A circular loop of wire has radius of 9.50 cmcm. A sinusoidal electromagnetic plane wave traveling in air passes through the loop, with the direction of the magnetic field of the wave perpendicular to the plane of the loop. The intensity of the wave at the location of the loop is 0.0215 W/m2W/m2, and the wavelength of the wave is 6.90 mm.What is the maximum emf induced in the loop?
Express your answer with the appropriate units.

Answer:

Chemical changes are recognized when a substance changes its properties permanently and it cannot be the same substance as before.

Instead the physical changes implies that if you can return to the same substance through a reverse process.

Explanation:

A chemical change is, by example, a combustion, if a sheet of paper burns, its result is ashes, the ashes cannot go back to being a sheet of paper because its properties changed, heat energy was generated that changed matter permanently.

A physical change, by example, is that of freezing water, the water becomes ice, but this can easily become water again if the temperature is increased, its properties do not change and the chemistry of the substance does not change.

A chemical change takes place when a chemical reaction takes place, while when a matter changes forms but not the chemical identity then a physical change takes place.  

• A product or a new compound formation takes place from a chemical change as the rearrangement of atoms takes place to produce novel chemical bonds.  

• In a chemical change always a chemical reaction takes place.  

• Some of the chemical changes examples are souring milk, burning wood, digesting food, mixing acid and base, cooking food, etc.  

• In a physical change no new chemical species form.  

• The changing of the state of a pure substance between liquid, gas, or solid is a physical change as there is no change in the identity of the matter.  

• Some of the physical changes are melting of ice, tempering of steel, breaking a bottle, crumpling a sheet of aluminum foil, boiling water, and shredding paper.  

Thus, a new substance is formed during a chemical change, while a physical change does not give rise to a new substance.  

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Answer:

1.125×10⁻⁹ J

Explanation:

Applying,

E = 1/2CV²................... Equation 1

Where E = Energy stored in the capacitor, C = capacitance of the capacitor, V = Voltage of the battery.

Given; C = 1.0 nF,  = 1.0×10⁻⁹ F, V = 1.5 V

Substitute into equation 1

E = 1/2(1.0×10⁻⁹×1.5²)

E = 1.125×10⁻⁹ J

Hence the energy stored by the capacitor is 1.125×10⁻⁹ J

Answer:

cost_cost = $ 96

Explanation:

In this exercise we have units in the groin system and the SI system, to avoid problems let's reduce everything to the SI system

         performance = 21 miles / gallon (1,609 km / 1 mile) (1 gallon / 3,785 l)

         perfomance= 8,927 km / l

now let's use a direct rule of proportions (rule of three). If a liter travels 8,927 km, how many liters are needed to travel the 2015 km

          #_gasoline = 2015 km (1l / 8.927 km) = 225.72 liters

Now let's find the total cost of fuel. Ns indicates that $ 0.11 = 1 peso and the liter of fuel costs 5.8 pesos

            cost_litre = 5.8 peso ($ 0.11 / 1 peso) = $ 0.638

             cost_gasoline = #_gasoline   cost_litro

             cost_gasoline = 225.72   0.638

             cost_gasoline = $ 144

This cost is for the one way trip, the total round trip cost is

             cost_total = 2 cost_gasoline

             cost_total = $ 288

Now let's look for the cost in the vehicle, you and two people will go, for which a total of 3 people will go, so the cost per person is

                cost_person = total_cost / #_people

                cost_person = 288/3

                cost_cost = $ 96

Answer:

The intensity at distance 2d from source is  [tex]I_1 = \frac{1}{4} * I[/tex]  

Explanation:

From the question we are told that

     The distance of the wave from point source is  d  

     The  intensity is  [tex]I[/tex]  

     The distance we are considering is  2d

Generally the intensity of a wave is mathematically represented as

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

Here P is power of point source      

Now when  d =  2d

          [tex]I_1 = \frac{ P }{\pi (2d)^2 }[/tex]        

           [tex]I_1 = \frac{ 1 }{4 } * \frac{ P }{\pi d^2 }[/tex]

    =>   [tex]I_1 = \frac{1}{4} * I[/tex]  

The intensity at a distance 2d from the source is equal to [tex]I'=\frac{I}{4}[/tex]

Given the following data:

To determine the intensity at a distance 2d from the source:

Mathematically, the intensity of a wave is given by the formula:

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

Where:

Since the distance is doubled (2d), we have:

[tex]I'=\frac{P}{\pi (2d)^2}\\\\I'=\frac{P}{4\pi (d)^2}\\\\I'=\frac{1}{4} \times \frac{P}{\pi (d)^2}\\\\I'=\frac{1}{4} \times I\\\\I'=\frac{I}{4}[/tex]

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Answer:

The linear velocity, v = 5.93 m/s

Explanation:

To find the linear velocity after 5 seconds, we find its angular velocity after 5 seconds using

ω' = ω + αt where ω = initial angular speed = 4.00 rev/s = 4.00 × 2π rad/s = 25.13 rad/s, ω' =  = final angular speed, α = angular acceleration = 2.25 rad/s² and t = time = 5.00 s

ω' = ω + αt

= 25.13 rad/s + 2.25 rad/s² × 5.00 s

= 25.13 rad/s + 11.25 rad/s

= 36.38 rad/s

The linear velocity v is gotten from v = rω' where r = radius of pulley = 326 mm/2 = 163 mm = 0.163 m

v = rω'

= 0.163 m × 36.38 rad/s

= 5.93 m/s

So, the linear velocity  v = 5.93 m/s

Answer:

0.37 cm

Explanation:

The image is formed on the retina which is at a constant distance of 2.70 cm to the lens. Therefore, image distance = 2.70 cm.

The object is at a distant of 265 cm to the lens of the eye.

From lens formula,

[tex]\frac{1}{f}[/tex] = [tex]\frac{1}{u}[/tex] + [tex]\frac{1}{v}[/tex]

where: f is the focal length, u is the object distance and v is the image distance.

Thus, u = 265.00 cm and v = 2.70 cm.

[tex]\frac{1}{f}[/tex] = [tex]\frac{1}{265}[/tex] + [tex]\frac{27}{10}[/tex]

  = [tex]\frac{10+7155}{2650}[/tex]

[tex]\frac{1}{f}[/tex]  = [tex]\frac{7165}{2650}[/tex]

⇒ f = [tex]\frac{2650}{7165}[/tex]

      = 0.37

The focal length of the eye is 0.37 cm.

Explanation:

The separation between the interference fringes on the screen increases because the distance of the screen from the slit is increased. Therefore option (a) is  correct.

The separation between the interference fringes on the screen increases because  the distance of the screen from the slit is increased, which is contradictory. Therefore option (b) is incorrect.

The separation between the interference fringes on the screen increases because  the distance of the screen from the slit is increased, which is contradictory. Therefore option (c) is incorrect.

The separation between the interference fringes on the screen increases because  the distance of the screen from the slit is increased, which is contradictory. Therefore option (d) is incorrect.

The separation between the interference fringes on the screen increases because  the distance of the screen from the slit is increased, which is contradictory. Therefore option (e) is incorrect.

Answer:

The rms current is 0.3112 A.

Explanation:

Given that,

Suppose, The capacitance is 170 μF and the inductance is 2.94 mH. The resistance in the top branch is 278 Ohms, and in the bottom branch is 151 Ohms. The potential of the power supply is 47 V .

We know that,

When the frequency is very large then the capacitance can be treated as a short circuit and inductance as open circuit.

So,

We need to calculate the rms current

Using formula of current

[tex]I=\dfrac{V}{R}[/tex]

Where, V = voltage

R = resistance

Put the value into the formula

[tex]I=\dfrac{47}{151}[/tex]

[tex]I= 0.3112 \ A[/tex]

Hence, The rms current is 0.3112 A.

The provided question is not correct as, there is more than one options are correct, however the explaining every correct option -

Answer:

The correct answer are - silver, copper and aluminium all three used as fuse material.

Explanation:

A safety device in any electric circuit of that prevents the electric system in case of short circuit by breaking the connection of electric system or circuit termed as the Fuse or fuse element. Normally the fuse are made up of wire or element of material that are low in melting point and high in resistance.

Zinc, lead, tin, silver, copper, aluminium, and alloy of tin and alloy are used as fuse element or material for their low melting point and high resistance these are easily breaks the electric path in case of short circuit.

Answer:

c. 70 Ω

Explanation:

The R and R resistors are in parallel.  The 2R and 2R resistors are in parallel.  The 4R and 4R resistors are in parallel.  Each parallel combination is in series with each other.  Therefore, the equivalent resistance is:

Req = 1/(1/R + 1/R) + 1/(1/2R + 1/2R) + 1/(1/4R + 1/4R)

Req = R/2 + 2R/2 + 4R/2

Req = 3.5R

Req = 70Ω

Answer:

a. 0.342 kg-m² b. 2.0728 kg-m²

Explanation:

a. Since the skater is assumed to be a cylinder, the moment of inertia of a cylinder is I = 1/2MR² where M = mass of cylinder and r = radius of cylinder. Now, here, M = 56.5 kg and r = 0.11 m

I = 1/2MR²

= 1/2 × 56.5 kg × (0.11 m)²

= 0.342 kgm²

So the moment of inertia of the skater is

b. Let the moment of inertia of each arm be I'. So the moment of inertia of each arm relative to the axis through the center of mass is (since they are long rods)

I' = 1/12ml² + mh² where m = mass of arm = 0.05M, l = length of arm = 0.875 m and h = distance of center of mass of the arm from the center of mass of the cylindrical body = R/2 + l/2 = (R + l)/2 = (0.11 m + 0.875 m)/2 = 0.985 m/2 = 0.4925 m

I' = 1/12 × 0.05 × 56.5 kg × (0.875 m)² + 0.05 × 56.5 kg × (0.4925 m)²

= 0.1802 kg-m² + 0.6852 kg-m²

= 0.8654 kg-m²

The total moment of inertia from both arms is thus I'' = 2I' = 1.7308 kg-m².

So, the moment of inertia of the skater with the arms extended is thus I₀ = I + I'' = 0.342 kg-m² + 1.7308 kg-m² = 2.0728 kg-m²

a) The moment of inertia as the skater pulled his/her arm inward by assuming he/she is a cylinder is 0.3418kg-m².

b) If the body of the skater is assumed to be a cylinder of the same size, and the arms are rods extending straight out from his/her body being rotated at the ends, the moment of inertia is 1.7495kg-m².

Given the data in the question;

Mass of skater; [tex]M = 56.5kg[/tex]

a)

When the skater has his arms pulled inward by assuming they are cylinder of radius; [tex]R = 0.11 m[/tex]

Moment of inertia; [tex]I = \ ?[/tex]

From Parallel axis theorem; Moment of Inertia for a cylindrical body is expressed:

[tex]I = \frac{1}{2}MR^2[/tex]

Where M is the mass and R is the radius

We substitute our given values into the equation

[tex]I = \frac{1}{2}\ *\ 56.5kg\ *\ (0.11m)^2\\\\I = \frac{1}{2}\ *\ 56.5kg\ *\ 0.0121m^2\\\\I = 0.3418kg.m^2[/tex]

Therefore, the moment of inertia as the skater pulled his/her arm inward by assuming he/she is a cylinder is 0.3418kg-m²

b)

With the skater's arms extended by assuming that each arm is 5% of the mass of their body

Mass of each arm; [tex]M_a = \frac{5}{100} * M = \frac{5}{100} * 56.5kg = 2.825kg[/tex]

Remaining mass; [tex]M_b = M - 2M_a = 56.5kg - 2(2.825kg) = 50.85kg[/tex]

Assume the body is a cylinder of the same size and the arms are 0.875 m long rods extending straight out from their body being rotated at the ends.

Length of arm; [tex]L = 0.875 m[/tex]

From Parallel axis theorem; Moment of Inertia about vertical axis is expressed as:

[tex]I = \frac{1}{2}M_bR^2 + \frac{2}{3}M_aL^2[/tex]

We substitute in our values

[tex]I = \frac{1}{2}*50.85kg*(0.11m)^2 + \frac{2}{3}*2.825kg*(0.875m)^2\\\\I = [\frac{1}{2}*50.85kg * 0.0121m^2] + [\frac{2}{3}*2.825kg*0.765625m^2]\\\\I = 0.3076kg.m^2 + 1.4419kg.m^2\\\\I = 1.7495kg.m^2[/tex]

Therefore, if the body of the skater is assumed to be a cylinder of the same size, and the arms are rods extending straight out from his/her body being rotated at the ends, the moment of inertia is 1.7495kg-m².

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Explanation:

1 inch = 25.4 mm

1 foot = 12 inches

1 mile = 5260 feet

1 cm = 0.01 m or 10 mm

Now Sammy's height is 5 feet and 5.3 inches.

(a) We need to find Sammy's height in inches.

Since, 1 foot = 12 inches

5 feet = 5 × 12 inches = 60 inches

Now, 5 feet and 5.3 inches = 60 inches + 5.3 inches = 65.3 inches

Sammy's height is 65.3 inches.

(b) We need to find Sammy's height in feet.

Since, 1 foot = 12 inches

[tex]1\ \text{inch}=\dfrac{1}{12}\ \text{feet}[/tex]

So,

[tex]5.3\ \text{inch}=\dfrac{5.3}{12}\ \text{feet}=0.4416\ \text{feet}[/tex]

5 feet and 5.3 inches = 5 feet + 0.4416 feet = 5.44 feet

Sammy's height is 5.44 feet.

Answer:

total current = 12.417 A

so it will not fuse as current is less than 15 A

Explanation:

given data

toaster = 1140-W

blender = 270-W

lamp = 80-W

voltage = 120 V

solution

we know that current is express as

current = power ÷ voltage   ......................1

here voltage is same in all three device

so

current by toaster is

I = [tex]\frac{1140}{120}[/tex]

I = 9.5 A

and

current by blender

I = [tex]\frac{270}{120}[/tex]

I = 2.25 A

and

current by lamp is

I = [tex]\frac{80}{120}[/tex]

I = 0.667 A

so here device in parallel so

total current is = 9.5 A + 2.25 A + 0.667 A

total current = 12.417 A

so it will not fuse as current is less than 15 A

Answer:

f. 80 and 90

Explanation:

1 x 10⁻¹² W/m² sound intensity falls within 0 sound level

1 x 10⁻¹¹ W/m² sound intensity falls within 10 sound level

1 x 10⁻¹⁰ W/m² sound intensity falls within 20 sound level

1 x 10⁻⁹ W/m² sound intensity falls within 30 sound level

1 x 10⁻⁸ W/m² sound intensity falls within 40 sound level

1 x 10⁻⁷ W/m² sound intensity falls within 50 sound level

1 x 10⁻⁶ W/m² sound intensity falls within 60 sound level

1 x 10⁻⁵ W/m² sound intensity falls within 70 sound level

1 x 10⁻⁴ W/m² sound intensity falls within 80 sound level

1 x 10⁻³ W/m² sound intensity falls within 90 sound level

Given sound intensity (3.66 x 10⁻⁴ W/m²) falls with 1 x 10⁻⁴ W/m² of intensity which is within 80 and 90 sound level.

f. 80 and 90

Answer:

The potential will be Va/b

Explanation:

So Let sphere A charged Q to potential V.

so, V= KQ/a. ....(1

Thus, spherical shell B is connected to the sphere A by a wire, so all charge always reside on the outer surface.

therefore, potential will be ,

V ′ = KQ/b = Va/b... That is from .....(1), KQ=Va]

Answer:

The three factors that can contribute to creating a delay in advanced care for the passengers in the overturned bus include:

1. Lack of communication: Since the accident happened on the remote hillside, there is a possibility that, there would be no communication network which could have afforded them the opportunity to call medical or technical team.

2. Steep Nature of the Hill: This is another factor which will affect the care which they could have received. Steeply area tends to be difficult for climbing in or out from.

3. Thunderstorm: This factor is another reason which could contribute to delay in receiving advance care. Thunderstorm create barriers for location f the area where the bus overturned or in other situation complicate the rescue efforts of the team sent out to rescue.

Explanation:

Answer:

The mas unit is the the 'Kilogram' written as 'kg'

Volume is 10 L

Explanation:

The complete question is

If you have a density of 100 kg/L, and a mass of 1000 units, tell me the following: First, what are the mass units?

Second, what is the volume

mass units is the 'Kilogram', written as 'kg'

density = mass/volume = 100 kg/L

the mass  = 1000 kg

volume = mass/density = 1000/100 = 10 L

Answer:

a

   [tex]m = 0.169 \ kg[/tex]

b

  [tex]|v_{max} |= 0.5653 \ m/s[/tex]

Explanation:

From the question we are told that

    The  spring constant is  [tex]k = 14 \ N/m[/tex]

     The  maximum extension of the spring is  [tex]A = 6.0 \ cm = 0.06 \ m[/tex]

     The number of oscillation is  [tex]n = 30[/tex]

      The  time taken is  [tex]t = 20 \ s[/tex]

Generally the the angular speed of this oscillations is mathematically represented as

           [tex]w = \frac{2 \pi}{T}[/tex]

where T is the period which is mathematically represented as

     [tex]T = \frac{t}{n}[/tex]

substituting values

     [tex]T = \frac{20}{30 }[/tex]

     [tex]T = 0.667 \ s[/tex]

Thus  

       [tex]w = \frac{2 * 3.142 }{ 0.667}[/tex]

       [tex]w = 9.421 \ rad/s[/tex]

this angular speed can also be represented mathematically as

       [tex]w = \sqrt{\frac{k}{m} }[/tex]

=>   [tex]m =\frac{k }{w^2}[/tex]

substituting values

      [tex]m =\frac{ 15 }{(9.421)^2}[/tex]

      [tex]m = 0.169 \ kg[/tex]

In SHM (simple harmonic motion )the equation for velocity is  mathematically represented as

        [tex]v = - Awsin (wt)[/tex]

The  velocity is maximum when  [tex]wt = \(90^o) \ or \ 1.5708\ rad[/tex]

     [tex]v_{max} = - A* w[/tex]

=>   [tex]|v_{max} |= A* w[/tex]

=>    [tex]|v_{max} |= 0.06 * 9.421[/tex]

=>   [tex]|v_{max} |= 0.5653 \ m/s[/tex]

CHECK THE COMPLETE QUESTION BELOW;

the intensity of an electromagnetic wave is 80 MW/m2, what is the amplitude of the magnetic field of this wave? (c=3.0×108m/s, μ0=4π×10−7T⋅m/A, ε0=8.85×10−12C2/N⋅m2)

Answer:

2.4×10^5 N/C

Explanation:

the amplitude can be explained as the maximum field strength of the electric and magnetic fields. Wave energy is proportional to its amplitude squared (E2 or B2).

We were told to calculate the amplitude of the magnetic field, which can be done using expresion below

S=ε²/2cμ

Where S is the intensity intensity of an electromagnetic wave given as 80 MW/m2

ε² is the Amplitude which we are looking for

c= speed of light given as 3×10^8m/s

Substitute the values into above formula we have,

S=ε²/2cμ

Making Amplitude subject of formula

ε²=S×2cμ

ε²=[80×10^6)(2×3×10^8)(4Π×10^-7)

= 245598.44

ε²=2.4×10^5 N/C

Therefore, amplitude of the magnetic field of this wave is S=2.4×10^5 N/C

Answer

1.07E22 Joules

Explanation;

We know that mass expands by a factor

=>>1/√[1-(v/c)²]

But v= 0.509c

So

1/√(1 - 0.509²)

=>>> 1/√(1 - 0.2591)

= >> 1/√(0.7409) = 1.16

But given that 121 tons is rest mass so 121- 1.16= 119.84 tons is kinetic energy

And we know that rest mass-energy equivalence is 9 x 10^19 joules per ton.

So Multiplying by 119.84

Kinetic energy will be 1.07x 10^22 joules

Answer:

The gauge pressure is  [tex]P = 687.4 \ Pa[/tex]

Explanation:

From the question we are told that

    The rate of flow is  [tex]Q = 0.00027 m^3 /s[/tex]

      The height is h  =  10 m

      The radius is  r =  0.01 m

     The  viscosity is  [tex]\eta = 1mPa \cdot s = 1 *10^{-3} \ Pa\cdot s[/tex]

Generally the gauge pressure according to Poiseuille's equation  is mathematically represented as  

               [tex]P = 8 \pi \eta * \frac{L * v }{ A}[/tex]

Here v is the velocity of the water which is mathematically represented according to continuity equation as

             [tex]v = \frac{Q}{A }[/tex]

Where A is the cross-sectional area which is mathematically represented as

            [tex]A = \pi r^2[/tex]

substituting values

          [tex]A = 3.142 *(0.01)^2[/tex]

           [tex]A = 3.142 *10^{-4} \ m^2[/tex]

So

      [tex]v = \frac{ 0.00027}{3.142*10^{-4}}[/tex]

       [tex]v = 0.8593 \ m/s[/tex]

So

       [tex]P = 8 * 3.142 * 1.0*10^{-3}* \frac{10 * 0.8593 }{ 3.142*10^{-4}}[/tex]

       [tex]P = 687.4 \ Pa[/tex]

Answer:

a. Moon around Earth.

Explanation:

Charon orbit takes around 6.4 earth days to complete its orbit. Charon does not rises or sets, it hovers over same spot around the Pluto. The same side of Charon faces the Pluto, this is called Tidal Locking.

The moon orbit takes around 27 days to complete its orbit. The moon has different sides that are faced with sun which creates light or dark face of moon on the earth. Moon has 384,400 km distance from the earth.

The object that should exhibit the longest orbital period is option a. Moon around Earth.

Charon's orbit takes around 6.4 earth days to finish its orbit. Charon does not rise or sets, it hovers over similar spot around Pluto. The same side of Charon faces Pluto, this we called Tidal Locking. Here the moon orbit should take approx 27 days to finish its orbit. The moon has various sides that are faced with the sun which developed the light or dark face of the moon on the earth. Also, Moon has 384,400 km distance from the earth.

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Answer:

D) diffraction

Explanation:

Corona is an optical phenomenon produced by the diffraction of sunlight or moonlight, as light moves through water droplets in the atmosphere.

This phenomenon produces one or more diffuse concentric rings of light around the Sun or Moon, usually seen as colored circles.

Therefore, the explanation for these phenomena of colored concentric circles, sometimes seen with the Sun or the Moon involves diffraction.

Answer:

Option A

Electrons are emitted if low intensity, high-frequency light hits a metal surface.

Explanation:

From the experiments conducted to study the photoelectric effect, conclusions were made that the key factor that contributes to the emission of electrons from the surface of the metal is the frequency of the beam of light. This frequency has to be beyond a minimum threshold, if not, there will be no emission of electrons from the metal surface no matter the intensity of the beam of light or the length of time it is incident upon the metal surface.

This makes option A correct because it highlights the contributions made by the threshold frequency to the photoelectric effect.

Answer:

Explanation:

change in magnetic flux = .16 x .16 x .510 - 0

= .013056 weber .

rate of change of flux = change in flux / time

= .013056 / 40.5 x 10⁻³

= .32237

voltage induced = .32237 V

electrical energy dissipated = v² / R where v is voltage , R is resistance

= .32237² / 6.35

= 16.36 x 10⁻³ J .

Answer:

Frequency

Explanation:

When waves travel from one medium to another, it is only the frequency of the wave that remains constant . when a wave is refracted at the boundary between two media, the wave will slow down and its wavelength decreases. The wave usually bends at the interface between the two media. The wavelength and speed of a wave may change at the boundary between two media but its frequency remains the same.

Hence the frequency of light is its only property that remains constant.

Answer:

The final temperature of both objects is 400 K

Explanation:

The quantity of heat transferred per unit mass is given by;

Q = cΔT

where;

c is the specific heat capacity

ΔT is the change in temperature

The heat transferred by the  object A per unit mass is given by;

Q(A) = caΔT

where;

ca is the specific heat capacity of object A

The heat transferred by the  object B per unit mass is given by;

Q(B) = cbΔT

where;

cb is the specific heat capacity of object B

The heat lost by object B is equal to heat gained by object A

Q(A) = -Q(B)

But heat capacity of object B is twice that of object A

The final temperature of the two objects is given by

[tex]T_2 = \frac{C_aT_a + C_bT_b}{C_a + C_b}[/tex]

But heat capacity of object B is twice that of object A

[tex]T_2 = \frac{C_aT_a + C_bT_b}{C_a + C_b} \\\\T_2 = \frac{C_aT_a + 2C_aT_b}{C_a + 2C_a}\\\\T_2 = \frac{c_a(T_a + 2T_b)}{3C_a} \\\\T_2 = \frac{T_a + 2T_b}{3}\\\\T_2 = \frac{300 + (2*450)}{3}\\\\T_2 = 400 \ K[/tex]

Therefore, the final temperature of both objects is 400 K.

Complete question is;

A circuit consists of four 100-W lamps connected in parallel across a 230-V supply. Inadvertently, a voltmeter has been connected in series with the lamps. The resistance of the voltmeter is 1500 Ω and that of the lamps under the conditions stated is six times their value then burning normally. What will be the reading of the voltmeter?

Answer:

150.42 V

Explanation:

We are told that the circuit consists of four 100W lamps.

We know that Power is given by the equation;

P = V²/R

Thus;

R = V²/P

Now, we are told that the four lamps are connected in parallel across a 230V supply.

Thus, V = 230 V

So resistance, R = 230²/100

R = 529 Ω

We are told that the resistance of the lamps under the conditions stated is six times their value when burning normally.

Thus, total resistance of each lamp under the conditions = 529 × 6 = 3174 Ω

So, since they are connected in parallel, equivalent resistance for each lamp = 3174/4 = 793.5 Ω

Now, since this resistance is connected in series with the voltmeter resistance of 1500 Ω

Therefore, total circuit resistance = 1500 + 793.5 = 2293.5 Ω

Thus;

circuit current = 230/2293.5 = 0.100283 A

Now, according to Ohm’s law, voltage drop across the voltmeter = 1500 × 0.100283 ≈ 150.42V

CHECK THE COMPLETE QUESTION BELOW

A circuit consists of four 100-W lamps connected in parallel across a 230-V supply. Inadvertently, a voltmeter has been connected in series with the lamps. The resistance of the voltmeter is 1500 Ω and that of the lamps under the conditions stated is six times their value then burning normally. What will be the reading of the voltmeter?

Answer:

the reading of the voltmeter=150.4V

Explanation:

We can determine the wattage of a lamp using below expression:

: W = I² R....................eqn(1)

But fro ohms law V=IR

then I= V/R

If we substitute I into equation (1)

We have W= V²/R

But W= 100W

V= 230V

Then

W=220²/R

100 = 2302/R

R = 529 Ω

We can as well calculate the Resistance of each lamp under given condition that they are sixtimes their value when burning normally.

R = 6 × 529 = 3174 Ω

We can also calculate quivalent resistance of the abovefour lamps connected in parallel then

R = 3174/4

= 793.5 Ω

total circuit resistance can be calculated since we know that resistance is connected to voltmeter of 1500 Ω resistance in series arrangement

Then

total circuit resistance = 1500 + 793.5

= 2293.5 Ω

Then from ohms law again

I= V/R

circuit current = 230/2293.5 A

The reading of the voltage drop across the voltmeter

= 1500 × 230/2293.5

= 150.4V

Answer:

In an adiabatic process we have

pV γ = const..

This explains that the pressure is a function of volume, p ( V ) ,

So can be written as:

p ( V ) × V γ = p 0 V γ 0 ,

or p ( V ) = p 0 V 0 / V γ

= p 0 V 0 / V ^(7 / 5)

Explanation:

Given that,

Separation between two long parallel conductors, r = 11 cm = 0.11 m

Current in first wire, [tex]I_1=3\ A[/tex]

Current in second wire, [tex]I_2=8\ A[/tex]

(a) The magnetic field created by I₁ at the location of I₂ is given by :

[tex]B_{12}=\dfrac{\mu_o I_1}{2\pi r}\\\\B_{12}=\dfrac{4\pi \times 10^{-7}\times 3I_1}{2\pi \times 0.11}\\\\B_{12}=5.45\times 10^{-6}\ T[/tex]

(b) Magnetic force per unit length exerted by [tex]I_1[/tex] on [tex]I_2[/tex] is given by :

[tex]\dfrac{F}{l}=\dfrac{\mu_o I_1I_2}{2\pi r}\\\\\dfrac{F}{l}=\dfrac{4\pi \times 10^{-7}\times 3\times 8}{2\pi \times 0.11}\\\\\dfrac{F}{l}=4.36\times 10^{-5}\ N/m[/tex]

(c) The magnetic field created by I₂ at the location of I₁ is given by :

[tex]B_{21}=\dfrac{\mu_o I_2} {2\pi r}\\\\B_{12}=\dfrac{4\pi \times 10^{-7}\times 8}{2\pi \times 0.11}\\\\B_{12}=1.45\times 10^{-5}\ T[/tex]

Hence, this is the required solution.