A parallel–plate capacitor is initially charged by connecting it to a battery. The battery is then disconnected. If the distance between the plates is increased, what happens to the charge on the capacitor and the voltage across it?

a. The charge remains fixed and the voltage decreases.
b. The charge decreases and the voltage remains fixed.
c. The charge remains fixed and the voltage increases.
d. The charge decreases and the voltage increases.

Answer:

B. over the symbol.

Explanation:

vectors are represented with a symbol carrying an arrow head with also indicates direction

Answer:

υ = 9.375 x 10¹⁴ Hz to 10.34 x 10¹⁴ Hz

Explanation:

The frequency of an electromagnetic radiation can be given by the following formula:

υ = c/λ

where,

υ = frequency of electromagnetic wave = ?

c = speed of light = 3 x 10⁸ m/s

λ = wavelength of electromagnetic wave = 290 nm to 320 nm

FOR LOWER LIMIT OF FREQUENCY:

λ = 320 nm = 3.2 x 10⁻⁷ m

Therefore,

υ = (3 x 10⁸ m/s)/(3.2 x 10⁻⁷ m)

υ = 9.375 x 10¹⁴ Hz

FOR UPPER LIMIT OF FREQUENCY:

λ = 290 nm = 3.2 x 10⁻⁷ m

Therefore,

υ = (3 x 10⁸ m/s)/(2.9 x 10⁻⁷ m)

υ = 10.34 x 10¹⁴ Hz

Therefore, the frequency range for UVB radiations is:

υ = 9.375 x 10¹⁴ Hz to 10.34 x 10¹⁴ Hz

Answer:

Q = 590,940 J

Explanation:

Given:

Specific heat (c) = 1.75 J/(g⋅°C)

Mass(m) = 2.01 kg = 2,010

Change in temperature (ΔT) = 191 - 23 = 168°C

Find:

Heat required (Q)

Computation:

Q = mcΔT

Q = (2,010)(1.75)(168)

Q = 590,940 J

Q = 590.94 kJ

Answer:

the frequency of revolution of the second particle is f

Explanation:

centripetal force is balanced by the magnetic force for object under magnetic field is given as

Mv²/r= qvB

But v= omega x r

Omega= 2pi x f

f= qB/2pi x M

So since frequency does not depend on the velocity.therefore the frequency of revolution of the second particle remains the same and its equal to f

Answer:

the planet Earth is a good example

Answer:

Explanation:

A.the direction of induced current will be clockwise

B: Changing 18cm and 6.8cm into 0.18m and 0.68

2.5

Divide them both by 2 to find the radius . Now we have 0.09 and .034m.

Now use Φ=(π*0.09^2)(.75 T)cos0 and the 0.019wb

(π*0.034^2)(.75 T)cos0 and the 0.00272wb

ow use ε=-N(ΔΦ/Δt)

For ΔΦ, 0.091-0.0027=0.0883

C.

To find the current, use I=ε/R

0.0883/2.5= 0.035A

Answer:

   h = 4 sin (314.15 t)

Explanation:

This is a kinematics exercise, as the system is rotating at a constant speed.

          w = θ / t

          θ = w t

in angular motion all angles are measured in radians, which is defined

         θ = s / R

   we substitute

          s / R = w t

          s = w R t

let's reduce the magnitude to the SI system

    w = 3000 rev / min (2π rad / 1rev) (1min / 60 s) = 314.16 rad / s

let's calculate

   s = 314.16 4 t

   s = 1,256.6 t

this is the value of the arc

Let's find the function of this system, let's use trigonometry to find the projection on the x axis

                  cos θ = x / R

                  x = R cos θ

                  x = R cos wt

projection onto the y-axis is

               sin θ = y / R

how is a uniform movement

               θ = w t

               y = R sin wt

In the case y = h

              h = R sin wt

              h = 4 sin (314.15 t)

Answer:

The electric field just inside the charged conductor is zero.

Explanation:

Electric field is defined as the region where electrical force is experienced by an electric charge usually as a result of the presence of another electric charge. A charged conductor is said to be in electrostatic equilibrium when it is in an electrostatically balanced state. This simply means a state in which the free electrical charges in the charged conductor have stopped moving.

For any charged conductor that has attained electrostatic equilibrium, the electric field at any point below the surface of the charged conductor falls to zero. Hence the electric field just inside the charged conductor is zero.

Answer:

As the mass increases, the moment of inertia(I) increases, therefore, the angular momentum(L) increases too.

Explanation:

friction can be defined as resistance in motion of bodies in relative to one another

momentum is the product of mass and velocity

torque is the time rate of change in momentum

τ = [tex]\frac{dL}{dt}[/tex]

where L = Iω = mvr

I = moment of inertia

ω=  angular frequency

if there is no external force(torque) acting on the system, then

[tex]\frac{dL}{dt}[/tex] = 0

dL = 0 = constant

moment of inertia I depends on the distribution of mass on the axis of rotation.

as the mass increases, the angular momentum(L) increases

angular frequency, ω, remains constant

Answer:

The voltage across the capacitor will remain constant

The capacitance of the capacitor will increase

The electric field between the plates will remain constant

The charge on the plates will increase

The energy stored in the capacitor will increase

Explanation:

First of all, if a capacitor is connected to a voltage source, the voltage or potential difference across the capacitor will remain constant. The electric field across the capacitor will stay constant since the voltage is constant, because the electric field is proportional to the voltage applied. Inserting a dielectric material into the capacitor increases the charge on the capacitor.

The charge on the capacitor is equal to

Q = CV

Since the voltage is constant, and the charge increases, the capacitance will also increase.

The energy in a capacitor is given as

E = [tex]\frac{1}{2}CV^{2}[/tex]

since the capacitance has increased, the energy stored will also increase.

Answer:

please brainliest!!!

Explanation:

V1/√T1 =V2/√T2

V1 = 331m/s

T1 = 0°C = 273k

V2 = ?

T2 = 35°c = 308k

Answer:

a. The primary turns is 60 turns

b. The secondary voltage will be 360 volts.

Explanation:

Given data

secondary turns N2= 40 turns

primary turns N1= ?

primary voltage V1= 120 volts

secondary voltage V2= 8 volts

Applying the transformer formula which is

[tex]\frac{N1}{N2} =\frac{V1}{V2}[/tex]

we can solve for N1 by substituting into the equation above

[tex]\frac{N1}{40} =\frac{120}{8} \\\ N1= \frac{40*120}{8} \\\ N1= \frac{4800}{8} \\\ N1= 60[/tex]

the primary turns is 60 turns

If the primary voltage is V1 240 volts hence the secondary voltage V2 will be (to get the voltage of the secondary coil using emf substitute the values of the previously gotten N1 and N2 using V1 as 240 volts)

[tex]\frac{40}{60} =\frac{240}{V2}\\\\V2= \frac{60*240}{40} \\\\V2=\frac{ 14400}{40} \\\\V2= 360[/tex]

the secondary voltage will be 360 volts.

(a) In the primary coil, you have "60 turns".

(b) The emf across the secondary coil would be "360 volts".

According to the question,

Primary voltage, V₁ = 120 volts

Secondary voltage, V₂ = 8 volts

Secondary turns, N₂ = 40 turns

(a) By applying transformer formula,

→ [tex]\frac{N_1}{N_2} = \frac{V_1}{V_2}[/tex]

or,

   N₁ = [tex]\frac{N_2\times V_1}{V_2}[/tex]

By substituting the values,

        = [tex]\frac{40\times 120}{8}[/tex]

        = [tex]\frac{4800}{8}[/tex]

        = 60

(2) Again by using the above formula,

→ V₂ = [tex]\frac{60\times 240}{40}[/tex]

       = [tex]\frac{14400}{40}[/tex]

       = 360 volts.

Thus the above approach is correct.  

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Answer:

72J

Explanation:

distance moved is equal to 3m.then just substitute x with 3m.

Fx = (14(3) - 3.0(3)2)) N

Fx =(42-18)N

Fx =24N

W=Fx *S

W=24N*3m

W=72J

The answer is 72J.

Distance moved is equal to 3m.

Then just substitute x with 3m.

Fx = (14(3) - 3.0(3)2)) N

Fx =(42-18)N

Fx =24N

W=Fx *S

W=24N*3m

W=72J

A force is a push or pulls upon an object resulting from the object's interaction with another object. Whenever there is an interaction between two objects, there is a force upon each of the objects.

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#SPJ2

Answer:

Explanation:

Change in entropy is expressed as ΔS = ΔQ/T where;

ΔQ is the total heat change during conversion of ice to water.

T is the room temperature

First we need to calculate the total change in heat using the conversion formulae;

ΔQ = mL + mcΔθ (total heat energy absorbed during phase change)

m is the mass of ice/water = 900g = 0.9kg

L is the latent heat of fusion of ice = 3.33 x 10⁵J/kg

c is the specific heat capacity of water = 4200J/kgK

Δθ is the change in temperature of water = 10°C - 0C = 10°C = 283K

Substituting the given values into ΔQ;

ΔQ = 0.9(333000)+0.9(4200)(283)

ΔQ = 299700 + 1069740

ΔQ = 1,369,440 Joules

Since Change in entropy ΔS = ΔQ/T

ΔS =  1,369,440/30+273

ΔS = 1,369,440/303

ΔS = 4519.60 J/K

Hence, the change in entropy of the universe is 4519.60 J/K

Answer:

B. Keeping your right hand flat, point your thumb in the direction of the current or the velocity of the charged particle, the remaining four fingers perpendicular to the thumb in the direction of magnetic field. The magnetic force, as the result of the magnetic field on the current, is the direction your palm is facing.

Explanation:

This is the Fleming's right hand rule, which was stated to explain the relationship or induction ability of the magnetic field, current or velocity of charged particles and magnetic force. These three variables are held mutually perpendicularly to one another.

The most suitable description of the right-hand rule is option B which clarifies the perpendicular mutual relationship of the thumb in the direction of the current or the velocity of the charged particle, the remaining four fingers perpendicular to the thumb in the direction of magnetic field. The magnetic force, as the result of the magnetic field on the current, is the direction your palm is facing.

Answer:

0.306mm

Explanation:

The radius of the conductor is 3mm, or 0.003m

The area of the conductor is:

A = π*r^2 = π*(.003)^2 = 2.8*10^-5 m^2

The current density is:

J = 130/2.8*10^-5 = 4.64*10^6 A/m

According to the listed reference:

Vd = J/(n*e) = 4.64*10^6 / ( 8.46*10^28 * 1.6*10^-19 ) = 0.34*10^-6 m/s = 0.34mm/s

The distance traveled is:

x = v*t = 0.34 * .90 = 0.306 mm

Answer:

Three halves of a wavelength I.e 7lambda/2

Explanation:

See attached file pls

Answer:effective

Explanation:

Answer:

The  fraction fraction of the final energy is stored in an initially uncharged capacitor after it has been charging for 3.0 time constants is  

      [tex]k = 0.903[/tex]

Explanation:

From the question we are told that

     The time  constant  [tex]\tau = 3[/tex]

The potential across the capacitor can be mathematically represented as

     [tex]V = V_o (1 - e^{- \tau})[/tex]

Where [tex]V_o[/tex] is the voltage of the capacitor when it is fully charged

    So   at  [tex]\tau = 3[/tex]

     [tex]V = V_o (1 - e^{- 3})[/tex]

     [tex]V = 0.950213 V_o[/tex]

   Generally energy stored in a capacitor is mathematically represented as

             [tex]E = \frac{1}{2 } * C * V ^2[/tex]

In this equation the energy stored is directly proportional to the the square of the potential across the capacitor

Now  since capacitance is  constant  at  [tex]\tau = 3[/tex]

        The  energy stored can be evaluated at as

         [tex]V^2 = (0.950213 V_o )^2[/tex]

       [tex]V^2 = 0.903 V_o ^2[/tex]

Hence the fraction of the energy stored in an initially uncharged capacitor is  

      [tex]k = 0.903[/tex]

Answer:

At 3.86K

Explanation:

The following data are obtained from a straight line graph of C/T plotted against T2, where C is the measured heat capacity and T is the temperature:

gradient = 0.0469 mJ mol−1 K−4 vertical intercept = 0.7 mJ mol−1 K−2

Since the graph of C/T against T2 is a straight line, the are related by the straight line equation: C /T =γ+AT². Multiplying by T, we get C =γT +AT³ The electronic contribution is linear in T, so it would be given by the first term: Ce =γT. The lattice (phonon) contribution is proportional to T³, so it would be the second term: Cph =AT³. When they become equal, we can solve these 2 equations for T. This gives: T = √γ A .

We can find γ and A from the graph. Returning to the straight line equation C /T =γ+AT². we can see that γ would be the vertical intercept, and A would be the gradient. These 2 values are given. Substituting, we f ind: T =

√0.7/ 0.0469 = 3.86K.

Answer:

Power = Force × Distance/time

Power = Force × Velocity

Power = 2,300.0 N × 28.0 m/s²

Power = 64400 Nm/s

Explanation:

First show the formula of Power

Re-arrange formula and used to work out Power

Pretty simple stuff!

Hope this Helps!!

Answer:

a) The final speed of the 0.400-kg puck after the collision is 2.254 meters per second, b) The negative sign of the solution found in part a) indicates that 0.400-kg puck is moving westwards, c) The speed of the 0.900-kg puck after the collision is 3.606 meters per second eastwards.

Explanation:

a) Since collision is perfectly elastic and there are no external forces exerted on pucks system, the phenomenon must be modelled after the Principles of Momentum and Energy Conservation. Changes in gravitational potential energy can be neglected. That is:

Momentum

[tex]m_{1}\cdot v_{1,o} + m_{2}\cdot v_{2,o} = m_{1}\cdot v_{1,f} + m_{2}\cdot v_{2,f}[/tex]

Energy

[tex]\frac{1}{2}\cdot (m_{1}\cdot v_{1,o}^{2}+ m_{2}\cdot v_{2,o}^{2})=\frac{1}{2}\cdot (m_{1}\cdot v_{1,f}^{2}+ m_{2}\cdot v_{2,f}^{2})[/tex]

[tex]m_{1}\cdot v_{1,o}^{2} + m_{2}\cdot v_{2,o}^{2} = m_{1}\cdot v_{1,f}^{2} + m_{2}\cdot v_{2,f}^{2}[/tex]

Where:

[tex]m_{1}[/tex], [tex]m_{2}[/tex] - Masses of the 0.400-kg and 0.900-kg pucks, measured in kilograms.

[tex]v_{1,o}[/tex], [tex]v_{2,o}[/tex] - Initial speeds of the 0.400-kg and 0.900-kg pucks, measured in meters per second.

[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Final speeds of the 0.400-kg and 0.900-kg pucks, measured in meters per second.

If [tex]m_{1} = 0.400\,kg[/tex], [tex]m_{2} = 0.900\,kg[/tex], [tex]v_{1,o} = +5.86\,\frac{m}{s}[/tex], [tex]v_{2,o} = 0\,\frac{m}{s}[/tex], the system of equation is simplified as follows:

[tex]2.344\,\frac{kg\cdot m}{s} = 0.4\cdot v_{1,f} + 0.9\cdot v_{2,f}[/tex]

[tex]13.736\,J = 0.4\cdot v_{1,f}^{2}+0.9\cdot v_{2,f}^{2}[/tex]

Let is clear [tex]v_{1,f}[/tex] in first equation:

[tex]0.4\cdot v_{1,f} = 2.344 - 0.9\cdot v_{2,f}[/tex]

[tex]v_{1,f} = 5.86-2.25\cdot v_{2,f}[/tex]

Now, the same variable is substituted in second equation and resulting expression is simplified and solved afterwards:

[tex]13.736 = 0.4\cdot (5.86-2.25\cdot v_{2,f})^{2}+0.9\cdot v_{2,f}^{2}[/tex]

[tex]13.736 = 0.4\cdot (34.340-26.37\cdot v_{2,f}+5.063\cdot v_{2,f}^{2})+0.9\cdot v_{2,f}^{2}[/tex]

[tex]13.736 = 13.736-10.548\cdot v_{2,f} +2.925\cdot v_{2,f}^{2}[/tex]

[tex]2.925\cdot v_{2,f}^{2}-10.548\cdot v_{2,f} = 0[/tex]

[tex]2.925\cdot v_{2,f}\cdot (v_{2,f}-3.606) = 0[/tex]

There are two solutions:

[tex]v_{2,f} = 0\,\frac{m}{s}[/tex] or [tex]v_{2,f} = 3.606\,\frac{m}{s}[/tex]

The first root coincides with the conditions before collision and the second one represents a physically reasonable solution.

Now, the final speed of the 0.400-kg puck is: ([tex]v_{2,f} = 3.606\,\frac{m}{s}[/tex])

[tex]v_{1,f} = 5.86-2.25\cdot (3.606)[/tex]

[tex]v_{1,f} = -2.254\,\frac{m}{s}[/tex]

The final speed of the 0.400-kg puck after the collision is 2.254 meters per second.

b) The negative sign of the solution found in part a) indicates that 0.400-kg puck is moving westwards.

c) The speed of the 0.900-kg puck after the collision is 3.606 meters per second eastwards.

Answer:

257 kN.

Explanation:

So, we are given the following data or parameters or information in the following questions;

=> "A jet transport with a landing speed

= 200 km/h reduces its speed to = 60 km/h with a negative thrust R from its jet thrust reversers"

= > The distance = 425 m along the runway with constant deceleration."

=> "The total mass of the aircraft is 140 Mg with mass center at G. "

We are also give that the "aerodynamic forces on the aircraft are small and may be neglected at lower speed"

Step one: determine the acceleration;

=> Acceleration = 1/ (2 × distance along runway with constant deceleration) × { (landing speed A)^2 - (landing speed B)^2 × 1/(3.6)^2.

=> Acceleration = 1/ (2 × 425) × (200^2 - 60^2) × 1/(3.6)^2 = 3.3 m/s^2.

Thus, "the reaction N under the nose wheel B toward the end of the braking interval and prior to the application of mechanical braking" = The total mass of the aircraft × acceleration × 1.2 = 15N - (9.8 × 2.4 × 140).

= 140 × 3.3× 1.2 = 15N - (9.8 × 2.4 × 140).

= 257 kN.

The reaction N under the nose wheel B towards the end of the braking interval =  257 kN

Given data :

Landing speed of Jet = 200 km/h

Distance = 425 m

Total mass of aircraft = 140 Mg  with mass center at G

  Jet acceleration = 1 / (2 *425) * (200²  - 60² ) *  1 / (3.6)²

                              = 3.3 m/s²

Reaction N = Total mass of aircraft * jet acceleration* 1.2 = 15N - (9.8*2.4* 140).   ----- ( 1 )

∴ Reaction N = 140 * 3.3 * 1.2 = 15 N - ( 9.8*2.4* 140 )  

 Hence Reaction N = 257 KN

We can conclude that the The reaction N under the nose wheel B towards the end of the braking interval =  257 kN

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Answer:

4 tonne/m³

Explanation:

ρ = m / V

ρ = 49 g / (π (17.4 mm / 2)² (50.3 mm))

ρ = 0.0041 g/mm³

Converting to tonnes/m³:

ρ = 0.0041 g/mm³ (1 kg / 1000 g) (1 tonne / 1000 kg) (1000 mm / m)³

ρ = 4.1 tonne/m³

Rounding to one significant figure, the density is 4 tonne/m³.

Answer:

The frequency increases by 4 because it is inversely proportional to the wavelength.

Answer:

P = cte / V

therefore pressure and volume are inversely proportional

Explanation:

For this exercise we can join the ideal gases equation

        PV = n R T

they indicate that the amount of matter and the temperature are constant, therefore

         PV = cte

        P = cte / V

therefore pressure and volume are inversely proportional

Answer:

The de Broglie wavelength of electron βe = 2.443422 × 10⁻⁹ m

The de Broglie wavelength of proton βp = 5.70 × 10⁻¹¹ m

Explanation:

Thermal kinetic energy of electron or proton = KE

∴ KE = 3kbT/2

given that; kb = 1.38 x 10⁻²³ J/K , T = 1950 K

so we substitute

KE = ( 3 × 1.38 x 10⁻²³ × 1950 ) / 2

kE = 4.0365 × 10⁻²⁰ (  is the kinetic energy for both electron and proton at temperature T )

Now we know that

mass of electron M'e = 9.109 ×  10⁻³¹

mass of proton M'p = 1.6726 ×  10⁻²⁷

We also know that

KE = p₂ / 2m

from the equation, p = √ (2mKE)

{ p is momentum, m is mass }

de Broglie wavelength = β

so β = h / p = h / √ (2mKE)

h = Planck's constant = 6.626 ×  10⁻³⁴

∴ βe =  h / √ (2m'e × KE)

βe = 6.626 ×  10⁻³⁴ / √ (2 × 9.109 ×  10⁻³¹ × 4.0365 × 10⁻²⁰ )

βe = 6.626 ×  10⁻³⁴ / √  7.3536957 × 10⁻⁵⁰

βe = 6.626 × 10⁻³⁴  / 2.71176984642871 × 10⁻²⁵

βe = 2.443422 × 10⁻⁹ m

βp =  h / √ (2m'p ×KE)

βp = 6.626 ×  10⁻³⁴ / √ (2 × 1.6726 ×  10⁻²⁷ × 4.0365 × 10⁻²⁰ )

βp = 6.626 ×  10⁻³⁴ / √ 1.35028998 × 10⁻⁴⁶

βp =  6.626 ×  10⁻³⁴ / 1.16201978468527 ×  10⁻²³

βp = 5.702140 × 10⁻¹¹ m

Answer:

its supposed to be (a) 1W

Answer:

   R = r_protón / 2

Explanation:

The alpha particle when entering the magnetic field experiences a force and with Newton's second law we can describe its movement

      F = m a

Since the magnetic force is perpendicular, the acceleration is centripetal.

       a = v² / R

the magnetic force is

       F = q v x B = q v B sin θ

the field and the speed are perpendicular so the sin 90 = 1

we substitute

          qv B = m v² / R

          R = q v B / m v²

in the exercise they indicate

the charge  q = 2 e

the mass     m = 4 m_protón

        R = 2e v B / 4m_protón v²

we refer the result to the movement of the proton

         R = (e v B / m_proton) 1/2

the data in parentheses correspond to the radius of the proton's orbit

         R = r_protón / 2

Answer:

E = 17.64 x 10⁶ N/C = 17.64 MN/C

Explanation:

The electric field is given by the following formula:

E = F/q

E= W/q

E = mg/q

where,

E = magnitude of electric field = ?

m = mass of plastic sphere = 5.4 g = 5.4 x 10⁻³ kg

g = acceleration due to gravity = 9.8 m/s²

= charge = 3 nC = 3 x 10⁻⁹ C

Therefore,

E = (5.4 x 10⁻³ kg)(9.8 m/s²)/(3 x 10⁻⁹ C)

E = 17.64 x 10⁶ N/C = 17.64 MN/C