A series LR circuit contains an emf source of 19 V having no internal resistance, a resistor, a 22 H inductor having no appreciable resistance, and a switch. If the emf across the inductor is 80% of its maximum value 4 s after the switch is closed, what is the resistance of the resistor

Answers

Answer 1

Answer: R = 394.36ohm

Explanation: In a LR circuit, voltage for a resistor in function of time is given by:

[tex]V(t) = \epsilon. e^{-t.\frac{L}{R} }[/tex]

ε is emf

L is indutance of inductor

R is resistance of resistor

After 4s, emf = 0.8*19, so:

[tex]0.8*19 = 19. e^{-4.\frac{22}{R} }[/tex]

[tex]0.8 = e^{-\frac{88}{R} }[/tex]

[tex]ln(0.8) = ln(e^{-\frac{88}{R} })[/tex]

[tex]ln(0.8) = -\frac{88}{R}[/tex]

[tex]R = -\frac{88}{ln(0.8)}[/tex]

R = 394.36

In this LR circuit, the resistance of the resistor is 394.36ohms.


Related Questions

g One of the harmonics in an open-closed tube has frequency of 500 Hz. The next harmonic has a frequency of 700 Hz. Assume that the speed of sound in this problem is 340 m/s. a. What is the length of the tube

Answers

Answer:

The length of the tube is 85 cm

Explanation:

Given;

speed of sound, v = 340 m/s

first harmonic of open-closed tube is given by;

N----->A , L= λ/₄

λ₁ = 4L

v = Fλ

F = v / λ

F₁ = v/4L

Second harmonic of open-closed tube is given by;

L = N-----N + N-----A, L = (³/₄)λ

[tex]\lambda = \frac{4L}{3}\\\\ F= \frac{v}{\lambda}\\\\F_2 = \frac{3v}{4L}[/tex]

Third harmonic of open-closed tube is given by;

L = N------N + N-----N + N-----A, L = (⁵/₄)λ

[tex]\lambda = \frac{4L}{5}\\\\ F= \frac{v}{\lambda}\\\\F_3 = \frac{5v}{4L}[/tex]

The difference between second harmonic and first harmonic;

[tex]F_2 -F_1 = \frac{3v}{4L} - \frac{v}{4L}\\\\F_2 -F_1 = \frac{2v}{4L} \\\\F_2 -F_1 =\frac{v}{2L}[/tex]

The difference between third harmonic and second harmonic;

[tex]F_3 -F_2 = \frac{5v}{4L} - \frac{3v}{4L}\\\\F_3 -F_2 = \frac{2v}{4L} \\\\F_3 -F_2 =\frac{v}{2L}[/tex]

Thus, the difference between successive harmonic of open-closed tube is

v / 2L.

[tex]700H_z- 500H_z= \frac{v}{2L} \\\\200 = \frac{v}{2L}\\\\L = \frac{v}{2*200} \\\\L = \frac{340}{2*200}\\\\L = 0.85 \ m\\\\L = 85 \ cm[/tex]

Therefore, the length of the tube is 85 cm

A circular coil of wire 8.40 cm in diameter has 17.0 turns and carries a current of 3.20 A . The coil is in a region where the magnetic field is 0.610 T.Required:a. What orientation of the coil gives the maximum torque on the coil ?b. What is this maximum torque in part (A) ?c. For what orientation of the coil is the magnitude of the torque 71.0 % of the maximum found in part (B)?

Answers

Answer:

a) for the torque to be maximum, sin should be maximum

i.e (sinФ)maximum = 1

b) therefore the Maximum torque is

Tmax = 0.1838 × 1 = 0.1838  N.m

c) Given the torque is 71.0% of its maximum value; Ф  = 45.24⁰ ≈ 45⁰

Explanation:

Given that; Diameter is 8.40 cm,

Radius (R) = D/2 = 8.40/2 = 4.20 cm = 0.042 m

Number of turns (N) = 17

Current in the loop (I) = 3.20 A

Magnetic field (B) = 0.610 T

Let the angle between the loop's area vector A and the magnetic field B be

Now. the area of the loop is;

A = πR²

A = 3.14 ( 0.042 )²

A =  0.005539 m²

Torque on the loop (t) = NIABsinФ

t = 17 × 3.20 ×0.005539 × 0.610 × sinФ

t = 0.1838sinФ N.m

for the torque to be maximum, sin should be maximum

i.e (sinФ)maximum = 1

therefore the Maximum torque is

Tmax = 0.1838 × 1 = 0.1838  N.m

Given the torque is 71.0% of its maximum value

t = 0.71 × tmax

t = 0.71 × 0.1838

t = 0.1305

Now

0.1305 N.m =  0.1838 sinФ N.m

sinФ = 0.1305 / 0.1838

sinФ = 0.71001

Ф = sin⁻¹ 0.71001

Ф  = 45.24⁰ ≈ 45⁰

Two waves are traveling in the same direction along a stretched string. The waves are 45.0° out of phase. Each wave has an amplitude of 7.00 cm. Find the amplitude of the resultant wave.

Answers

Answer:

The amplitude of the resultant wave is 12.93 cm.

Explanation:

The amplitude of resultant of two waves, y₁ and y₂, is given as;

Y = y₁ + y₂

Let y₁ = A sin(kx - ωt)

Since the wave is out phase by φ, y₂ is given as;

y₂ = A sin(kx - ωt + φ)

Y = y₁ + y₂ = 2A Cos (φ / 2)sin(kx - ωt + φ/2 )

Given;

phase difference, φ = 45°

Amplitude, A = 7.00 cm

Y = 2(7) Cos (45 /2) sin(kx - ωt + 22.5° )

Y = 12.93 cm

Therefore, the amplitude of the resultant wave is 12.93 cm.

A race-car drives around a circular track of radius RRR. The race-car speeds around its first lap at linear speed v_iv i ​ v, start subscript, i, end subscript. Later, its speed increases to 4v_i4v i ​ 4, v, start subscript, i, end subscript. How does the magnitude of the car's centripetal acceleration change after the linear speed increases

Answers

Answer:

The magnitude of the centripetal acceleration increases by 16 times when the linear speed increases by 4 times.

Explanation:

The initial centripetal acceleration, a of the race-car around the circular track of radius , R with a linear speed v is a = v²/R.

When the linear speed of the race-car increases to v' = 4v, the centripetal acceleration a' becomes a' = v'²/R = (4v)²/R = 16v²/R.

So the centripetal acceleration, a' = 16v²/R.

To know how much the magnitude of the car's centripetal acceleration changes, we take the ratio a'/a = 16v²/R ÷ v²/R = 16

a'/a = 16

a' = 16a.

So the magnitude of the centripetal acceleration increases by 16 times when the linear speed increases by 4 times.

UVC light used in sterilizers, has wavelengths between 100 to 280 nm. If a certain UVC wave has a wavelength of 142.9 nm, what is the energy of one of its photons in J

Answers

Answer:

The energy of one of its photons is 1.391 x 10⁻¹⁸ J

Explanation:

Given;

wavelength of the UVC light, λ = 142.9 nm = 142.9 x 10⁻⁹ m

The energy of one photon of the UVC light is given by;

E = hf

where;

h is Planck's constant = 6.626 x 10⁻³⁴ J/s

f is frequency of the light

f = c / λ

where;

c is speed of light = 3 x 10⁸ m/s

λ  is wavelength

substitute in the value of f into the main equation;

E = hf

[tex]E = \frac{hc}{\lambda} \\\\E = \frac{6.626*10^{-34} *3*10^{8}}{142.9*10^{-9}} \\\\E = 1.391*10^{-18} \ J[/tex]

Therefore, the energy of one of its photons is 1.391 x 10⁻¹⁸ J

Warm blooded animals are homeothermic; that is, they maintain an approximately constant body temperature. (Forhumans it's about 37 oC.) When they are in an environment that is below their optimum temperature, they use energy derived from chemical reactions within their bodies to warm them up. One of the ways that animals lose energy to their environment is through radiation. Every object emits electromagnetic radiation that depends on its temperature. For very hot objects like the sun, that radiation is visible light. For cooler objects, like a house or a person, that radiation is in the infrared and is invisible. Nonetheless, it still carries energy. Other ways that energy is lost by a warm animal to a cool environment includes conduction (direct touching of a cooler object) and convection (cooler air moving and carrying thermal energy away). See Heat Transfer for a discussion of all three.

For this problem, we'll just consider how much energy an animal needs to burn (obtain from internal chemical reactions) in order to stay warm just from radiation losses. The rate at which an object loses energy through radiation is given by the Stefan-Boltzmann equation:

Rate of energy loss = AεσT4



where T is the absolute (Kelvin) temperature, A is the area of the object, ε is the emissivity (unitless and =1 for a perfect emitter, less for anything else), and σ is the Stefan-Boltzmann constant:

σ = 5.67 x 10-8 J/(s m2 K4)



Consider a patient trying to sleep naked in a cool room (55 oF = 13 oC). Assume that the person being considered is a perfect emitter and absorber of radiation (ε = 1), has a surface area of about 2.5 m2, and a mass of 80 kg.

a. A person emits thermal radiation at a rate corresponding to a temperature of 37 oC and absorbs radiation at a rate (from the air and walls) corresponding to a temperature of 13 oC. Calculate the individual's net rate of energy loss due to radiation (in Watts = Joules/second).
net rate of energy loss = Watts

b. Assume the patient produces no energy to keep warm. If they have a specific heat about equal to that of water (1 Cal/kg-oC) how much would their temperature fall in one hour? (1 Cal = 1kcal = 103 cal)
ΔT = oC

c. Given that the energy density of fat is about 9 Cal/g, how many grams of fat would the person have to utilize to maintain their body temperature in that environment for one hour?
amount of fat needed = g

Answers

Answer:

a) 360.7 J/s

b) 16.23 °C

c) 34.48 g

Explanation:

The mass of the person = 80 kg

The person is a perfect emitter, ε = 1

surface area of the person = 2.5 m^2

a) If he emits radiation at 37 °C, [tex]T_{out}[/tex] = 37 + 273 = 310 K

and receives radiation at 13 °C, [tex]T_{in}[/tex] = 13 + 273 = 286 K

Rate of energy loss E = Aεσ([tex]T^{4} _{out}[/tex] - [tex]T^{4} _{in}[/tex] )

where σ = 5.67 x 10^-8 J/(s m^2 K^4)

substituting values, we have

E = 2.5 x 1 x 5.67 x 10^-8 x ([tex]310^{4}[/tex] - [tex]286^{4}[/tex]) = 360.7 J/s

b) If they have specific heat about equal to that of water = 1 Cal/kg-°C

but 1 Cal = 1 kcal = 10^3 cal

specific heat of person is therefore = 10^3 cal/kg-°C

heat loss = 360.7 J/s = 360.7 x 3600 = 1298520 J/hr

heat lost in 1 hour = 1 x 1298520 = 1298520 J

This heat lost = mcΔT

where ΔT is the temperature fall

m is the mass

c is the specific heat equivalent to that of water

the specific heat is then = 10^3 cal/kg-°C

equating, we have

1298520 = 80 x 10^3 x ΔT

1298520 = 80000ΔT

ΔT = 1298520/80000 = 16.23 °C

c) 1298520 J = 1298520/4184 = 310.35 Cal

density of fat = 9 Cal/g

gram of fat = 310.35/9 = 34.48 g

What is the thinnest soap film (excluding the case of zero thickness) that appears black when illuminated with light with a wavelength of 580 nm

Answers

Answer:

Explanation:

In case of soap film , light gets reflected from denser medium , hence interference takes place between two waves , one reflected from upper and second from lower surface . For destructive interference the condition is

2μt = nλ where μ is refractive index of water , t is thickness , λ is wavelength of light and n is an integer .

2 x 1.34 x t = 1  x 580

t = 216.42 nm .

Thickness must be 216.42 nm .

What happens to the deflection of the galvanometer needle (due to moving the magnet) when you increase the number of loops

Answers

Answer:

If the magnet is moved, the galvanometer needle will deflect, showing that current is flowing through the coil which will increase total induced electromotive force

Explanation:

galvanometer is an instrument that can detect and measure small current in an electrical circuit.

If the magnet is moved, the galvanometer needle will deflect, showing that current is flowing through the coil. If it is move in a way into the coil,the needle deflect in that way and if it move in another way, it will deflect in the other way.

The total induced emf is equal to the emf induced in each loop by the changing magnetic flux, then multiplied by the number of loops and an increase in the number of loops will cause increase in the total induced emf.

In a double‑slit interference experiment, the wavelength is lambda=487 nm , the slit separation is d=0.200 mm , and the screen is D=48.0 cm away from the slits. What is the linear distance Δx between the eighth order maximum and the fourth order maximum on the screen?

Answers

Answer:

Δx = 4.68 x 10⁻³ m = 4.68 mm

Explanation:

The distance between the consecutive maxima, in Young's Double Slit Experiment is given bu the following formula:

Δx = λD/d

So, the distance between the eighth order maximum and the fourth order maximum on the screen will be given as:

Δx = 4λD/d

where,

Δx = distance between eighth order maximum and fourth order maximum=?

λ = wavelength = 487 nm = 4.87 x 10⁻⁷ m

d = slit separation = 0.2 mm = 2 x 10⁻⁴ m

D = Distance between slits and screen = 48 cm = 0.48 m

Therefore,

Δx = (4)(4.87 x 10⁻⁷ m)(0.48 m)/(2 x 10⁻⁴ m)

Δx = 4.68 x 10⁻³ m = 4.68 mm

A vertical spring stretches 3.8 cm when a 13-g object is hung from it. The object is replaced with a block of mass 20 g that oscillates in simple harmonic motion. Calculate the period of motion.

Answers

Answer:

The period of motion is 0.5 second.

Explanation:

Given;

extension of the spring, x = 3.8 cm = 0.038 m

mass of the object, m = 13 g = 0.013 kg

Determine the force constant of the spring, k;

F = kx

k = F / x

k = mg / x

k = (0.013 x 9.8) / 0.038

k = 3.353 N/m

When the object is replaced with a block of mass 20 g, the period of motion is calculated as;

[tex]T = 2\pi\sqrt{\frac{m}{k} } \\\\T = 2\pi\sqrt{\frac{0.02}{3.353} } \\\\T = 0.5 \ second[/tex]

Therefore, the period of motion is 0.5 second.

How much energy is required to accelerate a spaceship with a rest mass of 121 metric tons to a speed of 0.509 c?

Answers

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

A string of holiday lights has 15 bulbs with equal resistances. If one of the bulbs
is removed, the other bulbs still glow. But when the entire string of bulbs is
connected to a 120-V outlet, the current through the bulbs is 5.0 A. What is the
resistance of each bulb?

Answers

Answer:

Resistance of each bulb = 360 ohms

Explanation:

Let each bulb have a resistance r .

Since, even after removing one of the bulbs, the circuit is closed and the other bulbs glow. Therfore, the bulbs are connected in Parallel connection.

[tex] \frac{1}{r(equivalent)} = \frac{1}{r1} + \frac{1}{r2} + + + + \frac{1}{r15} [/tex]

[tex] \frac{1}{r(equivalent)} = \frac{15}{r} [/tex]

R(equivalent) = r/15

Now, As per Ohms Law :

V = I * R(equivalent)

120 V = 5 A * r/15

r = 360 ohms

If one could transport a simple pendulum of constant length from the Earth's surface to the Moon's, where acceleration due to gravity is one-sixth (1/6) that on the Earth, by what factor would be the pendulum frequency be changed

Answers

Answer:

The frequency will change by a factor of 0.4

Explanation:

T = 2(pi)*sqrt(L/g)

Since g(moon) = (1/6)g(earth), the period would change by sqrt[1/(1/6)] = sqrt(6) ~ 2.5 times longer on the moon. Since the period & frequency are inverses, the frequency would be 1/2.5 or 0.4 times shorter on the moon.

You're conducting an experiment on another planet. You drop a rock from a height of 1 m and it hits the ground 0.4 seconds later. What is acceleration due to gravity on the planet ?

Answers

Answer:

Here,

v (final velocity) = 0

u (initial velocity) = u

a = ?

s = 1m

t = 0.4s

using the first equation of motion,

0 = u + 0.4a

= -0.4a = u

using the second equation of motion:

1 = 0.4u + 0.08a

from the bold equation

1 = 0.4(-0.4a) + 0.08a

1 = -0.16a + 0.08a

1 = -0.08a

a = -1/0.08

a = -100/8

a = -12.5 m/s/s

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A competitive diver leaves the diving board and falls toward the water with her body straight and rotating slowly. She pulls her arms and legs into a tight tuck position. What happens to her rotational kinetic energy

Answers

Answer: her rotational kinetic energy increases

During the spin cycle of your clothes washer, the tub rotates at a steady angular velocity of 31.7 rad/s. Find the angular displacement Δθ of the tub during a spin of 98.3 s, expressed both in radians and in revolutions.

Answers

Answer:

[tex]\Delta \theta = 3116.11\,rad[/tex] and [tex]\Delta \theta = 495.944\,rev[/tex]

Explanation:

The tub rotates at constant speed and the kinematic formula to describe the change in angular displacement ([tex]\Delta \theta[/tex]), measured in radians, is:

[tex]\Delta \theta = \omega \cdot \Delta t[/tex]

Where:

[tex]\omega[/tex] - Steady angular speed, measured in radians per second.

[tex]\Delta t[/tex] - Time, measured in seconds.

If [tex]\omega = 31.7\,\frac{rad}{s}[/tex] and [tex]\Delta t = 98.3\,s[/tex], then:

[tex]\Delta \theta = \left(31.7\,\frac{rad}{s} \right)\cdot (98.3\,s)[/tex]

[tex]\Delta \theta = 3116.11\,rad[/tex]

The change in angular displacement, measured in revolutions, is given by the following expression:

[tex]\Delta \theta = (3116.11\,rad)\cdot \left(\frac{1}{2\pi} \frac{rev}{rad} \right)[/tex]

[tex]\Delta \theta = 495.944\,rev[/tex]

hat a 15 kg body is pulled along a horizontal fictional table by a force of 4N what is the acceleration of the body ​

Answers

Answer:

Acceleration of the body is:

[tex]a=0.27\,\,m/s^2[/tex]

Explanation:

Use Newton's second Law to solve for the acceleration:

[tex]F=m\,\,a\\a=\frac{F}{m} \\a=\frac{4\,N}{15\,\,kg} \\a=0.27\,\,m/s^2[/tex]

Which of
of
these
following material is
used as fuse material?
carbon,
silver
Copper
Aluminium

Answers

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.

What will be the nature of the image formed from both a convex lens and a concave
lens of 20 centimeter focus distance, when the object is placed at a distance of
10 centimeters?​

Answers

Answer:

Explanation:

Using the lens formula

1//f = 1/u+1/v

f is the focal length of the lens

u is the object distance

v is the image distance

For convex lens

The focal length of a convex lens is positive and the image distance can either be negative or positive.

Given f = 20cm and u = 10cm

1/v = 1/f - 1/u

1/v = 1/20-1/10

1/v = (1-2)/20

1/V = -1/20

v = -20/1

v = -20 cm

Since the image distance is negative, this shows that the nature of the image formed by the convex lens is a virtual image

For concave lens

The focal length of a concave lens is negative and the image distance is negative.

Given f = -20cm and u = 10cm

1/v = 1/f - 1/u

1/v = -1/20-1/10

1/v = (-1-2)/20

1/V = -3/20

v = -20/3

v = -6.67 cm

Since the image distance is negative, this shows that the nature of the image formed by the concave lens is a virtual image

If 50 mL of each of the liquids in the answer choices were poured into a 250 mL beaker, which layer would be directly above a small rubber ball with a density of 0.960 g/mL? A. sea water – density of 1.024 g/mL B. mineral oil – density of 0.910 g/mL C. distilled water – density of 1.0 g/mL D. petroleum oil – density of 0.820 g/mL

Answers

Answer:

B. mineral oil – density of 0.910 g/mL.

Explanation:

Hello,

In this case, since the density is known as the degree of compactness a body has (mass in the occupied volume), the higher the density, the higher the weight of the body, therefore, if submerged into a liquid it could float if less dense than the liquid or sink if more dense than the liquid.

In such a way, since the rubber is more dense than mineral (0.960 g/mL > 0.910 g/mL) oil but less dense than distilled water (0.960 g/mL < 1.0 g/mL) we can say that B. mineral oil – density of 0.910 g/mL is directly above it when submerged.

Best regards.

A small omnidirectional stereo speaker produces waves in all directions that have an intensity of 8.00 at a distance of 4.00 from the speaker.

At what rate does this speaker produce energy?

What is the intensity of this sound 9.50 from the speaker?

What is the total amount of energy received each second by the walls (including windows and doors) of the room in which this speaker is located?

Answers

Answer:

A. We have that radius r = 4.00m intensity I = 8.00 W/m^

total power = power/ Area ( 4πr2)= 8.00 w/m^2( 4π ( 4.00 m)2=1607.68 W

b) I = total power/ 4πr2= 8.00 W/m2 ( 4.00 m/ 9.5 m)2= 1.418 W/m2

c) E = total power x time= 1607 . 68 W x 1s= 1607.68 J


A load of 1 kW takes a current of 5 A from a 230 V supply. Calculate the power factor.

Answers

Answer:

Power factor = 0.87 (Approx)

Explanation:

Given:

Load = 1 Kw = 1000 watt

Current (I) = 5 A

Supply (V) = 230 V

Find:

Power factor.

Computation:

Power factor = watts / (V)(I)

Power factor = 1,000 / (230)(5)

Power factor = 1,000 / (1,150)

Power factor = 0.8695

Power factor = 0.87 (Approx)

A toroidal solenoid has 590 turns, cross-sectional area 6.20 cm^2 , and mean radius 5.00 cm .Part A. Calcualte the coil's self-inductance.Part B. If the current decreases uniformly from 5.00 A to 2.00 A in 3.00 ms, calculate the self-induced emf in the coil.Part C. The current is directed from terminal a of the coil to terminal b. Is the direction of the induced emf froma to b or from b to a?

Answers

Complete Question

A toroidal solenoid has 590 turns, cross-sectional area 6.20 cm^2 , and mean radius 5.00 cm .

Part A. Calculate  the coil's self-inductance.

Part B. If the current decreases uniformly from 5.00 A to 2.00 A in 3.00 ms, calculate the self-induced emf in the coil.

Part C. The current is directed from terminal a of the coil to terminal b. Is the direction of the induced emf from a to b or from b to a?

Answer:

Part A  

       [tex]L = 0.000863 \ H[/tex]

Part B  

       [tex]\epsilon = 0.863 \ V[/tex]

Part C

    From terminal a to terminal b

Explanation:

From the question we are told that

      The  number of turns is  [tex]N = 590 \ turns[/tex]

      The cross-sectional area is  [tex]A = 6.20 cm^2 = 6.20 *10^{-4} \ m[/tex]

      The  radius is [tex]r = 5.0 \ cm = 0.05 \ m[/tex]

       

Generally the coils self -inductance is mathematically represented as

              [tex]L = \frac{ \mu_o N^2 A }{2 \pi * r }[/tex]

Where [tex]\mu_o[/tex] is the permeability of  free space with value [tex]\mu_o = 4\pi * 10^{-7} N/A^2[/tex]

substituting values

             [tex]L = \frac{ 4\pi * 10^{-7} * 590^2 6.20 *10^{-4} }{2 \pi * 0.05 }[/tex]

             [tex]L = \frac{ 2 * 10^{-7} * 590^2 6.20 *10^{-4} }{ 0.05 }[/tex]

             [tex]L = 0.000863 \ H[/tex]

Considering the Part B

      Initial current is [tex]I_1 = 5.00 \ A[/tex]

      Current at time t is [tex]I_t = 3.0 \ A[/tex]

       The  time taken is  [tex]\Delta t = 3.00 ms = 0.003 \ s[/tex]

The self-induced emf is mathematically evaluated as

          [tex]\epsilon = L * \frac{\Delta I}{ \Delta t }[/tex]          

=>         [tex]\epsilon = L * \frac{ I_1 - I_t }{ \Delta t }[/tex]

substituting values

             [tex]\epsilon = 0.000863 * \frac{ 5- 2 }{ 0.003 }[/tex]  

             [tex]\epsilon = 0.863 \ V[/tex]

The direction of the induced emf is  from a to b because according to Lenz's law the induced emf moves in the same direction as the current

This question involves the concepts of the self-inductance, induced emf, and Lenz's Law

A. The coil's self-inductance is "0.863 mH".

B. The self-induced emf in the coil is "0.58 volts".

C. The direction of the induced emf is "from b to a".

A.

The self-inductance of the coil is given by the following formula:

[tex]L=\frac{\mu_oN^2A}{2\pi r}[/tex]

where,

L = self-inductance = ?

[tex]\mu_o[/tex] = permeability of free space = 4π x 10⁻⁷ N/A²

N = No. of turns = 590

A = Cross-sectional area = 6.2 cm² = 6.2 x 10⁻⁴ m²

r = radius = 5 cm = 0.05 m

Therefore,

[tex]L=\frac{(4\pi\ x\ 10^{-7}\ N/A^2)(590)^2(6.2\ x\ 10^{-4}\ m^2)}{2\pi(0.05\ m)}[/tex]

L = 0.863 x 10⁻³ H = 0.863 mH

B.

The self-induced emf is given by the following formula:

[tex]E=L\frac{\Delta I}{\Delta t}\\\\[/tex]

where,

E = self-induced emf = ?

ΔI = change in current = 2 A

Δt = change in time = 3 ms = 0.003 s

Therefore,

[tex]E=(0.000863\ H)\frac{2\ A}{0.003\ s}[/tex]

E = 0.58 volts

C.

According to Lenz's Law, the direction of the induced emf always opposes the change in flux that causes it. Hence, the direction of the induced emf will be from b to a.

Learn more about Lenz's Law here:

https://brainly.com/question/12876458?referrer=searchResults

An object on a level surface experiences a horizontal force of 12.7 N due to kinetic friction. The coefficient of kinetic friction is 0.42.
What is the mass of the object? (Express your answer to two significant figures)kg

Answers

Answer:

The mass of the object is 3.08 kg.

Explanation:

The horizontal force is12.7 N and the coefficient of the kinetic fraction are 0.42. Now we have to compute the mass of the object. Thus, use the below formula to find the mass of the object.

Let the mass of the object = m.

The coefficient of kinetic friction, n = 0.42

Therefore,  

Force, F = n × mg

12.7 = 0.42 × 9.8 × m

m = 3.08 kg

The mass of the object is 3.08 kg.

A wire of 5.8m long, 2mm diameter carries 750ma current when 22mv potential difference is applied at its ends. if drift speed of electrons is found then:_________.
(a) The resistance R of the wire(b) The resistivity p, and(c) The number n of free electrons per unit volume.​

Answers

Explanation:

According to Ohms Law :

V = I * R

(A) R (Resistance) = 0.022 / 0.75 = 0.03 Ohms

Also,

[tex]r = \alpha \frac{length}{area} = \alpha \frac{5.8}{3.14 \times 0.001 \times 0.001} [/tex]

(B)

[tex] \alpha(resistivity) = 1.62 \times {10}^{ - 8} [/tex]

Drift speed is missing. It is given as;

1.7 × 10^(-5) m/s

A) R = 0.0293 ohms

B) ρ = 1.589 × 10^(-8)

C) n = 8.8 × 10^(28) electrons

This is about finding, resistance and resistivity.

We are given;

Length; L = 5.8 m

Diameter; d = 2mm = 0.002 m

Radius; r = d/2 = 0.001 m

Voltage; V = 22 mv = 0.022 V

Current; I = 750 mA = 0.75 A

Area; A = πr² = 0.001²π

Drift speed; v_d = 1.7 × 10^(-5) m/s

A) Formula for resistance is;

R = V/I

R = 0.022/0.75

R = 0.0293 ohms

B) formula for resistivity is given by;

ρ = RA/L

ρ = (0.0293 × 0.001²π)/5.8

ρ = 1.589 × 10^(-8)

C) Formula for current density is given by;

J = n•e•v_d

Where;

J = I/A = 0.75/0.001²π A/m² = 238732.44 A/m²

e is charge on an electron = 1.6 × 10^(-19) C

v_d = 1.7 × 10^(-5) m/s

n is number of free electrons per unit volume

Thus;

238732.44 = n(1.6 × 10^(-19) × 1.7 × 10^(-5))

238732.44 = (2.72 × 10^(-24))n

n = 238732.44/(2.72 × 10^(-24))

n = 8.8 × 10^(28)

Read more at; brainly.com/question/17005119

g Two point sources emit sound waves of 1.0-m wavelength. The source 1 is at x = 0 and source 2 is at x = 2.0 m along x-axis. The sources, 2.0 m apart, emit waves which are in phase with each other at the instant of emission. Where, along the line between the sources, are the waves out of phase with each other by π radians?

Answers

Answer:

constructive interferencia  0, 1 , 2 m

destructive inteferencia   1/4, 3/4. 5/4, 7/4 m

Explanation:

This exercise is equivalent to the double slit experiment, the two sources are in phase and separated by a distance, therefore the waves observed in the line between them have an optical path difference and a phase difference, given by the expression

            Δr / λ = Φ / 2π

            Δr = Φ/2π   λ

let's apply this expression to our case

λ = 1 m

            Δr = Φ 1 / 2π

We have constructive interference for angle of  Φ = 0, 2π, ...

let's find the values ​​where they occur

  Φ         Δr

   0          0

  2π         1

  4π        2

Destructive interference occurs by    Φ = π /2, 3π / 2, ...

 Φ          Δr

 π/2       ¼ m

 3π /2    ¾ m

5π /2     5/4 m

7π /2      7/4 m

Sammy is 5 feet and 5.3 inches tall. What is Sammy's height in inches? ​

Answers

Answer:

[tex]\boxed{\sf 65.3 \ inches}[/tex]

Explanation:

1 foot = 12 inches

Sammy is 5 feet tall.

5 feet = ? inches

Multiply the feet value by 12 to find in inches.

5 × 12

= 60

Add 5.3 inches to 60 inches.

60 + 5.3

= 65.3

65.3 Inches.
12 (1 Foot) X 5= 60 + 5.3 = 65.3

A lamp in a child's Halloween costume flashes based on an RC discharge of a capacitor through its resistance. The effective duration of the flash is 0.220 s, during which it produces an average 0.520 W from an average 3.00 V.
A. How much charge moves through the lamp (C)?
B. Find the capacitance (F).
C. What is the resitance of the lamo?

Answers

Answer:

A. 0.0374C

B. 0.012F

C. 18 ohms

Explanation:

See attached file

If two identical wires carrying a certain current in the same direction are placed parallel to each other, they will experience a force of repulsion.
a) true
b) false

Answers

Answer:

The answer is B.  false

Explanation:

Current in the same direction

 When current flow through to parallel conductors of a given length, when the current flows in the same direction

1. A force of attraction between the wires occurs and this tends to draw the wires inward

2. A magnetic field in the same direction is produced.

Current in opposite direction

when the current is in opposite direction

1. Force of repulsion between the two wires occurs, draws the wire outward

2. A magnetic field in opposite direction occurs

A 70 kg man floats in freshwater with 3.2% of his volume above water when his lungs are empty, and 4.85% of his volume above water when his lungs are full.

Required:
a. Calculate the volume of air he inhales - called his lung capacity - in liters.
b. Does this lung volume seem reasonable?

Answers

Answer:

Explanation:

A) Vair = 1.3 L

B) Volume is not reasonable

Explanation:

A)

Assume

m to be total mass of the man

mp be the mass of the man that pulled out of the water

m1 be the mass above the water with the empty lung

m2 be the mass above the water with full lung

wp be the weight that the buoyant force opposes as a result of the air.

Va be the volume of air inside man's lungs

Fb be the buoyant force due to the air in the lung

given;

m = 78.5 kg

m1 = 3.2% × 78.5 = 2.5 kg

m2 = 4.85% × 78.5 = 3.8kg

But, mp = m2- m1

mp = 3.8 - 2.5

mp = 1.3kg

So using

Archimedes principle, the relation for formula for buoyant force as;

Fb = (m_displaced water)g = (ρ_water × V_air × g)

Where ρ_water is density of water = 1000 kg/m³

Thus;

Fb = wp = 1.3× 9.81

Fb = 12.7N

But

Fb = (ρ_water × V_air × g)

So

Vair = Fb/(ρ_water × × g)

Vair = 12.7/(1000 × 9.81)

V_air = 1.3 × 10^(-3) m³

convert to litres

1 m³ = 1000 L

Thus;

V_air = 1.3× 10^(-3) × 1000

V_air = 1.3 L

But since the average lung capacity of an adult human being is about 6-7litres of air.

Thus, the calculated lung volume is not reasonable

Explanation:

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