / (x² -a²),  V = k q 2x / (a² -x²)

Explanation:

A dipole is a system formed by two charges of equal magnitude, but different sign, separated by a distance 2a; let's look for the electrical potential in an axial line

       V = k (q / √(a² + y²) - q /√ (a² + y²))

        V = 0

the potential on the equator

we place the positive charge to the left and perform the calculation for a point outside the dipole

    V = k (q / (x-a) - q / (x + a))

    V = k q 2a / (x² -a²)

 we perform the calculation for a point between the dipo charges

     V = k (q / (a-x) - q / (a ​​+ x))

     V = k q 2x / (a² -x²)

4 0
4 years ago
Consider 3 polarizers. Polarizer 1 has a vertical transmission axis and polarizer 3 has a horizontal transmission axis. Taken to
kirill [66]

Answer:

Option D.) 0.09 I₀

Explanation:

The intensity, I, of a polarized light after passing through a polarizing filter is given by I = I_{0} cos^{2}  \theta

I_{0} = Original Intensity

\theta = Angle between the direction of polarization and the axis of filter

After passing through the first polarizer, the initial intensity, I_{o} is halved, i.e I_{1} = \frac{I_{0} }{2}

After passing through the second polarizer, the angle of polarization, θ = 30⁰

I_{2} = I_{1} cos^{2} \theta

I_{2} = \frac{I_{0} }{2} cos^{2} 30\\I_{2} =\frac{3}{8} I_{0} \\I_{2} = 0.375 I_{0}

After passing through the third polarizer, the angle of polarization, θ = 90⁰-30⁰ = 60⁰

I_{3} = I_{2} cos^{2} \theta

I_{3} = 0.375I_{0} cos^{2} 60

I_{2} =0.09 I_{0}

7 0
4 years ago
How is this formula ( Rt= r1 x r2 / r1 + r2) which is used to calculate total resistance of two parallel resistors found from th
Maru [420]

Explanation:

The total resistance used to calculate the total resistance of the two parallel resistor is given by :

R_t=\dfrac{r_1\times r_2}{r_1+r_2}

Taking reciprocal of the above equation.

\dfrac{1}{R_t}=\dfrac{r_1+r_2}{r_1\times r_2}

We can also write the above equation as :

\dfrac{1}{R_t}=\dfrac{r_1}{r_1\times r_2}+\dfrac{r_2}{r_1\times r_2}

On simplification of above equation,

\dfrac{1}{R_t}=\dfrac{1}{r_2}+\dfrac{1}{r_1}

or

\dfrac{1}{R_t}=\dfrac{1}{r_1}+\dfrac{1}{r_2}

Hence, proved.

4 0
3 years ago
a stone is thrown vertically upwards with a velocity of 20 m per second determine the total time of flight of stone in air​
miv72 [106K]

Answer:

Explanation:

The best way to do this is to remember the rule about the halfway mark in a parabolic path. At a trajectory's half way point in its travels, it will be at its max height. To get the total time in the air, we take that time at half way and double it. Here's what we know that we are told:

initial velocity is 20 m/s

Here's what we know that we are NOT told:

a = -9.8 m/s/s and

final velocity is 0 at an object's max height in parabolic motion.

We will use the equation:

v=v_0+at where v is final velocity and v0 is initial velocity. Filling in:

0 = 20 + (-9.8)t and

-20 = -9.8t so

t = 2 seconds. The stone reaches its max height 2 seconds after it is thrown; that means that after another 2 seconds it will be on the ground. Total air time is 4 seconds.

8 0
3 years ago
A block of mass 0.84 kg is suspended by a string which is wrapped so that it is at a radius of 0.061 m from the center of a pull
lora16 [44]

Answer:

E_l = 1.713 J

Explanation:

Given data:

mass of block is M_b = 0.84 kg

radius of block = 0.061 m

moment of inertia is 6.20 \times 10^{-3} kg m^2

D is distance covered by block = 0.65 m

speed of block is 1.705 m/s

From conservation of momentum  we have

M_b g D = \frac{1}{2} M_b v^2 + \frac{1}{2} I \omega^2 +  E_{loss}

0.84 \times 9.81 \times 0.65 = \frac{1}{2}\times  0.84 \times 1.705^2 +\frac{1}{2} \times 6.2 \times 10^{-3} [\frac{1.705}{0.061}]^2 + E_l

solving for energy loss

E_l = 1.713 J

3 0
3 years ago
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