In an oscillating LC circuit, the total stored energy is U and the maximum current in the inductor is I. When the current in the inductor is I/2, the energy stored in the capacitor is

Answers

Answer 1

Answer:

The definition of that same given problem is outlined in the following section on the clarification.

Explanation:

The Q seems to be endless (hardly any R on the circuit). So energy equations to describe and forth through the inducer as well as the condenser.  

Presently take a gander at the energy stored in your condensers while charging is Q.

⇒  [tex]U =\frac{Qmax^2}{C}[/tex]

So conclude C doesn't change substantially as well as,

When,

⇒  [tex]Q=\frac{Qmax}{2}[/tex]

⇒  [tex]Q^2=\frac{Qmax^2}{4}[/tex]

And therefore only half of the population power generation remains in the condenser that tends to leave this same inductor energy at 3/4 U.


Related Questions

Two objects attract each other with a gravitational force of magnitude 1.02 10-8 N when separated by 19.7 cm. If the total mass of the two objects is 5.14 kg, what is the mass of each

Answers

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

A 5.0-Ω resistor and a 9.0-Ω resistor are connected in parallel. A 4.0-Ω resistor is then connected in series with this parallel combination. An ideal 6.0-V battery is then connected across the series-parallel combination of the three resistors. What is the current through (a) the 4.0-Ω resistor? (b) the 5.0-Ω resistor? (c) the 9.0-Ω resistor?

Answers

Answer:

Explanation:

The current through the  resistor is 0.83 A

.

Part b

The current through  resistor is 0.53 A

.

Part c

The current through  resistor is 0.30 A

An 88.0 kg spacewalking astronaut pushes off a 645 kg satellite, exerting a 110 N force for the 0.450 s it takes him to straighten his arms. How far apart are the astronaut and the satellite after 1.40 min?

Answers

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.

Two carts are connected by a loaded spring on a horizontal, frictionless surface. The spring is released and the carts push away from each other. Cart 1 has mass M and Cart 2 has mass M/3.

a) Is the momentum of Cart 1 conserved?

Yes

No

It depends on M

b) Is the momentum of Cart 2 conserved?

Yes

No

It depends on M

c) Is the total momentum of Carts 1 and 2 conserved?

Yes

No

It depends on M

d) Which cart ends up moving faster?

Cart 1

Cart 2

They move at the same speed

e) If M = 6 kg and Cart 1 moves with a speed of 16 m/s, what is the speed of Cart 2?

0 m/s

4.0 m/s

5.3 m/s

16 m/s

48 m/s

64 m/s

Answers

Answer:

a) yes

b) no

c) yes

d)Cart 2 with mass [tex]\frac{M}{3}[/tex]   is expected to be more faster

e) u₂ = 48 m/s

Explanation:

a) the all out linear momentum of an arrangement of particles of Cart 1 not followed up on by external forces is constant.

b) the linear momentum of Cart 2 will be acted upon by external force by Cart 1 with mass M, thereby it's variable and the momentum is not conserved

c) yes, the momentum is conserved because no external force acted upon it and both Carts share the same velocity after the reaction

note: m₁u₁ + m₂u₂ = (m₁ + m₂)v

d) Cart 2 with mass [tex]\frac{M}{3}[/tex] will be faster than Cart 1 because Cart 2 is three times lighter than Cart 1.

e) Given

m₁=  M

u₁ = 16m/s

m₂ =[tex]\frac{M}{3}[/tex]

u₂ = ?

from law of conservation of momentum

m₁u₁= m₂u₂

M× 16 = [tex]\frac{M}{3}[/tex] × u₂(multiply both sides by 3)

therefore, u₂ = [tex]\frac{3(M .16)}{M}[/tex] ("." means multiplication)

∴u₂ = 3×16 = 48 m/s

A 100 kg lead block is submerged in 2 meters of salt water, the density of which is 1096 kg / m3. Estimate the value of the hydrostatic pressure.

Answers

Answer:

21,920 Pascals

Explanation:

P = ρgh

P = (1096 kg/m³) (10 m/s²) (2 m)

P = 21,920 Pa

2. A pair of narrow, parallel slits sep by 0.25 mm is illuminated by 546 nm green light. The interference pattern is observed on a screen situated at 1.3 m away from the slits. Calculate the distance from the central maximum to the

Answers

Answer:

for the first interference m = 1   y = 2,839 10-3 m

for the second interference m = 2   y = 5,678 10-3 m

Explanation:

The double slit interference phenomenon, for constructive interference is described by the expression

                d sin θ = m λ

where d is the separation between the slits, λ the wavelength and m an integer that corresponds to the interference we see.

In these experiments in general the observation screen is L >> d, let's use trigonometry to find the angles

           tan θ = y / L

with the angle it is small,

          tan θ = sin θ / cos θ = sin θ

   

we substitute

         sin θ = y / L

         d y / L = m λ

the distance between the central maximum and an interference line is

        y = m λ L / d

let's reduce the magnitudes to the SI system

     λ = 546 nm = 546 10⁻⁹ m

     d = 0.25 mm = 0.25 10⁻³ m

let's substitute the values

      y = m 546 10⁻⁹ 1.3 / 0.25 10⁻³

      y =  m 2,839 10⁻³

the explicit value for a line depends on the value of the integer m, for example

for the first interference m = 1

the distance from the central maximum to the first line is y = 2,839 10-3 m

for the second interference m = 2

the distance from the central maximum to the second line is y = 5,678 10-3 m

A skater on ice with arms extended and one leg out spins at 3 rev/s. After he draws his arms and the leg in, his moment of inertia is reduced to 1/2. What is his new angular speed

Answers

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]

Julie is playing with a toy car and is pushing it around on the floor. The little car has a mass of 6.3 g. The car has a velocity of 2.5 m/s. What is the car's momentum?

Answers

Answer:

Momentum of the car = [tex]1.575\times 10^{-2}[/tex] kg meter per second

Explanation:

Julie is playing with a car which has mass = 6.3 g = [tex]6.3\times 10^{-3}[/tex] kg

Velocity of the car is 2.5 meter per second

Since formula to calculate the momentum of an object is,

p = mv

Where, p = momentum of the object

m = mass of the object

v = velocity of the object

By substituting these values in the formula,

p = [tex](6.3\times 10^{-3})\times 2.5[/tex]

  = [tex]1.575\times 10^{-2}[/tex] Kg meter per second

Therefore, momentum of the car will be [tex]1.575\times 10^{-2}[/tex] Kg meter per second.


When looking at the chemical symbol, the charge of the ion is displayed as the
-superscript
-subscript
-coefficient
-product

Answers

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:

A wave with a frequency of 1200 Hz propagates along a wire that is under a tension of 800 N. Its wavelength is 39.1 cm. What will be the wavelength if the tension is decreased to 600 N and the frequency is kept constant

Answers

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

Please Help!!!! I WILL GIVE BRAINLIEST!!!!!!!!!!!!!

Upon using Thomas Young’s double-slit experiment to obtain measurements, the following data were obtained. Use these data to determine the wavelength of light being used to create the interference pattern. Do this using three different methods.

The angle to the eighth maximum is 1.12°.

The distance from the slits to the screen is 302.0 cm.

The distance from the central maximum to the fifth minimum is 3.33 cm.

The distance between the slits is 0.000250 m.



The 3 equations I used were 1). d sin θ_m =(m)λ 2). delta x =λL/d and 3.) d(x_n)/L=(n-1/2)λ
but all my answers are different.
DID I DO SOMETHING WRONG!!!!!!!

Answers

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.

mention two similarities of citizen and aliens​

Answers

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:

A lens is designed to work in the visible, near-infrared, and near-ultraviolet. The best resolution of this lens from a diffraction standpoint is

Answers

In the near ultraviolet

The  lens is designed to work in the visible, near-infrared, and near-ultraviolet. The best resolution of this lens from a diffraction standpoint is: in the near-ultraviolet.

What is diffraction?

The act of bending light around corners such that it spreads out and illuminates regions where a shadow is anticipated is known as diffraction of light. In general, since both occur simultaneously, it is challenging to distinguish between diffraction and interference. The diffraction of light is what causes the silver lining we see in the sky. A silver lining appears in the sky when the sunlight penetrates or strikes the cloud.

Longer wavelengths of light are diffracted at a greater angle than shorter ones, with the amount of diffraction being dependent on the wavelength of the light. Hence, among the light waves of  the visible, near-infrared, and near-ultraviolet range, near-ultraviolet waves have the shortest wavelengths. So,  The best resolution of this lens from a diffraction standpoint is in the near-ultraviolet, where diffraction is minimum.

Learn more about diffraction here:

https://brainly.com/question/11176463

#SPJ5

Five identical cylinders are each acted on by forces of equal magnitude. Which force exerts the biggest torque about the central axes of the cylinders

Answers

Answer:

From the image, the force as shown in option A will exert the biggest torque on the cylinder about its central axes.

Explanation:

The image is shown below.

Torque is the product of a force about the center of rotation of a body, and the radius through which the force acts. For a given case such as this, in which the cylinders are identical, and the forces are of equal magnitude, the torque at the maximum radius away from the center will exert the maximum torque. Also, the direction of the force also matters. To generate the maximum torque, the force must be directed tangentially away from the circle formed by the radius through which the force acts away from the center. Option A satisfies both condition and hence will exert the most torque on the cylinder.

A "laser cannon" of a spacecraft has a beam of cross-sectional area A. The maximum electric field in the beam is 2E. The beam is aimed at an asteroid that is initially moving in the direction of the spacecraft. What is the acceleration of the asteroid relative to the spacecraft if the laser beam strikes the asteroid perpendicularly to its surface, and the surface is not reflecting

Answers

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.

A solid conducting sphere is placed in an external uniform electric field. With regard to the electric field on the sphere's interior, which statement is correct

Answers

Complete question:

A solid conducting sphere is placed in an external uniform electric field. With regard to the electric field on the sphere's interior, which statement is correct?

A. the interior field points in a direction parallel to the exterior field

B. There is no electric field on the interior of the conducting sphere.

C. The interior field points in a direction perpendicular to the exterior field.

D. the interior field points in a direction opposite to the exterior field.

Answer:

B. There is no electric field on the interior of the conducting sphere.

Explanation:

Conductors are said to have free charges that move around easily. When the conductor is now placed in a static electric field, the free charges react to attain electrostatic equilibrium (steady state).

Here, a solid conducting sphere is placed in an external uniform electric field. Until the lines of the electric field are perpendicular to the surface, the free charges will move around the spherical conductor, causing polarization. There would be no electric field in the interior of the spherical conductor because there would be movement of  free charges in the spherical conductor in response to any field until its neutralization.

Option B is correct.

There is no electric field on the interior of the conducting sphere.

can I get help please?​

Answers

Answer shown on photo

The Law of Biot-Savart shows that the magnetic field of an infinitesimal current element decreases as 1/r2. Is there anyway you could put together a complete circuit (any closed path of current-carrying wire) whose field exhibits this same 1/r^2 decrease in magnetic field strength? Explain your reasoning.

Answers

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.

The voltage between the cathode and the screen of a television set is 30 kV. If we assume a speed of zero for an electron as it leaves the cathode, what is its speed (m/s) just before it hits the screen

Answers

Answer:

The speed is  [tex]v =10.27 *10^{7} \ m/s[/tex]

Explanation:

From the question we are told that

      The  voltage  is  [tex]V = 30 kV = 30*10^{3} V[/tex]

      The  initial velocity of the electron is  [tex]u = 0 \ m/s[/tex]

Generally according to the law of energy conservation

    Electric potential Energy  =  Kinetic energy of the electron

So  

      [tex]PE = KE[/tex]

Where  

      [tex]KE = \frac{1}{2} * m* v^2[/tex]

Here  m is the mass of the electron with a value of  [tex]m = 9.11 *10^{-31} \ kg[/tex]

     and  

         [tex]PE = e * V[/tex]

      Here  e is the charge on the electron with a value  [tex]e = 1.60 *10^{-19} \ C[/tex]

=>    [tex]e * V = \frac{1}{2} * m * v^2[/tex]

=>      [tex]v = \sqrt{ \frac{2 * e * V}{m} }[/tex]

substituting values  

           [tex]v = \sqrt{ \frac{2 * (1.60*10^{-19}) * 30*10^{3}}{9.11 *10^{-31}} }[/tex]

          [tex]v =10.27 *10^{7} \ m/s[/tex]

When a particular wire is vibrating with a frequency of 6.3 Hz, a transverse wave of wavelength 53.3 cm is produced. Determine the speed of wave pulses along the wire.

Answers

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

An ice skater spinning with outstretched arms has an angular speed of 5.0 rad/s . She tucks in her arms, decreasing her moment of inertia by 11 % . By what factor does the skater's kinetic energy change? (Neglect any frictional effects.)

Answers

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

The interference of two sound waves of similar amplitude but slightly different frequencies produces a loud-soft-loud oscillation we call __________.
a. the Doppler effect
b. vibrato
c. constructive and destructive interference
d. beats

Answers

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

Monochromatic coherent light shines through a pair of slits. If the wavelength of the light is decreased, which of the following statements are true of the resulting interference pattern? (There could be more than one correct choice.)
a. The distance between the maxima decreases.
b. The distance between the minima decreases.
c. The distance between the maxima stays the same.
d. The distance between the minima increases.
e. The distance between the minima stays the same.

Answers

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

Use Coulomb’s law to derive the dimension for the permittivity of free space.



Answers

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.

An aging coyote cannot run fast enough to catch a roadrunner. He purchases on eBay a set of jet-powered roller skates, which provide a constant horizontal acceleration of 15.0 m/s2. The coyote starts at rest 70.0 m from the edge of a cliff at the instant the roadrunner zips past in the direction of the cliff.

Required:
a. Determine the minimum constant speed the roadrunner must have to reach the cliff before the coyote. At the edge of the cliff, the roadrunner escapes by making a sudden turn, while the coyote continues straight ahead. The coyote’s skates remain horizontal and continue to operate while he is in flight.
b. The cliff is 100 m above the flat floor of the desert. Determine how far from the base of the cliff the coyote lands.
c. Determine the components of the coyote’s impact velocity

Answers

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

A very long, solid cylinder with radius R has positive charge uniformly distributed throughout it, with charge per unit volume \rhorho.
(a) Derive the expression for the electric field inside the volume at a distance r from the axis of the cylinder in terms of the charge density \rhorho.
(b) What is the electric field at a point outside the volume in terms of the charge per unit length \lambdaλ in the cylinder?
(c) Compare the answers to parts (a) and (b) for r = R.
(d) Graph the electric-field magnitude as a function of r from r = 0 to r = 3R.

Answers

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

A uniform thin rod of mass ????=3.41 kg pivots about an axis through its center and perpendicular to its length. Two small bodies, each of mass m=0.249 kg , are attached to the ends of the rod. What must the length L of the rod be so that the moment of inertia of the three-body system with respect to the described axis is ????=0.929 kg·m2 ?

Answers

Answer:

The length of the rod for the condition on the question to be met is [tex]L = 1.5077 \ m[/tex]

Explanation:

The  Diagram for this  question is  gotten from the first uploaded image  

From the question we are told that

          The mass of the rod is [tex]M = 3.41 \ kg[/tex]

           The mass of each small bodies is  [tex]m = 0.249 \ kg[/tex]

           The moment of inertia of the three-body system with respect to the described axis is   [tex]I = 0.929 \ kg \cdot m^2[/tex]

             The length of the rod is  L  

     Generally the moment of inertia of this three-body system with respect to the described axis can be mathematically represented as

        [tex]I = I_r + 2 I_m[/tex]

Where  [tex]I_r[/tex] is the moment of inertia of the rod about the describe axis which is mathematically represented as  

        [tex]I_r = \frac{ML^2 }{12}[/tex]

And   [tex]I_m[/tex] the  moment of inertia of the two small bodies which (from the diagram can be assumed as two small spheres) can be mathematically represented  as

           [tex]I_m = m * [\frac{L} {2} ]^2 = m* \frac{L^2}{4}[/tex]

Thus  [tex]2 * I_m = 2 * m \frac{L^2}{4} = m * \frac{L^2}{2}[/tex]

Hence

       [tex]I = M * \frac{L^2}{12} + m * \frac{L^2}{2}[/tex]

=>   [tex]I = [\frac{M}{12} + \frac{m}{2}] L^2[/tex]

substituting vales  we have  

        [tex]0.929 = [\frac{3.41}{12} + \frac{0.249}{2}] L^2[/tex]

       [tex]L = \sqrt{\frac{0.929}{0.40867} }[/tex]

      [tex]L = 1.5077 \ m[/tex]

     

An empty parallel plate capacitor is connected between the terminals of a 9.0-V battery and charged up. The capacitor is then disconnected from the battery, and the spacing between the capacitor plates is doubled. As a result of this change, what is the new voltage between the plates of the capacitor

Answers

Answer:

The new voltage between the plates of the capacitor is 18 V

Explanation:

The charge on parallel plate capacitor is calculated as;

q = CV

Where;

V is the battery voltage

C is the capacitance of the capacitor, calculated as;

[tex]C = \frac{\epsilon _0A}{d} \\\\q =CV = (\frac{\epsilon _0A}{d})V = \frac{\epsilon _0A V}{d}[/tex]

[tex]q = \frac{\epsilon _0A V}{d}[/tex]

where;

ε₀ is permittivity of free space

A is the area of the capacitor

d is the space between the parallel plate capacitors

If only the space between the capacitors is doubled and every other parameter is kept constant, the new voltage will be calculated as;

[tex]q = \frac{\epsilon _0A V}{d} \\\\\frac{\epsilon _0A V}{d} = \frac{\epsilon _0A V}{d} \\\\\frac{V_1}{d_1} = \frac{V_2}{d_2} \\\\V_2 = \frac{V_1d_2}{d_1} \\\\(d_2 = 2d_1)\\\\V_2 = \frac{V_1*2d_1}{d_1} \\\\(V_1 = 9V)\\\\V_2 = \frac{9*2d_1}{d_1} \\\\V_2 = 9*2\\\\V_2 = 18 \ V[/tex]

Therefore, the new voltage between the plates of the capacitor is 18 V

An asteroid that has an orbit with a semi-major axis of 4 AU will have an orbital period of about ______ years.

Answers

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.

A 1500 kg car drives around a flat 200-m-diameter circular track at 25 m/s. What are the magnitude and direction of the net force on the car

Answers

Answer:

9,375

Explanation:

Data provided

The mass of the car m = 1500 Kg.

The diameter of the circular track D = 200 m.

For the computation of magnitude and direction of the net force on the car first we need to find out the radius of the circular path which is shown below:-

The radius of the circular path is

[tex]R = \frac{D}{2}[/tex]

[tex]= \frac{200}{2}[/tex]

= 100 m

after the radius of the circular path we can find the magnitude of the centripetal force with the help of below formula

[tex]Force F = \frac{mv^2}{R}[/tex]

[tex]= \frac{1500\times (25)^2}{100}[/tex]

= 9,375

Therefore for computing the magnitude of the centripetal force we simply applied the above formula.

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