Equal but opposite charges Q are placed on the square plates of an air-filled parallel-plate capacitor. The plates are then pulled apart to twice their original separation, which is small compared to the dimensions of the plates. Which of the following statements about this capacitor are true?
A) The energy stored in the capacitor has doubled.
B) The energy density in the capacitor has increased.
C) The electric field between the plates has increased.
D) The potential difference across the plates has doubled.
E) The capacitance has doubled.

Answer:

Explanation:

Carton cycle consists of four thermodynamic processes . The first is isothermal expansion at higher temperature , then adiabatic expansion which lowers the temperature of gas . The third process is isothermal compression at lower temperature and the last process is adiabatic compression which increases the temperature of the gas to its original temperature .

So the given process of isothermal compression must have been done at the temperature of 300K  , keeping the temperature constant .

Work done on gas at isothermal compression is equal to heat transfer .

work done on gas = 80 x 10³ J

work done on gas = n RT ln v₁ / v₂

n is number of moles v₁ and v₂ are initial and final volume

molecular weight of gas = 28.97 g

1.5 kg = 1500 / 28.97 moles

= 51.77 moles

work done on gas = n RT ln v₁ / v₂

Putting the values in the equation above

80 x 10³ = 51.78 x 8.31 x 300 x ln v₁ / .2

ln v₁ / .2 = .62

v₁ / .2 = 1.8589

v₁ = 0.37 m³

Answer:

6.8%

Explanation:

According to Stefan-Boltzmann law, radiation is directly proportional with temperature raised to the fourth power:

P ∝ T⁴

Writing a proportion:

P₁ / P₂ = T₁⁴ / T₂⁴

1.3P / P = (T₁ / T₂)⁴

T₁ / T₂ = ∜1.3

T₁ = 1.068 T₂

The temperature increased by 6.8%.

Answer:

c. 66 V

Explanation:

p =IV

I =P/V

P1/V1=P2/V2

60/110=36/V2

0.55 = 36/V2

V2 =36/0.55 = 65.5V

V2 = 66V

Answer:

Explanation:

coefficient of linear expansion α = 2.9 x 10⁻⁵

coefficient of volume expansion γ = 3 x 2.9 x 10⁻⁵ = 8.7 x 10⁻⁵

[tex]d_t = d_0( 1 - \gamma t )[/tex]

[tex]d_{82} = 1.13\times 10^4( 1 - 8.7\times 10^{-5}\times82 )[/tex]

= 1.13 x 10⁴ - 806.14 x 10⁻¹

= 1.13 x 10⁴ - 0.00806 x 10⁴

= 1.1219 x 10⁴ kg / m³

b ) mass of the sample will remain the same as mass does not increase or decrease with temperature .

Answer:

Explanation:

a = F / m

where a is acceleration , F is thrust and m is mass

taking log and differentiating

da / a = dF / F - dm / m

(da / a)x 100 = (dF / F)x100 - (dm / m) x100

percentage increase in a = percentage increase in F - percentage increase in m

= percentage increase in acceleration a   = 39 - 13 = 26 %

required increase = 26 %.

Answer:

Convection and Radiation mechanisms carry most of the heat

Explanation:

This is because Convection proceeds strongy as heated air rises from the hot element while Radiation is also strong, although the material of the cooking pots will how effective it is.

Complete Question

One arm of a U-shaped tube (open at both ends) contains water, and the other alcohol.

If the two fluids meet at exactly the bottom of the U, and the alcohol is at a height of 18 cm, at what height will the water be?Assume the density of alcohol is [tex]\rho_a = 790\ kg/m^3[/tex]

Answer:

The  height of water is  [tex]h_w = 0.142 \ m[/tex]

Explanation:

From the question we are told that

    The height of the alcohol is  [tex]h_a =18 \ cm = 0.18 \ m[/tex]

     The  density of the alcohol is  [tex]\rho_a = 790\ kg/m^3[/tex]

Generally the pressure on both arm of the tube are equal given that they are both open

i,e    [tex]P_a = P_w[/tex]

Where  [tex]P_a[/tex] is pressure of alcohol and  [tex]P_w[/tex] is pressure of water

   So the pressure on the arm of the tube containing the alcohol is mathematically evaluated as

         [tex]P_a = g * h * \rho[/tex]

substituting values

         [tex]P_a =9.8 * 0.18 * 790[/tex]

         [tex]P_a = 1394 \ Pa[/tex]

Generally the pressure on the arm of the tube containing the water is mathematically evaluated as

       [tex]P_w = g * h_w * \rho_w[/tex]

where  [tex]\rho_w[/tex] is the density of water which has  a value [tex]\rho _w = 1000 \ kg/m^3[/tex]

So  

      [tex]1394 = 9.8 * h_w * 1000[/tex]

=>    [tex]h_w = \frac{1394}{9800}[/tex]

=>  [tex]h_w = 0.142 \ m[/tex]

Hahahahaha. Okay.

So basically , force is equal to mass into acceleration.

F=ma

so when F=ma , we get acceleration=6m/s/s

Force is doubled.

Mass is 1/3 times original.

2F=1/3ma

Now , we rearrange , and we get 6F=ma

So , now for 6 times the original force , we get 6 times the initial acceleration.

So new acceleration = 6*6= 36m/s/s

Answer:

9.6×10⁹ A

Explanation:

From the question above,

P = VI.................... Equation 1

Where P = Electric power, V = Voltage, I = current.

make I the subject of the equation

I = P/V............. Equation 2

Given: P = 200 kW = 200×10³ W, V = 48000 V.

Substitute these vales into equation 2

I = 200×10³×48000

I = 9.6×10⁹ A.

Hence the current in the line is 9.6×10⁹ A.

Answer:

D. Metallic atoms have valence shells that are mostly empty, which

means these atoms are more likely to give up electrons and allow

them to move freely.

Explanation:

Metals usually contain very few electrons in their valence shells hence they easily give up these few valence electrons to yield metal cations.

In the metallic bond, metal cations are held together by electrostatic attraction between the metal ions and a sea of mobile electrons.

Since metals give up their electrons easily, it is very easy for them to participate in metallic bonding. They give up their electrons easily because their valence shells are mostly empty, metal valence shells usually contain only a few electrons.

Complete question:

if two point charges are separated by 1.5 cm and have charge values of +2.0 and -4.0 μC, respectively, what is the value of the mutual force between them.

Answer:

The mutual force between the two point charges is 319.64 N

Explanation:

Given;

distance between the two point charges, r = 1.5 cm = 1.5 x 10⁻² m

value of the charges, q₁ and q₂ = 2 μC and - μ4 C

Apply Coulomb's law;

[tex]F = \frac{k|q_1||q_2|}{r^2}[/tex]

where;

F is the force of attraction between the two charges

|q₁| and |q₂| are the magnitude of the two charges

r is the distance between the two charges

k is Coulomb's constant = 8.99 x 10⁹ Nm²/C²

[tex]F = \frac{k|q_1||q_2|}{r^2} \\\\F = \frac{8.99*10^9 *4*10^{-6}*2*10^{-6}}{(1.5*10^{-2})^2} \\\\F = 319.64 \ N[/tex]

Therefore, the mutual force between the two point charges is 319.64 N

Answer:

Explanation:

[tex]mass = 2.5kg\\height = 4.0\\Acceleration \: due \:to\:gravity = 10m/s^2\\\\P.E = mgh\\P.E = 2.5kg\times10\times4\\\\Potential \: Energy = 100 joules[/tex]

The potential energy if g = 10m/s² is 98 J

Potential energy is the energy held by an object because of its position relative to other objects, stresses within itself, its electric charge, or other factors.

Formula for potential energy :

P.E. = mgh

Given :

The book is held from the ground of a distance  = 4.0 m,

so

h = 4.0 m

we know that the book weighs,

2.5 kg

so

m = 2.5 kg.

Now we just put it in the formula  ;

PE = (2.5kg) × (9.8 m/s²) × ( 4.0 m)

P E = 98 J

Therefore, The potential energy if g = 10m/s² is 98 J

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Answer:

The value of the inductance is 0.955 mH

Explanation:

Given;

frequency of the oscillator, f = 18 kHz = 18,000 Hz

the peak current, I₀ = 70 mA = 0.07 A

the root mean square voltage, [tex]V_{rms}[/tex] = 5.4 V

The root mean square current is given as;

[tex]I_{rms}= \frac{I_o}{\sqrt{2} }[/tex]

[tex]I_{rms} = \frac{0.07}{\sqrt{2} } \\\\I_{rms} = 0.05 \ A[/tex]

Inductive reactance is given by;

[tex]X_L =\frac{V_{rms}}{I_{rms}} \\\\X_L = \frac{5.4}{0.05} \\\\X_L = 108 \ ohms[/tex]

Inductance is given by;

[tex]L = \frac{X_L}{2\pi f} \\\\L = \frac{108}{2\pi *18,000} \\\\L = 9.55 *10^{-4} \ H[/tex]

L = 0.955 mH

Therefore, the value of the inductance is 0.955 mH

The value of the inductance (L) for this oscillating circuit is equal to [tex]9.55 \times 10^{-4}[/tex] Henry.

Given the following data:

To determine the value of the inductance (L):

First of all, we would find the root mean square (rms) current by using the formula:

[tex]I_{rms} = \frac{I_o}{\sqrt{2} }\\\\I_{rms} = \frac{70 \times 10^{-3}}{1.4142} \\\\I_{rms} = 0.050 \;A[/tex]

Next, we would calculate the inductive reactance of the oscillator by using the formula:

[tex]X_L = \frac{V_{rms}}{I_{rms}} \\\\X_L = \frac{5.4}{0.050} \\\\X_L = 108 \; Ohms[/tex]

Now, we can solve for the value of the inductance (L):

[tex]L = \frac{X_L}{2\pi f}[/tex]

Where:

Substituting the parameters into the formula, we have;

[tex]L = \frac{108}{2 \times 3.142 \times 18 \times 10^3} \\\\L = \frac{108}{113112}[/tex]

L = [tex]9.55 \times 10^{-4}[/tex] Henry.

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Answer:

Maximum acceleration is 800m/s^2

Explanation:

See attached file

Answer:

Explanation:

Given the formula for calculating the depth in metres expressed as

depth in meters = ½ (1500 m/sec × Echo travel time in seconds)

Given depth of the challenger = 10, 994 meters, we will substitute this given value into the formula given to calculate the time take for the echo to travel.

10, 994 = depth in meters = ½ * 1500 m/sec × Echo travel time in seconds

10,994 = 750 * Echo travel time in seconds

Dividing both sides by 750;

Echo travel time in seconds = 10,994 /750

Echo travel time in seconds ≈ 14.66secs (to two decimal places)

Therefore, it would take an echo sounder’s ping 14.66secs to make the trip from a ship to the Challenger Deep and back

Answer:

f = 40Hz

Explanation:

v=f x wavelength

f =v / wavelength

f = 2/5 x 10-²= 40 Hz

f = 40Hz

f = 40Hz

In physics, the term frequency refers to the number of waves that pass a fixed point in unit time.

It also describes the number of cycles or vibrations undergone during one unit of time by a body in periodic motion.

Wavelength (λ) - The wavelength of light is defined as the distance between the crests or troughs of a wave motion.

The wave equation: v = fλ

As per question,

Wavelength = 5.00 cm

v = 2.00 m/s.

v=fλ

f =v / λ

f = 2/5 x 10⁻² = 40 Hz

f = 40Hz

Therefore,

The frequency is 40Hz.

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Answer:

7.68×10^25kg

Explanation:

The formula for energy used per year is calculated as

Energy used per year =12 x Energy used per month

By substituting Energy used per month in the above formula, we get

Energy used per year =12 x 1600kWh

= 19200kWh

Conversion:

From kWh to J:

1 kWh=3.6 x 10^6 J

Therefore, it is converted to J as

19200 kWh =19200 x 3.6 x 10^6 J

= 6.912×10^10 J

Hence, energy used per year is 6.912×10^10 J

To find the mass that is converted to energy per year.

E = MC^2 ............1

E is the energy used per year

C is the speed of light = 3.0× 10^8m/s

Where E= 6.912×10^10 J

Substituting the values into equation 1

6.912×10^10 J = M × 3.0× 10^8m/s

M = 6.912×10^10 J / (3.0× 10^8m/s)^2

M = 6.912×10^10 J/9×10^16

M = 7.68×10^25kg

Hence the mass to be converted is

7.68×10^25kg

Answer:

The  percentage of the kinetic energy that is left in the system after collision to that before is 0.174 %

Explanation:

Given;

mass of bullet, m₁ = 4g = 0.004kg

initial velocity of bullet, u₁ = 589 m/s

mass of block of wood, m₂ = 2.3 kg

initial velocity of the block of wood, u₂ = 0

let the final velocity of the system after collision = v

Apply the principle of conservation of linear momentum

m₁u₁ + m₂u₂ = v(m₁+m₂)

0.004(589) + 2.3(0) = v(0.004 + 2.3)

2.356 = 2.304v

v = 2.356 / 2.304

v = 1.0226 m/s

Initial kinetic energy of the system

K.E₁ = ¹/₂m₁u₁² + ¹/₂m₂u₂²

K.E₁ = ¹/₂(0.004)(589)² = 693.842 J

Final kinetic energy of the system

K.E₂ = ¹/₂v²(m₁ + m₂)

K.E₂ = ¹/₂ x 1.0226² x (0.004 + 2.3)

K.E₂ = 1.209 J

The kinetic energy left in the system = final kinetic energy of the system

The percentage of the kinetic energy that is left in the system after collision to that before = (K.E₂ / K.E₁) x 100%

                       = (1.209 / 693.842) x 100%

                        = 0.174 %

Therefore, the  percentage of the kinetic energy that is left in the system after collision to that before is 0.174 %

Answer:

Power=50.17dioptre

Power=50.17D

Explanation:

P=1/f = 1/d₀ + 1/d₁

Where d₀ = the eye's lens and the object distance= 5.70m=

d₁= the eye's lens and the image distance= 0.02m

f= focal length of the lense of the eye

We know that the object can be viewed clearly by the person ,then image and lens of the eye's distance needs to be equal with the retinal and the eye lens distance and this distance is given as 0.02m

Therefore, we can calculate the power using above formula

P= 1/5.70 + 1/0.02

Power=50.17dioptre

Therefore, the power the eye's is using to see the object from distance is 5.70D

Answer:

Doubling the wavelength of the diffracting doubles the angle of diffraction. So, the width of the central bright spot pattern formed on the screen will also be doubled.

Explanation:

For a single slit diffraction, the path length difference is related to the wavelength of the light leaving the slit onto the screen by

D sin θ = mλ

where D sin θ is the path length of the waves, each.

mλ is the wavelength of the wavelet

where m is the the order of each minimum

m = m = 1,−1,2,−2,3, . . .

The wavelength of each wavelet is always a multiple of the wavelength of the light source, and from the equation, we can see that the angle of diffraction depend on the wavelength of the light. From this we can see that increasing the wavelength of the light increases the angle of diffraction, and hence we can say that doubling the wavelength will double the diffraction angle. Also, the width of the central bright spot of the screen will spread or increase with the angle of diffraction, so doubling the wavelength doubles the central bright spot on the screen.

Answer:

B=uonI/2

Explanation:

See attached file

Answer:

ω(t) = 0.4 + 0.036 t²

Explanation:  

The angular displacement of the disk is given as the function of time:

β(t) = Ct + B t³

where,

C = 0.4 rad/s

B = 0.012 rad/s³

Therefore,

β(t) = 0.4 t + 0.012 t³

Now, for angular velocity ω(t), we must take derivative of angular displacement with respect to t:

ω(t) = dβ/dt = (d/dt)(0.4 t + 0.012 t³)

ω(t) = 0.4 + 0.036 t²

Answer:

100 degrees Celsius

Explanation:

Water starts to boil at 100 degrees celcius or 212 degrees fahrenheit.

At 100 degrees Celsius water begin to boil and turns to steam.

The melting point for water is 0 degrees C (32 degrees F). The boiling point of water varies with atmospheric pressure. At lower pressure or higher altitudes, the boiling point is lower. At sea level, pure water boils at 212 °F (100°C).

If the temperature is much above 212°F, the water will boil. That means that it won't just evaporate from the surface but will form vapor bubbles, which then grow, inside the liquid itself. If the water has very few dust flecks etc.

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Answer:

The wave will be travelling in the y-axis

Explanation:

An e-m wave has a spatially varying electric field that is always associated with a magnetic field that changes over time and vice versa. The electric field and the magnetic field oscillates perpendicularly to each other, and together form a wave that travels in a perpendicular direction to the magnetic and the electric field in space. The movement of the e-m wave through space is usually away from the source where it is generated. So, if the electric field travels in the z-axis, and the magnetic field travels through along the x-axis, then the e-m wave generated will travel in the y-axis direction.

Answer:

f1 / f2 = n2 / n1  

Explanation:

To solve this problem, we should remember that the formula for index of refraction is defined as:

n = c / v

or

n v = c

Where,

n = index of refraction

c = speed of light

v = speed of light in the medium

Since speed of light is constant, then we can simply equate the materials 1 and 2:

n1 v1 = n2 v2

Where the speed of light in the medium (v) can be expressed as:

v = w * f

Where,

w = wavelength of light

f = frequency of light

Therefore substituting this back into the relating equation:

n1  w1 f1 = n1  w2 f1

Since it is given that the light is monochromatic, w1 = w2, this further simplifies the equation to:

n1 f1 = n2 f2

f1 / f2 = n2 / n1                  (ANSWER)

Answer:

Explanation:

Work required = q x V

where q is charge on electron and V is potential difference

= 1.6 x 10⁻¹⁹  x 12

= 19.2 x 10⁻¹⁹ J

Answer:

2.1 rad(anticlockwise).

Explanation:

So, we are given the following data or parameters or information in the question above:

=> "The torsional stiffness of the spring support is k = 50 N m/rad. "

=> "If a concentrated torque of mag- nitude Ta = 500 Nm is applied in the center of the bar"

=> "L = 300 mm Assume a shear modu- lus G = 10 kN/mm2 and polar monnent of inertia J = 2000 mln"

Hence;

G × J = 10 kN/mm2 × 2000 mln = 20 Nm^2.

Also, L/2 = 300 mm /2 = 0.15 m (converted to metre).

==> 0.15/20 (V - w) + θ = 0.

==> 0.15/20 (V - w ) = -θ.

Where V = k = 50 N m/rad

w = 183.3 θ.

Therefore, w + Vθ = 500 Nm.

==> 183.3 + 50 θ = 500 Nm.

= 6.3

Anticlockwise,

θ = 2.1 rad.

Answer:

a) using

R=V/I =117/12.3

R=9.5 ohms

b)

E=V-I*R =117-3.38*9.5

E=84.8Volts

c)

at (1/3)rd of normal speed ,back emf is (1/3) of its maximum

value

E=(1/3)*84.8=28.3Volts

Current drawn

I=V-Eback/R =117-28.3/9.5

I=9.33A

Explanation:

The resistance is = 9.5 ohms

The back emf developed at normal speed is = 84.90 volts

The current drawn at one-third normal speed =9.33 A.

To calculate the resistance of the armature coil this formula is used;

V = IR

make R the subject of formula,

R = V/I

where R = resistance

V = voltage

I = Current

R = 117/12.3

R = 9.5 ohms

To calculate the back emf developed at normal speed, this formula is used;

E = V + Ir ( for normal emf)

But for back emf which is the difference between the supplied voltage and the loss from the current through the resistance, this formula is used;

E = V - Ir

where V = 117v

I = 3.38

r = 9.5

E = 117 - ( 3.38 × 9.5)

= 117 - 32.11

= 84.90 volts

To calculate the current drawn at one-third normal speed;

1/3 of Emf = 1/3 × 84.90

= 28.3volts

Therefore current (I) = V - E/ R

= 117- 28.3/9.5

= 88.7/9.5

= 9.33 A

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Answer:

408N at 89.89°

Explanation:

This problem requires that we resolve the force vectors into

x- and y

-componentsOnce this is done, we can add the components easily, as the one 2-dimensional problem will be two 1-dimensional problems.

Finally, we will convert the resultant force into standard form and find the equilibrant.

Resolve into components:

F1x =F1cos 180°= 232(−1)=−232N

F1y=F1sin180°=0N

F2x=F2cos(−140°)=194(−0.766)=−148.6N

F1y=F1sin(−140°)=232(−0.643)=−149.17N

Note the change of the angle used to give the direction of

F2. Standard angles (rotation from thex

-axis; counterclockwise is +) should be used to avoid sign errors in the results.

Now, we add the components:

Fx=F1x+F2x=−380.6N

Fy=F1y+F1y=−148.17N

Technically, this is the resultant force. However, it should be changed back into standard form. Here's how:

F=√(Fx)2(Fy)2=√(−380.6)^2(−148.17)^2=408N

θ=tan−1(−148.17−380.6)

=89.89°