A rock is thrown into a swimming pool that is filled with water at a uniform temperature. As the rock moves beneath the pool's surface and sinks to the bottom of the pool, which of the following statements is true?

a. The buoyant force on the rock increases as it sinks.
b. The buoyant force on the rock decreases as it sinks.
c. The buoyant force on the rock is zero as it sinks.
d. The buoyant force on the rock is constant as it sinks.
e. The buoyant force on the rock as it sinks is nonzero at first but becomes zero once the terminal velocity is reached.

Answer:

v₀ = 0.5058 m/s

Explanation:

From the question, for the block to hit the bottle, the elastic potential energy of the spring at the bottle (x = 0.08 m) should be equal to the sum of the elastic potential energy of the spring at x = 0.05 m and the kinetic energy of block at x = 0.05 m

Now, the potential energy of the block at x = 0.08 m is ½kx²

where;

k is the spring constant given by; k = ω²m

ω is the angular velocity of the oscillation

m is the mass of the block.

Thus, potential energy of the spring at the bottle(x = 0.08 m) is;

U = ½ω²m(0.08m)²

Also, potential energy of the spring at the bottle(x = 0.05 m) is;

U = ½ω²m(0.05m)²

and the kinetic energy of the block at x = 0.05 m is;

K = ½mv₀²

Thus;

½ω²m(0.08)² = ½ω²m(0.05)² + ½mv₀²

Inspecting this, ½m will cancel out to give;

ω²(0.08)² = ω²(0.05)² + v₀²

Making v₀ the subject, we have;

v₀ = ω√((0.08)² - (0.05)²)

So,

v₀ = 8.1√((0.08)² - (0.05)²)

v₀ = 0.5058 m/s

Answer:

Final magnification = -375

Explanation:

The total magnification of a compound microscope is expressed mathematically as the product of the magnifying power of each of the lenses that are combined in the compound microscope.

Final magnification = (magnifying power of the objective lens) × (magnifying power of the eyepiece lens) = m₁ × m₂

Magnifying power of a lens is defined as the ratio of the least distance of distinct vision to the focal length of the lens.

For the eyepiece, the least distance of distinct vision is the distance from the object to the near point = 25 + 1 + 0.15 = 26.15 cm

Focal length = 1 cm

Magnifying power of the eyepiece length = (26.15/1) = 26.15

For the objective lens,

The focal length = 0.14 cm

Least distance of distinct vision = -(1 + 1.00 + 0.15 - 0.14) = -2.01 cm (the negative sign means the image seen is upside down)

Magnifying power of the objective lens = (-2.01/0.14) = -14.357

Final magnification = -14.357 × 26.15 = -375

Hope this Helps!!!

Answer:

Dear user,

Answer to your query is provided below

1) Endocrine System is a collection of glands which secretes different kinds of hormones which regulate the body functioning.

2) These hormones regulate metabolism, growth and development, tissue function, sexual functions, sleep and reproduction.

3) The important endocrine glands present in the body are:

Pituitary, thyroid, adrenal, thymus, pancreas, ovary, testes etc

Explanation:

The endocrine system is the system of ductless glands, each of which secretes different types of hormone directly into the bloodstream to regulate the body functions like growth, metabolism, and sexual development.

Answer:

a.     F = 2.32*10^-18 N

b.     The force F is 2.59*10^11 times the weight of the electron

Explanation:

a. In order to calculate the magnitude of the force exerted on the electron you first calculate the acceleration of the electron, by using the following formula:

[tex]v^2=v_o^2+2ax[/tex]         (1)

v: final speed of the electron = 6.60*10^5 m/s

vo: initial speed of the electron = 4.00*10^5 m/s

a: acceleration of the electron = ?

x: distance traveled by the electron = 5.40cm = 0.054m

you solve the equation (2) for a and replace the values of the parameters:

[tex]a=\frac{v^2-v_o^2}{2x}=\frac{(6.60*10^5m/s)^2-(4.00*10^5m/s)^2}{2(0.054m)}\\\\a=2.55*10^{12}\frac{m}{s^2}[/tex]

Next, you use the second Newton law to calculate the force:

[tex]F=ma[/tex]

m: mass of the electron = 9.11*10^-31kg

[tex]F=(9.11*10^{-31}kg)(2.55*10^{12}m/s^2)=2.32*10^{-18}N[/tex]

The magnitude of the force exerted on the electron is 2.32*10^-18 N

b. The weight of the electron is given by:

[tex]F_g=mg=(9.11*10^{-31}kg)(9.8m/s^2)=8.92*10^{-30}N[/tex]

The quotient between the weight of the electron and the force F is:

[tex]\frac{F}{F_g}=\frac{2.32*10^{-18}N}{8.92*10^{-30}N}=2.59*10^{11}[/tex]

The force F is 2.59*10^11 times the weight of the electron

Answer:

(b) Torque will increase.

Explanation:

Torque is given as the product of force and moment arm (radius).

τ = F x r

F = τ / r

where;

F is force

τ  is torque

r is radius (moment arm)

Keeping force constant, we will have the following;

τ ∝ r

This shows that torque is directly proportional moment arm (radius), thus increase in moment arm, will cause increase in torque.

For instance;

let the constant force = 5 N

let the initial moment arm, r = 2m

Torque, τ  = 5 N x 2m = 10 Nm

When the moment arm is increased to 4 m

Torque, τ  = 5 N x 4m = 20 Nm

Therefore, at a constant force, increasing in the Moment arm, will cause increase in torque.

Coorect option is "(b) Torque will increase."

Answer: 40.2 hp

Explanation:

The mass of the car is 2000kg.

in 21 seconds, it accelerates from 0m/s to 25m/s, so the average acceleration is:

a = (25m/s)/(21s) = 1.2 m/s^2

We know that force = mass*acceleration, then:

F = 2000kg*1.2m/s^2 = 2400N.

Now, the Power can be written as:

P = W/t where W is work = F*d (force per distance) and t is time.

then we have:

P = (F*d)/t and d/t  is the average velocity, and we know the velocity v = 25m/s, so the average velocity will be (1/2)*25m/s = 12.5m/s

P = F*12.5m/s

and the force is F = 2400N

P = 2400 N*12.5m/s = 30,000 W

and we know that 1hp = 746 W

then, P = (30,000/746) hp = 40.2 hp

Answer:

[tex]P=40hp[/tex]

Explanation:

Hello,

In this cased, for the given velocities, we can compute the work done by the car as shown below, considering the kinetic energy only:

[tex]W=\frac{1}{2}m(v_f^2-v_0^2)[/tex]

Whereas the mass is 2000 kg, the final velocity is 25 m/s and the initial velocity is 0 m/s, therefore the work is:

[tex]W=\frac{1}{2}*2000kg*[(25\frac{m}{s})^2 -(0\frac{m}{s})^2]\\\\W=\frac{1}{2}*2000kg*625\frac{m^2}{s^2} \\\\W=6.25x10^5J[/tex]

Now, the power:

[tex]P=\frac{W}{t}=\frac{6.25x10^5}{21.0s}=29762W[/tex]

And in horsepower:

[tex]P=29762W*\frac{1hp}{746W} \\\\P=40hp[/tex]

Best regards.

Answer:

More current will be loss through the metal wire strands if the force on them was repulsive, and more stress will be induced on the wire strands due to internal and external flexing.

Explanation:

A wire bundle is made up of wire strands bunched together to increase flexibility that is not always possible in a single solid metal wire conductor. In the strands of wire carrying a high voltage power, each strand carries a certain amount of current, and the current through the strands all travel in the same direction. It is know that for two conductors or wire, separated by a certain distance, that carries current flowing through them in the same direction, an attractive force is produced on these wires, one on the other. This effect is due to the magnetic induction of a current carrying conductor. The forces between these strands of the high voltage wire bundle, pulls the wire strands closer, creating more bond between these wire strands and reducing internal flex induced stresses.

If the case was the opposite, and the wires opposed themselves, the effect would be that a lot of cost will be expended in holding these wire strands together. Also, stress within the strands due to the repulsion, will couple with external stress from the flexing of the wire, resulting in the weakening of the material.

The biggest problem will be that more current will be lost in the wire due to increased surface area caused by the repulsive forces opening spaces between the strand. This loss is a s a result of the 'skin effect' in wire transmission, in which current tends to flow close to the surface of the metal wire. The skin effect generates power loss as heat through the exposed surface area.

Answer:

H = 45 m

Explanation:

First we find the launch velocity of the ball by using the following formula:

v₀ = √(v₀ₓ² + v₀y²)

where,

v₀ = launching velocity = ?

v₀ₓ = Horizontal Component of Launch Velocity = 15 m/s

v₀y = Vertical Component of Launch Velocity = 30 m/s

Therefore,

v₀ = √[(15 m/s)² + (30 m/s)²]

v₀ = 33.54 m/s

Now, we find the launch angle of the ball by using the following formula:

θ = tan⁻¹ (v₀y/v₀ₓ)

θ = tan⁻¹ (30/15)

θ = tan⁻¹ (2)

θ = 63.43°

Now, the maximum height attained by the ball is given by the formula:

H = (v₀² Sin² θ)/2g

H = (33.54 m/s)² (Sin² 63.43°)/2(10 m/s²)

H = 45 m

The maximum height of the ball is 45 m.

Let u represent the initial velocity and θ represent the angle. Hence:

Vertical component is 30 m/s, hence:

u*sinθ = 30  (1)

Horizontal component is 15 m/s, hence:

u*cosθ = 15   (2)

Divide eqn 1 by 2:

tan(θ) = 2

θ = 63.4°

u*sin(63.4°) = 30

u = 33.5 m/s

The maximum height (h) is:

[tex]h=\frac{u^2sin^2\theta}{2g} \\\\h=\frac{33.5^2*sin^2(63.4)}{2*10} =45\ m[/tex]

The maximum height of the ball is 45 m.

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Answer:

v =   11 m/s   is her final speed

Explanation:

work done by gravity = m g Δh =   40×9.8×10   = 3920 Joules

Work done by friction = - force×distance =   - 20×100   =   - 2000 Joules

[minus sign because friction force is opposite to the direction of motion]

Initial K.E. = (1/2) m u^2 = (1/2) × 40 × 5^2   = 500 Joules

Now, by work energy theorem

Work done = change in kinetic energy.

Final K.E. = initial K.E. + total work =    500 + 3920 - 2000  = 2420 Joules

Now, Final K.E. = (1/2) m v^2  [final speed being v= speed at the bottom]

⇒  2420 = (1/2)×40×v^2

   ⇒  121 = v^ 2

  v =   11 m/s   is her final speed

Answer:

The energy of a single photon of X‑rays is  [tex]E_v = 52898.9 \ eV[/tex]

Explanation:

From the question we are told that

    The mass of the tissue is [tex]m = 175 \ g = 0.175 \ kg[/tex]

    The amount of energy delivered is  [tex]E_T = 4.05 \mu J = 4.05 *10^{-6} \ J[/tex]

    The wavelength is  [tex]\lambda = 0.0235 \ nm = 0.0235 *10^{-9} \ m[/tex]

Generally the energy of a single photon is mathematically represented as

         [tex]E = \frac{hc}{\lambda }[/tex]

Where h is the Planck's  constant with values [tex]h = 6.63 *10^{-34}\ J\cdot s[/tex]

and c is the speed of light with values [tex]c = 3.0 *10^{8} \ m/s[/tex]

      Substituting values

       [tex]E = \frac{6.63 *10^{-34} * 3.0*10^{8}}{0.0235 *10^{-9} }[/tex]

      [tex]E = 8.463 *10^{-15} \ J[/tex]

Converting this to electron  volt we have

       [tex]E_v = 8.464 *10^{-15} * [\frac{1 }{1.6*10^{-19}} ][/tex]

      [tex]E_v = 52898.9 \ eV[/tex]

Answer:

Both attempt to explain human behavior

Explanation:

Psychology is generally regarded as the science of human behavior. Behaviourism is the psychological theory which holds that behaviour can be fully understood in terms of conditioning, without actually considering thoughts or feelings. The theory holds that psychological disorders can be aptly handled by simply altering the behavioural patterns of the individual. It involves the study of stimulus and responses.

Cognitive psychology attempts to decipher what is going on in people's minds. That is, it looks at the mind as a processor of information. Hence we can define cognitive psychology as the study of the internal mental processes. This according to behaviorists, cannot be studied in measurable terms as in behaviourism (stimulus response approach) even though mental processes are known to influence human behavior significantly.

Hence, both behaviourism and cognitive psychology attempt to study human behavior from different perspectives.

Answer:

I belive its THAT HELPS US BE A BETTER PERSON

Answer:

Understands the purpose of moral standards and how to best fulfill that.

Explanation:

Answer:

The pressure is [tex]p_1 = 4051.4 \ Pa[/tex]

Explanation:

From the question we are told that

     The gauge pressure at the mouth is  [tex]p_1[/tex]

     The radius of the column is  [tex]r_2 = 4 \ mm = 0.004 \ m[/tex]

    The speed of the liquid outside the body is  [tex]v_2 = 3.1 \ m/s[/tex]

      The area of the column is  [tex]A_2[/tex]

       The area inside the mouth [tex]A_1 = 10 A_2[/tex]

Generally according to continuity equation

       [tex]v_1 A_1 = v_2 A_2[/tex]

=>       [tex]v_ 1 = v_2 * \frac{A_2}{A_1}[/tex]

=>      [tex]v_ 1 = 3.1 * \frac{1}{10}[/tex]

=>        [tex]v_ 1 = 0.31 \ m/s[/tex]

So

      [tex]A_1 = 10A_2[/tex]

=>   [tex]\pi * r_1^2 = 10(\pi * r_2^2)[/tex]

=>   [tex]r_1 = 10 * r_2[/tex]

substituting values

        [tex]r_1 = 10 * 0.004[/tex]

        [tex]r_1 =0.04 \ m[/tex]

Now the height of inside the mouth is  [tex]h_1 = d = 2r_1 = 2* 0.04 = 0.08\ m[/tex]

Now the height of the column is  [tex]h_2 = d = 2r_2 = 2* 0.004 = 0.008\ m[/tex]

Generally according to Bernoulli's  equation

        [tex]p_1 = [\frac{1}{2} \rho v_2^2 + h_2 \rho g +p_2] -[\frac{1}{2} \rho * v_1^2 + h_1 \rho g ][/tex]

Now  [tex]\rho = 1000 \ kg m^{-3}[/tex] which is the density of water

        [tex]p_2[/tex] is the gauge pressure of the atmosphere which is  zero

 So

       [tex]p_1 = [(0.5 * 1000 * (3.1)^2) +(0.008 * 1000 * 9.8) + 0]-[/tex]

                                                  [tex][(0.5 * 1000 * 0.31^2) + (0.08*1000 * 9.8)][/tex]                          

       [tex]p_1 = 4051.4 \ Pa[/tex]

The initial pressure of the gauge pressure at the mouth is 4757 pascals.

The principle of continuity equation asserts that in a steady flow state, the quantity of fluid flowing at the inlet is equivalent to the quantity of fluid at the outlet given that there is a constant mass flow rate.

It can be expressed by using the formula:

[tex]\mathbf{A_1v_1=A_2v_2}[/tex]

where;

∴

Making v₁ the subject of the formula:

[tex]\mathbf{v_1 = \dfrac{A_2}{A_1}\times v_2}[/tex]

[tex]\mathbf{v_1 = \dfrac{1}{10}\times3.1}[/tex]

[tex]\mathbf{v_1 =0.31 \ m/s}[/tex]

Since the area are equivalent to each other

[tex]\mathbf{A_1 = 10A_2}[/tex]

[tex]\mathbf{\pi r^2_1 = 10\times \pi r^2_2}[/tex]

[tex]\mathbf{ r^2_1 = 10\times r^2_2}[/tex]

where;

∴

[tex]\mathbf{ r_1 = 10\times 0.004}[/tex]

[tex]\mathbf{ r^2_1 = 0.04 \ m}[/tex]

However, in fluid dynamics, Bernoulli's equation can be applied for the estimation of the initial gauge pressure by using the expression:

[tex]\mathbf{\dfrac{1}{2} \rho v_1^2 + h_1 \rho g + p_1 = \dfrac{1}{2}\rhov_2^2 + h_2 \rho g +p_2}[/tex]

Since the gauge pressure of the atmosphere [tex]\mathbf{p_2}[/tex] = 0

∴

The initial gauge pressure applied at the mouth can be determined as:

[tex]\mathbf{p_1 = \dfrac{1}{2} \rho ( v_2^2-v_1^2)}[/tex]

here;

[tex]\mathbf{\rho = 1000 \ kg/m^3}[/tex]

[tex]\mathbf{p_1 = \dfrac{1}{2} \times 1000 \ kg/m^3 ( 3.1^2-0.31^2)}[/tex]

[tex]\mathbf{p_1 =500 \ kg/m^3 (9.5139)}[/tex]

[tex]\mathbf{p_1 =4756.95 \ Pa}[/tex]

[tex]\mathbf{p_1 \simeq4757 \ Pa}[/tex]

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Answer:

d because a double covalent is a bond  ;p

Explanation:

in a Lewis structure, a double line represent two covalent bond. The correct option is D.

A covalent bond is formed between the atoms of two different elements by mutual sharing of one or more pairs of electrons.

To form a covalent bond, one atom loses the electrons in its valence shell to become stable and called as donor. Another atom of different element gains the electrons from the first atom and become an acceptor.

They both share to be stable and from a stable compound. Bond can be single, double or triple. As, the bond increases, the strength of holding the atoms increases.

Thus,  the correct option is D.

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Answer:

One method to solve this problem is to take the ball and the floor as the system

second method to solve this exercise and we take the system as formed by the ball only

Explanation:

One method to solve this problem is to take the ball and the floor as the system, therefore the forces that occur during the collision are driving force and reaction, internal to the system, therefore the moment is conserved.

A second method to solve this exercise and we take the system as formed by the ball only, in this case there is a change at the moment and this is not preserved, This second case is a little more difficult to solve

Answer:

4.197 N

Explanation:

current through the wires I = 700 A

distance between wires r = 70 cm = 0.7 m

length of each wire L = 30 m

force per unit length on the wires is given as

[tex]\frac{F}{L}[/tex] = [tex]\frac{u*I^{2} }{2\pi r }[/tex]

where u = permeability of vacuum = 1.256 [tex]10^{-6}[/tex] m-kg-[tex]s^{-2}[/tex][tex]A^{-2}[/tex]

F =  [tex]\frac{u*I^{2}L }{2\pi r }[/tex]

F =  [tex]\frac{1.256*10^{-6} *700^{2}*30 }{2*3.142*0.7 }[/tex] = 4.197 N

Answer:

Specific gravity of other fluid = .854 (Approx)

Explanation:

Given:

Mass of water = 35 g

Mass of filled bottle with water = 98.44 g

Mass of filled bottle with fluid = 89.22 g

Computation:

Mass of water = 98.44g - 35g = 63.44g

Density of water = 1000 g/L

Volume of bottle = 63.44/1000 = 0.06344L

Mass of other liquid = 89.22g - 35g = 54.22g

Density of other liquid = 54.22g/0.06344L = 854.665826 g/L

Water has a specific gravity = 1

So ,  specific gravity of other fluid

1000 / 854.665826 = 1 / specific gravity of other fluid

Specific gravity of other fluid = .854 (Approx)

Answer:

The auroras C.

Explanation:

Answer:

Intensity of an earthquake (passes 1 km) = 8,748 × 10⁶

Explanation:

Given:

Distance (r1) = 54 km

Distance (r2) = 1 Km

Note:

Intensity of an earthquake with 54 km distance is 3.0 × 10⁶ J/m²s is not given.

So,

Intensity of an earthquake with 54 km distance (I₁) = 3.0 × 10⁶ J/m²s

Find:

Intensity when it passed a point only 1.0 km .

Computation:

Intensity of an earthquake = (r1 / r2)² (I₁)

Intensity of an earthquake (passes 1 km) = (54 / 1)²(3.0 × 10⁶)

Intensity of an earthquake (passes 1 km) = (54)²(3.0 × 10⁶)

Intensity of an earthquake (passes 1 km) = 8,748 × 10⁶

Answer:

V₂ = 0.0016 m³ = 1.6 L

Explanation:

The work done by the system against a constant external pressure is given by:

W = PΔV

W = P(V₂ - V₁)

V₂ = (W/P) + V₁

where,

W = Work done by the piston = 217 J

P = Constant External Pressure = (1.5 atm)(101325 Pa/ 1 atm) = 151988 Pa

V₁ = Initial Volume = (0.15 L)(0.001 m³/1 L) = 0.00015 m³

Therefore,

V₂ = (217 J/151988 Pa) + 0.00015 m³

V₂ = 0.0016 m³ = 1.6 L

Answer:

(e) 3.2

Explanation:

We are given that vector C and D.

Let R be the magnitude of C+D.

According to question

R=3D

We have to find the ratio of the magnitude of C to that of D.

By using right triangle property

[tex]C^2=R^2+D^2[/tex]

[tex]C^2=(3D)^2+D^2[/tex]

[tex]C^2=9D^2+D^2[/tex]

[tex]C^2=10D^2[/tex]

[tex]C=\sqrt{10D^2}=3.2D[/tex]

[tex]\frac{C}{D}=3.2[/tex]

Hence, the ratio of the magnitude of C to that of D=3.2

(e) 3.2

Answer:

12 m/s²

Explanation:

Maximum acceleration is a = Aω², where A is the amplitude and ω is the frequency.

a = (0.0030 m) (10 rev/s × 2π rad/rev)²

a = 12 m/s²

Answer:

aₓ = 0 ,       ay = -6.8125 m / s²

Explanation:

This is an exercise that we can solve with kinematics equations.

Initially the rabbit moves on the x axis with a speed of 1.10 m / s and after seeing the predator acceleration on the y axis, therefore its speed on the x axis remains constant.

x axis

          vₓ = v₀ₓ = 1.10 m / s

          aₓ = 0

y axis

initially it has no speed, so v₀_y = 0 and when I see the predator it accelerates, until it reaches the speed of 10.6 m / s in a time of t = 1.60 s. let's calculate the acceleration

         [tex]v_{y}[/tex]= v_{oy} -ay t

          ay = (v_{oy} -v_{y}) / t

          ay = (0 -10.9) / 1.6

          ay = -6.8125 m / s²

the sign indicates that the acceleration goes in the negative direction of the y axis

Answer:

Water emerge from the hole at 2.97 m/s

Explanation:

There exist and hydrostatic pressure at the base of the trough which is due to force exerted by the water at the top.

Given that the dept of the trough is 0.45 m

g = acceleration due to gravity

v = velocity

Then;

pgh = pv²/2

gh = v²/2

v = √2gh

v = √2*9.81*0.45

v = 2.97 m/s

Velocity of the water is 2.97 m/s

Answer:

The centripetal acceleration of the car will be 12.32 m/s² .

Explanation:

Given that

radius ,R= 57 m

Velocity , V=26.5 m/s

We know that centripetal acceleration given as follows

[tex]a_c=\dfrac{V^2}{R}[/tex]

Now by putting the values in the above equation we get

[tex]a_c=\dfrac{26.5^2}{57}=12.32\ m/s^2[/tex]

Therefore the centripetal acceleration of the car will be 12.32 m/s² .

Complete Question

A satellite in geostationary orbit is used to transmit data via electromagnetic radiation. The satellite is at a height of 35,000 km above the surface of the earth, and we assume it has an isotropic power output of 1 kW (although, in practice, satellite antennas transmit signals that are less powerful but more directional).

Reception devices pick up the variation in the electric field vector of the electromagnetic wave sent out by the satellite. Given the satellite specifications listed in the problem introduction, what is the amplitude E0 of the electric field vector of the satellite broadcast as measured at the surface of the earth? Use ϵ0=8.85×10^−12C/(V⋅m) for the permittivity of space and c=3.00×10^8m/s for the speed of light.

Answer:

The electric field vector of the satellite broadcast as measured at the surface of the earth is  [tex]E_o = 6.995 *10^{-6} \ V/m[/tex]

Explanation:

From the question we are told that

     The height of the satellite is  [tex]r = 35000 \ km = 3.5*10^{7} \ m[/tex]

      The power output of the satellite is [tex]P = 1 \ KW = 1000 \ W[/tex]

Generally the intensity of the electromagnetic radiation of the satellite at the surface of the earth is  mathematically represented as  

     [tex]I = \frac{P}{4 \pi r^2}[/tex]

substituting values

      [tex]I = \frac{1000}{4 * 3.142 (3.5*10^{7})^2}[/tex]

      [tex]I = 6.495*10^{-14} \ W/m^2[/tex]

This intensity of the electromagnetic radiation of the satellite at the surface of the earth can also be   mathematically represented as  

          [tex]I = c * \epsilon_o * E_o^2[/tex]

Where [tex]E_o[/tex] is the amplitude of the electric field vector of the satellite broadcast so

         [tex]E_o = \sqrt{\frac{2 * I}{c * \epsilon _o} }[/tex]

substituting values

          [tex]E_o = \sqrt{\frac{2 * 6.495 *10^{-14}}{3.0 *10^{8} * 8.85*10^{-12}} }[/tex]

           [tex]E_o = 6.995 *10^{-6} \ V/m[/tex]

Answer:

Φ = 361872 N.m^2 / C

Explanation:

Given:-

- The area of the two plates, [tex]A_p = 180 cm^2[/tex]

- The charge on each plate, [tex]q = 17 * 10^-^6 C[/tex]

- Permittivity of free space, [tex]e_o = 8.85 * 10^-^1^2 \frac{C^2}{N.m^2}[/tex]

- The radius for the flux region, [tex]r = 3.3 cm[/tex]

- The angle between normal to region and perpendicular to plates, θ = 4°

Find:-

Find the flux (in N · m2/C) through a circle of radius 3.3 cm between the plates.

Solution:-

- First we will determine the area of the region ( Ar ) by using the formula for the area of a circle as follows. The region has a radius of r = 3.3 cm:

                             [tex]A_r = \pi *r^2\\\\A_r = \pi *(0.033)^2\\\\A_r = 0.00342 m^2[/tex]

- The charge density ( σ ) would be considered to be uniform for both plates. It is expressed as the ratio of the charge ( q ) on each plate and its area ( A_p ):

                           σ = [tex]\frac{q}{A_p} = \frac{17*10^-^6}{0.018} \\[/tex]

                           σ = 0.00094 C / m^2

- We will assume the electric field due to the positive charged plate ( E+ ) / negative charged plate ( E- ) to be equivalent to the electric field ( E ) of an infinitely large charged plate with uniform charge density.

                         [tex]E+ = E- = \frac{sigma}{2*e_o} \\\\[/tex]

- The electric field experienced by a region between two infinitely long charged plates with uniform charge density is the resultant effect of both plates. So from the principle of super-position we have the following net uniform electric field ( E_net ) between the two plates:

                        [tex]E_n_e_t = (E+) + ( E-)\\\\E_n_e_t = \frac{0.00094}{8.85*10^-^1^2} \\\\E_n_e_t = 106214689.26553 \frac{N}{C} \\[/tex]

- From the Gauss-Law the flux ( Φ ) through a region under uniform electric field ( E_net ) at an angle of ( θ ) is:

                        Φ = E_net * Ar * cos ( θ )

                        Φ = (106214689.26553) * (0.00342) * cos ( 5 )

                        Φ = 361872 N.m^2 / C