Which compound is composed of oppositely charged ions?

Answer:

The new angular speed is [tex]w = 6 \ rev/s[/tex]

Explanation:

From the  question we are told that

      The angular velocity of the spin is  [tex]w_o = 3 \ rev/s[/tex]

       The  original moment of inertia is  [tex]I_o[/tex]

        The new moment of inertia is  [tex]I =\frac{I_o}{2}[/tex]    

Generally angular momentum is mathematically represented as

      [tex]L = I * w[/tex]

Now according to the law of conservation of momentum, the initial momentum is equal to the final momentum hence the angular momentum is constant so

         [tex]I * w = constant[/tex]

=>       [tex]I_o * w _o = I * w[/tex]

where w is the new angular speed  

  So  

          [tex]I_o * 3 = \frac{I_o}{2} * w[/tex]

=>        [tex]w = \frac{3 * I_o}{\frac{I_o}{2} }[/tex]

=>         [tex]w = 6 \ rev/s[/tex]

Answer:

the correct answer is d Beats

Explanation:

when two sound waves interfere time has different frequencies, the result is the sum of the waves is

       y = 2A cos 2π (f₁-f₂)/2    cos 2π (f₁ + f₂)/2

where in this expression the first part represents the envelope and the second part represents the pulse or beatings of the wave.

When examining the correct answer is d Beats

Answer:

Explanation:

Heat absorbed by oil

= mass x specific heat x rise in temperature

= 30 x 10⁻³ x 950 x 2.2 x 10³ x ( 50-10 )

= 25.08 x 10⁵ J  

Heat absorbed by air

= 50 x 1.2 x 1.0054 x 10³ x ( 20-10 )

= 6.03 x 10⁵ J

Total heat absorbed = 31.11 x 10⁵ J

If time required = t

heat lost from room

= .35 x 10³ t

Total heat generated in time t

= 1.8 x 10³ t

Heat generated = heat used

1.8 x 10³ t =  .35 x 10³ t  + 31.11 x 10⁵

1.45 x 10³ t = 31.11 x 10⁵

t = 31.11 x 10⁵ / 1.45 x 10³

t = 2145.5 s

Answer:

a)  v_correcaminos = 22.95 m / s ,  b)  x = 512.4 m ,

c) v = (45.83 i ^ -109.56 j ^) m / s

Explanation:

We can solve this exercise using the kinematics equations

a) Let's find the time or the coyote takes to reach the cliff, let's start by finding the speed on the cliff

         v² = v₀² + 2 a x

they tell us that the coyote starts from rest v₀ = 0 and its acceleration is a=15 m / s²

         v = √ (2 15 70)

         v = 45.83 m / s

with this value calculate the time it takes to arrive

        v = v₀ + a t

        t = v / a

        t = 45.83 / 15

        t = 3.05 s

having the distance to the cliff and the time, we can find the constant speed of the roadrunner

         v_ roadrunner = x / t

         v_correcaminos = 70 / 3,05

         v_correcaminos = 22.95 m / s

b) if the coyote leaves the cliff with the horizontal velocity v₀ₓ = 45.83 m / s, they ask how far it reaches.

Let's start by looking for the time to reach the cliff floor

            y = y₀ + [tex]v_{oy}[/tex] t - ½ g t²

in this case y = 0 and the height of the cliff is y₀ = 100 m

          0 = 100 + 45.83 t - ½ 9.8 t²

          t² - 9,353 t - 20,408 = 0

we solve the quadratic equation

         t = [9,353 ±√ (9,353² + 4 20,408)] / 2

         t = [9,353 ± 13] / 2

         t₁ = 11.18 s

        t₂ = -1.8 s

Since time must be a positive quantity, the answer is t = 11.18 s

we calculate the horizontal distance traveled

        x = v₀ₓ t

        x = 45.83 11.18

        x = 512.4 m

c) speed when it hits the ground

         vₓ = v₀ₓ = 45.83 m / s

we look for vertical speed

         v_{y} = [tex]v_{oy}[/tex] - gt

         v_{y} = 0 - 9.8 11.18

         v_{y} = - 109.56 m / s

         v = (45.83 i ^ -109.56 j ^) m / s

Answer:

The rubber band will be stretched 0.02 m.

The work done in stretching is 0.11 J.

Explanation:

Force 1 = 44 N

extension of rubber band = 0.080 m

Force 2 = 11 N

extension = ?

According to Hooke's Law, force applied is proportional to the extension provided elastic limit is not extended.

F = ke

where k = constant of elasticity

e = extension of the material

F = force applied.

For the first case,

44 = 0.080K

K = 44/0.080 = 550 N/m

For the second situation involving the same rubber band

Force = 11 N

e = 550 N/m

11 = 550e

extension e = 11/550 = 0.02 m

The work done to stretch the rubber band this far is equal to the potential energy stored within the rubber due to the stretch. This is in line with energy conservation.

potential energy stored = [tex]\frac{1}{2}ke^{2}[/tex]

==> [tex]\frac{1}{2}* 550* 0.02^{2}[/tex] = 0.11 J

Answer:

he correct answers are a, b

Explanation:

In the two-slit interference phenomenon, the expression for interference is

          d sin θ= m λ                       constructive interference

          d sin θ = (m + ½) λ             destructive interference

in general this phenomenon occurs for small angles, for which we can write

           tanθ = y / L

           tan te = sin tea / cos tea = sin tea

           sin θ = y / La

un

derestimate the first two equations.

Let's do the calculation for constructive interference

         d y / L = m λ

the distance between maximum clos is and

         y = (me / d) λ

this is the position of each maximum, the distance between two consecutive maximums

         y₂-y₁ = (L   2/d) λ - (L 1 / d) λ₁          y₂ -y₁ = L / d λ

examining this equation if the wavelength decreases the value of y also decreases

the same calculation for destructive interference

         d y / L = (m + ½) κ

         y = [(m + ½) L / d] λ

again when it decreases the decrease the distance

the correct answers are a, b

Answer:

16 years.

Explanation:

Using Kepler's third Law.

P2=D^3

P=√d^3

Where P is the orbital period and d is the distance from the sun.

From the question the semi major axis of the asteroid is 4 AU= distance. The distance is always express in astronomical units.

P=?

P= √4^3

P= √256

P= 16 years.

Orbital period is 16 years.

Answer:

335.79cm/s

Explanation:

When a transverse wave of wavelength λ is produced during the vibration of a wire, the frequency(f), and the speed(v) of the wave pulses are related to the wavelength as follows;

v = fλ        ------------------(ii)

From the question;

f = 6.3Hz

λ = 53.3cm

Substitute these values into equation (i) as follows;

v = 6.3 x 53.3

v = 335.79cm/s

Therefore, the speed of the wave pulses along the wire is 335.79cm/s

Answer:

[tex]v_{2}=3.5 m/s[/tex]

Explanation:

Using the conservation of energy we have:

[tex]\frac{1}{2}mv^{2}=mgh[/tex]

Let's solve it for v:

[tex]v=\sqrt{2gh}[/tex]

So the speed at the lowest point is [tex]v=7 m/s[/tex]

Now, using the conservation of momentum we have:

[tex]m_{1}v_{1}=m_{2}v_{2}[/tex]

[tex]v_{2}=\frac{1*7}{2}[/tex]

Therefore the speed of the block after the collision is [tex]v_{2}=3.5 m/s[/tex]

I hope it helps you!

Answer:

The astronaut and the satellite are 53.718 m apart.

Explanation:

Given;

mass of spacewalking astronaut, = 88 kg

mass of satellite, = 645 kg

force exerts by the satellite, F = 110N

time for this action, t = 0.45 s

Determine the acceleration of the satellite after the push

F = ma

a = F / m

a = 110 / 645

a = 0.171 m/s²

Determine the final velocity of the satellite;

v = u + at

where;

u is the initial velocity of the satellite = 0

v = 0 + 0.171 x 0.45

v = 0.077 m/s

Determine the displacement of the satellite after 1.4 m

d₁ = vt

d₁ = 0.077 x (1.4 x 60)

d₁ = 6.468 m

According to Newton's third law of motion, action and reaction are equal and opposite;

Determine the backward acceleration of the astronaut after the push;

F = ma

a = F / m

a = 110 / 88

a = 1.25 m/s²

Determine the final velocity of the astronaut

v = u + at

The initial velocity of the astronaut = 0

v = 1.25 x 0.45

v = 0.5625 m/s

Determine the displacement of the astronaut after 1.4 min

d₂ = vt

d₂ = 0.5625 x (1.4 x 60)

d₂ = 47.25 m

Finally, determine the total separation between the astronaut and the satellite;

total separation = d₁ + d₂

total separation = 6.468 m + 47.25 m

total separation = 53.718 m

Therefore, the astronaut and the satellite are 53.718 m apart.

Answer:

Explanation:

Given data

length of cube= 0.2015 m

density = 19300 kg/m^3.

But the volume of cube is given as [tex]l*l*l= l^3[/tex]

[tex]volume -of- cube= 0.2015*0.2015*0.2015= 0.00818 m^3[/tex]

The density is expressed as = mass/volume

[tex]mass=19300*0.00818= 157.87m^3[/tex]

Given info

d = 0.000250 meters = distance between slits

L = 302 cm = 0.302 meters = distance from slits to screen

[tex]\theta_8 = 1.12^{\circ}[/tex] = angle to 8th max (note how m = 8 since we're comparing this to the form [tex]\theta_m[/tex])

[tex]x_n = x_5 = 3.33 \text{ cm} = 0.0333 \text{ meters}[/tex] (n = 5 as we're dealing with the 5th minimum )

---------------

Method 1

[tex]d\sin(\theta_m) = m\lambda\\\\0.000250\sin(\theta_8) = 8\lambda\\\\8\lambda = 0.000250\sin(1.12^{\circ})\\\\\lambda = \frac{0.000250\sin(1.12^{\circ})}{8}\\\\\lambda \approx 0.000 000 61082633\\\\\lambda \approx 6.1082633 \times 10^{-7} \text{meters}\\\\ \lambda \approx 6.11 \times 10^{-7} \text{ meters}\\\\ \lambda \approx 611 \text{ nm}[/tex]

Make sure your calculator is in degree mode.

-----------------

Method 2

[tex]\Delta x = \frac{\lambda*L*m}{d}\\\\L*\tan(\theta_m) = \frac{\lambda*L*m}{d}\\\\\tan(\theta_m) = \frac{\lambda*m}{d}\\\\\tan(\theta_8) = \frac{\lambda*8}{0.000250}\\\\\tan(1.12^{\circ}) = \frac{\lambda*8}{0.000250}\\\\\lambda = \frac{1}{8}*0.000250*\tan(1.12^{\circ})\\\\\lambda \approx 0.00000061094306 \text{ meters}\\\\\lambda \approx 6.1094306 \times 10^{-7} \text{ meters}\\\\\lambda \approx 611 \text{ nm}\\\\[/tex]

-----------------

Method 3

[tex]\frac{d*x_n}{L} = \left(n-\frac{1}{2}\right)\lambda\\\\\frac{0.000250*3.33}{302.0} = \left(5-\frac{1}{2}\right)\lambda\\\\0.00000275662251 \approx \frac{9}{2}\lambda\\\\\frac{9}{2}\lambda \approx 0.00000275662251\\\\\lambda \approx \frac{2}{9}*0.00000275662251\\\\\lambda \approx 0.00000061258279 \text{ meters}\\\\\lambda \approx 6.1258279 \times 10^{-7} \text{ meters}\\\\\lambda \approx 6.13 \times 10^{-7} \text{ meters}\\\\\lambda \approx 613 \text{ nm}\\\\[/tex]

There is a slight discrepancy (the first two results were 611 nm while this is roughly 613 nm) which could be a result of rounding error, but I'm not entirely sure.

Answer:

x=22.57 m

Explanation:

Given that

35 m in W of S

angle = 40 degrees

25 m in east

From the diagram

The angle

[tex]\theta=90-40=50^o[/tex]

From the triangle OAB

[tex]cos40^o=\frac{35^2+25^2-x^2}{2\times 35\times 25}[/tex]

[tex]1340.57=35^2+25^2-x^2[/tex]

x=22.57 m

Therefore the answer of the above problem will be 22.57 m

Answer:

The two masses are 3.39 Kg and 1.75 Kg

Explanation:

The gravitational force of attraction between two bodies is given by the formula;

F = Gm₁m₂/d²

where G is the gravitational force constant = 6.67 * 10⁻¹¹ Nm²Kg⁻²

m₁ = mass of first object; m₂ = mass of second object; d = distance of separation between the objects

Further calculations are provided in the attachment below

Answer: after 1.75 seconds

Explanation:

The only force acting on the ball is the gravitational force, so the acceleration will be:

a = -9.8 m/s^2

the velocity can be obtained by integrating over time:

v = -9.8m/s^2*t + v0

where v0 is the initial velocity; v0 = -7.95 m/s.

v = -9.8m/s^2*t - 7.95 m/s.

For the position we integrate again:

p = -4.9m/s^2*t^2 - 7.95 m/s*t + p0

where p0 is the initial position: p0 = 29m

p =  -4.9m/s^2*t^2 - 7.95 m/s*t + 29m

Now we want to find the time such that the position is equal to zero:

0 = -4.9m/s^2*t^2 - 7.95 m/s*t + 29m

Then we solve the Bhaskara's equation:

[tex]t = \frac{7.95 +- \sqrt{7.95^2 +4*4.9*29} }{-2*4.9} = \frac{7.95 +- 25.1}{9.8}[/tex]

Then the solutions are:

t = (7.95 + 25.1)/(-9.8) = -3.37s

t = (7.95 - 25.1)/(-9.8) = 1.75s

We need the positive time, then the correct answer is 1.75s

Answer and Explanation:

There is no probability of obtaining such a circuit of closed track current carrying wire whose field of magnitude displays i.e.  [tex]B \alpha \frac{1}{r^2}[/tex]

The magnetic field is a volume of vectors

And [tex]\phi\ bds = 0[/tex]. This ensures isolated magnetic poles or magnetic charges would not exit

Therefore for a closed path,  we never received magnetic field that followed the [tex]B \alpha \frac{1}{r^2}[/tex] it is only for the simple current-carrying wire for both finite or infinite length.

Answer:

The gray spheres is negatively charged while the red is positively charged

Explanation:

This is because theelectrophorus becomes less positive once it pulls some electrons away from the red sphere, but, the electrophorus is replaced on the slab and recharged by grounding it before it proceeds to charge the grey sphere, thereby giving it electrons and making it negatively charged

Answer:

The gray sphere has a positive charge and the red sphere has a positive charge.

Answer:

Coulomb's law is:

[tex]F = \frac{1}{4*pi*e0} *(q1*q2)/r^2[/tex]

First, force has units of Newtons, the charges have units of Coulombs, and r, the distance, has units of meters, then, working only with the units we have:

N = (1/{e0})*C^2/m^2

then we have:

{e0} = C^2/(m^2*N)

And we know that N = kg*m/s^2

then the dimensions of e0 are:

{e0} = C^2*s^2/(m^3)

(current square per time square over cubed distance)

And knowing that a Faraday is:

F = C^2*S^2/m^2

The units of e0 are:

{e0} = F/m.

Answer:

The main points of difference between a citizen and alien are: (a) A citizen is a permanent resident of a state, while an alien is a temporary resident, who comes for a specific duration of time as a tourist or on diplomatic assignment. ... Aliens do not possess such rights in the state where they reside temporarily

Explanation:

Answer:

superscript

Explanation:

When looking at the chemical symbol, the charge of the ion is displayed as the Superscript. This is because the charge of ions is usually written up on the chemical symbol while the atom/molecule is usually written down the chemical symbol. The superscript refers to what is written up on the formula while the subscript is written down on the formula.

An example is H2O . The 2 present represents two molecule of oxygen and its written as the subscript while Fe2+ in which the 2+ is written up is known as the superscript.

Answer:

superscript

Explanation:

Answer:

The angular momentum of the crate is [tex]M_{C} V_{1} d[/tex]

Explanation:

mass of the crate = [tex]M_{C}[/tex]

speed of forklift = [tex]V_{1}[/tex]

The distance between the center of the mass and the point A = d

Recall that the angular moment is the moment of the momentum.

[tex]L = P*d[/tex]    ..... equ 1

where L is the angular momentum,

P is the momentum of the system,

d is the perpendicular distance between the crate and the point on the axis about which the momentum acts. It is equal to d from the image

Also, we know that the momentum P is the product of mass and velocity

P = mv      ....equ 2

in this case, the mass = [tex]M_{C}[/tex]

the velocity = [tex]V_{1}[/tex]

therefore, the momentum P = [tex]M_{C}[/tex][tex]V_{1}[/tex]

we substitute equation 2 into equation 1 to give

[tex]L = M_{C} V_{1} d[/tex]

Answer:

The wavelength will be 33.9 cm

Explanation:

Given;

frequency of the wave, F = 1200 Hz

Tension on the wire, T = 800 N

wavelength, λ = 39.1 cm

[tex]F = \frac{ \sqrt{\frac{T}{\mu} }}{\lambda}[/tex]

Where;

F is the frequency of the wave

T is tension on the string

μ is mass per unit length of the string

λ is wavelength

[tex]\sqrt{\frac{T}{\mu} } = F \lambda\\\\\frac{T}{\mu} = F^2\lambda^2\\\\\mu = \frac{T}{F^2\lambda^2} \\\\\frac{T_1}{F^2\lambda _1^2} = \frac{T_2}{F^2\lambda _2^2} \\\\\frac{T_1}{\lambda _1^2} = \frac{T_2}{\lambda _2^2}\\\\T_1 \lambda _2^2 = T_2\lambda _1^2\\\\[/tex]

when the tension is decreased to 600 N, that is T₂ = 600 N

[tex]T_1 \lambda _2^2 = T_2\lambda _1^2\\\\\lambda _2^2 = \frac{T_2\lambda _1^2}{T_1} \\\\\lambda _2 = \sqrt{\frac{T_2\lambda _1^2}{T_1}} \\\\\lambda _2 = \sqrt{\frac{600* 0.391^2}{800}}\\\\\lambda _2 = \sqrt{0.11466} \\\\\lambda _2 =0.339 \ m\\\\\lambda _2 =33.9 \ cm[/tex]

Therefore, the wavelength will be 33.9 cm

Answer:

Velocity.

Explanation:

Projectile motion is characterized as the motion that an object undergoes when it is thrown into the air and it is only exposed to acceleration due to gravity.

As per the question, 'any change in the initial velocity of the projectile(object having gravity as the only force) would lead to a change in the range as well as the maximum height of the projectile.' To illustrate numerically:

Horizontal range: As per expression:

R= ([tex]u^{2}[/tex]*sin2θ)/g

the range depending on the square of the initial velocity.

Maximum height: As per expression:

H= ([tex]u^{2}[/tex] * [tex]sin^{2}[/tex]θ )/2g

the maximum distance also depends upon square of the initial velocity.

​

​

​

Answer:

A. Her total angular momentum has decreased

Explanation:

Total angular momentum is the product of her moment of inertia and angular velocity. In this scenario it doesn’t decrease but rather remains constant as the movement of the arms doesn’t have any effect on the total angular momentum.

The movement of the arm under certain conditions however has varying effects and changes on parameters such as the moment of inertia and the angular speed.

Answer:

 θ  = 180

Explanation:

When an electric dipole is placed in an electric field, there is a torque due to the electric force

           τ = p x E

by rotating the dipole there is a change in potential energy

        ΔU = ∫ τ dθ

        ΔU = p E (cos θ₂ - cos θ₁)

when the dipole starts from an angle to the equilibrium position for θ = 0

          ΔU = pE (cos θ  - cos 0)

           cos θ  = 1 + DU / pE)

let's apply this expression to our case, the change in potential energy is ΔU = -0.2J

let's calculate

          cos θ  = 1 -0.2 / 0.1

          cos θ  = -1

           θ  = 180

Answer:

Acceleration of the asteroid relative to the spacecraft = 2ε[tex]E^{2}[/tex]A/m

Explanation:

The maximum electric field in the beam = 2E

cross-sectional area of beam = A

The intensity of an electromagnetic wave with electric field is

I = cε[tex]E_{0} ^{2}[/tex]/2

for [tex]E_{0}[/tex] = 2E

I = 2cε[tex]E^{2}[/tex]    ....equ 1

where

I is the intensity

c is the speed of light

ε is the permeability of free space

[tex]E_{0}[/tex]  is electric field

Radiation pressure of an electromagnetic wave on an absorbing surface is given as

P = I/c

substituting for I from above equ 1. we have

P = 2cε[tex]E^{2}[/tex]/c = 2ε[tex]E^{2}[/tex]    ....equ 2

Also, pressure P = F/A

therefore,

F = PA    ....equ 3

where

F is the force

P is pressure

A is cross-sectional area

substitute equ 2 into equ 3, we have

F = 2ε[tex]E^{2}[/tex]A

force on a body = mass x acceleration.

that is

F = ma

therefore,

a = F/m

acceleration of the asteroid will then be

a = 2ε[tex]E^{2}[/tex]A/m

where m is the mass of the asteroid.

Answer:

  K_{f} / K₀ =1.12

Explanation:

This problem must work using the conservation of angular momentum (L), so that the moment is conserved in the system all the forces must be internal and therefore the torque is internal and the moment is conserved.

Initial moment. With arms outstretched

         L₀ = I₀ w₀

the wo value is 5.0 rad / s

final moment. After he shrugs his arms

         [tex]L_{f}[/tex] = I_{f}  w_{f}

indicate that the moment of inertia decreases by 11%

        I_{f} = I₀ - 0.11 I₀ = 0.89 I₀

        L_{f} = L₀

        I_{f} w_{f}  = I₀ w₀

        w_{f} = I₀ /I_{f}    w₀

let's calculate

        w_{f} = I₀ / 0.89 I₀   5.0

        w_{f} = 5.62 rad / s

Having these values ​​we can calculate the change in kinetic energy

         [tex]K_{f}[/tex] / K₀ = ½ I_{f} w_{f}² (½ I₀ w₀²)

         K_{f} / K₀ = 0.89 I₀ / I₀ (5.62 / 5)²

         K_{f} / K₀ =1.12

Question:

A 10 gauge copper wire carries a current of 15 A. Assuming one free electron per copper atom, calculate the drift velocity of the electrons. (The cross-sectional area of a 10-gauge wire is 5.261 mm².)

Answer:

3.22 x 10⁻⁴ m/s

Explanation:

The drift velocity (v) of the electrons in a wire (copper wire in this case) carrying current (I) is given by;

v = [tex]\frac{I}{nqA}[/tex]

Where;

n = number of free electrons per cubic meter

q =  electron charge

A =  cross-sectional area of the wire

First let's calculate the number of free electrons per cubic meter (n)

Known constants:

density of copper, ρ = 8.95 x 10³kg/m³

molar mass of copper, M = 63.5 x 10⁻³kg/mol

Avogadro's number, Nₐ = 6.02 x 10²³ particles/mol

But;

The number of copper atoms, N, per cubic meter is given by;

N = (Nₐ x ρ / M)          -------------(ii)

Substitute the values of Nₐ, ρ and M into equation (ii) as follows;

N = (6.02 x 10²³ x 8.95 x 10³) / 63.5 x 10⁻³

N = 8.49 x 10²⁸ atom/m³

Since there is one free electron per copper atom, the number of free electrons per cubic meter is simply;

n = 8.49 x 10²⁸ electrons/m³

Now let's calculate the drift electron

Known values from question:

A = 5.261 mm² = 5.261 x 10⁻⁶m²

I = 23A

q = 1.6 x 10⁻¹⁹C

Substitute these values into equation (i) as follows;

v = [tex]\frac{I}{nqA}[/tex]

v = [tex]\frac{23}{8.49*10^{28} * 1.6 *10^{-19} * 5.261*10^{-6}}[/tex]

v = 3.22 x 10⁻⁴ m/s

Therefore, the drift electron is 3.22 x 10⁻⁴ m/s

Answer:

the answers are provided in the attachments below

Explanation:

Gauss law state that the net electric field coming out of a closed surface is directly proportional to the charge enclosed inside the closed surface

Applying Gauss law to the long solid cylinder

A) E ( electric field ) = p*r / 2 * [tex]e_{0}[/tex]

B) E = 2K λ / r

C) Answers from parts a and b are the same

D) attached below

Applying Gauss's law which states that the net electric field in an enclosed surface is directly ∝ to the charge found in the enclosed surface.

A ) The expression for the electric field inside the volume at a distance r

Gauss law :  E. A = [tex]\frac{q}{e_{0} }[/tex]  ----- ( 1 )

where : A = surface area = 2πrL ,  q = p(πr²L)

back to equation ( 1 )

E ( electric field ) = p*r / 2 * [tex]e_{0}[/tex]

B) Electric field at point Outside the volume in terms of charge per unit length  λ

Given that:  linear charge density = area * volume charge density

                                            λ    =  πR²P

from Gauss's law : E ( 2πrL) = [tex]\frac{q}{e_{0} }[/tex]

∴ E = [tex]\frac{\pi R^{2}P }{2e_{0}r\pi }[/tex]    ----- ( 2 )

where : πR²P = λ

Back to equation ( 2 )

E = λ  / 2e₀π*r              where : k = 1 / 4πe₀

∴ The electric field ( E ) at point outside the volume in terms of charge per unit Length λ

E = 2K λ / r

C) Comparing answers A and B

Answers to part A and B are similar

Hence we can conclude that Applying Gauss law to the long solid cylinder

E ( electric field ) = p*r / 2 * [tex]e_{0}[/tex], E = 2K λ / r also Answers from parts a and b are the same.

Learn more about Gauss's Law : https://brainly.com/question/15175106

Answer:

 θ₁ = 0.04º , θ₂ = 0.00118º

Explanation:

The equation that describes the diffraction pattern of a network is

             d sin θ = m λ

where the diffraction order is, in this case they indicate that the order

m = 1

           θ = sin⁻¹ (λ / d)

Trfuvsmod ls inrsd fr ll red s SI units

           d = 12000 line / inc (1 inc / 2.54cm) = 4724 line / cm

the distance between two lines we can look for it with a direct proportions rule

If there are 4724 lines in a centimeter, the distance for two hundred is

            d = 2 lines (1 cm / 4724 line) = 4.2337 10⁻⁴ cm

let's calculate the angles

λ = 300 10-9 m

            θ₁ = sin⁻¹ (300 10-9 / 4,2337 10-4)

            θ₁ = sin⁻¹ (7.08 10-4)

            θ₁ = 0.04º

λ = 5000

          θ₂ = sin-1 (500 10-9 / 4,2337 10-4)

          θ₂ = 0.00118º