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sergeinik [125]
3 years ago
11

a. How many excess electrons must be distributed uniformly within the volume of an isolated plastic sphere 25.0 cm in diameter t

o produce an electric field of 1350 N/C just outside the surface of the sphere.b. What is the electric field at a point 10.0 cm outside the surface of the sphere?
Physics
1 answer:
Agata [3.3K]3 years ago
5 0

Answer:

Explanation:

The electric field outside the sphere is given as,

E = k Q /r²

here Q = n x 1.6 x 10⁻¹⁹ C

where n is the number of electons

if the dimeter of sphere d= 25 cm= 0.25 m

then the radius r = 0.125 m

we get

n= E r²/ k x 1.6 x 10⁻¹⁹ C

n = 1350N/C x (0.125m)² / (8.99 x 10⁹ N m²/C² x 1.6 x 10⁻¹⁹ C)

n = 14664731646

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Looking straight downward into a rain puddle whose surface is covered with a thin film of gasoline, you notice a swirling patter
Ivanshal [37]

Answer:

 

Explanation:

Point beneath you forms a beautiful iridescent green

refractive index of Gasoline n=1.38

Wavelength of Green light is \lambda =540\ nm

Here light first traverse from air(n=1) to gasoline , it reflects from  front surface of gasoline(n=1.38) so it suffers a phase change. After this light reflect from rear surface of gasoline and there is a decrease in refractive index(n=1.38 to n=1.33), so there is no phase change occurs .

For constructive interference

2t=(m+\frac{1}{2})\cdot \frac{\lambda }{n}

here t= thickness of gasoline film

n=refractive index

for m=0

t=\frac{\lambda }{4n}

t=\frac{540}{4\times 1.38}

t=97.82\approx 98\ nm                                                      

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garik1379 [7]
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Is a rock that cools quickly after lava rusted through a volcano what is the best description of its texture
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Read 2 more answers
A car is designed to get its energy from a rotating flywheel with a radius of 1.50 m and a mass of 430 kg. Before a trip, the fl
fiasKO [112]

Answer:

a

  KE  =  7.17 *10^{7} \ J

b

 t = 6411.09 \ s

Explanation:

From the question we are told that

    The radius of the flywheel is  r =  1.50 \ m

      The mass of the flywheel is m  = 430 \ kg

          The rotational speed of the flywheel is w  =  5,200 \ rev/min = 5200 *  \frac{2 \pi }{60} =544.61 \ rad/sec

      The power supplied by the motor is  P  =  15.0 hp =  15 * 746 =  11190 \ W

         

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

       I  = \frac{1}{2} mr^2

substituting values

       I  = \frac{1}{2}  ( 430)(1.50)^2

       I  = 483.75 \  kgm^2

The kinetic energy that is been stored is  

       KE  =  \frac{1}{2} * I * w^2

substituting values

        KE  =  \frac{1}{2} * 483.75 * (544.61)^2

        KE  =  7.17 *10^{7} \ J

Generally power is mathematically represented as

          P =  \frac{KE}{t}

=>      t =  \frac{KE}{P}

substituting the value

        t = \frac{7.17 *10^{7}}{11190}

        t = 6411.09 \ s

3 0
3 years ago
Projectile Motion: A hobbyist launches a projectile from ground level on a horizontal plain. It reaches a maximum height of 72.3
marin [14]

Answer:

The angle of launch from the horizontal direction is 20.99° .

Explanation:

Let u and θ be the initial speed and angle of projection from the horizontal axis of the object respectively.

The equations for projectile motion are :

H = ( u² sin²θ)/ 2g      ......(1)

Here H is maximum height of the projectile motion and g is acceleration due to gravity.

R = ( u² sin2θ)/g         .......(2)

Here R is the maximum horizontal displacement of the object.

Rearrange equation (1) in terms of u².

u² = (2gH)/sin²θ

Substitute this equation in equation (2).

R = (2gH sin2θ) / (sin²θ x g)

R = (2H sin2θ)/sin²θ

Using trigonometry property, sin2θ = 2 cosθ sinθ

So, above equation becomes,

R = (2H x 2 cosθ sinθ)/sin²θ

R = (4H cosθ)/sinθ

tanθ = R/4H

θ = tan⁻¹(R/4H)

Substitute 111 m for R and 72.3 m for H in the above equation.

θ = tan⁻¹( 111/ 4 x 72.3 )

θ = tan⁻¹(0.38)

θ = 20.99°

3 0
3 years ago
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