# Suppose The Electric Company Charges 10 Fils Per KW H. How Much Does It Cost To Use A 125 Watt Lamp 4 (2023)

Physics College

150 fils

Explanation:

Power = 125 Watt

Time 4 hours a day

Energy per day = Power x time per day = 125 x 4 = 500 watt hour

Energy for 30 days = 500 x 30 watt hour = 15000 watt hour = 15 KWH

Cost of 1 KWH = 10 fils

Cost of 15 KWH = 15 x 10 = 150 fils

## Related Questions

A 0.45 m radius, 500 turn coil is rotated one-fourth of a revolution in 4.01 ms, originally having its plane perpendicular to a uniform magnetic field. Find the magnetic field strength in T needed to induce an average emf of 10,000 V.

B = 0.126 T

Explanation:

As per Faraday's law we know that rate of change in magnetic flux will induce EMF in the coil

So here we can say that EMF induced in the coil is given as

initially the coil area is perpendicular to the magnetic field

and after one fourth rotation of coil the area vector of coil will be turned by 90 degree

so we can say

now we will have

Suppose a 50 turn coil lies in the plane of the page in a uniform magnetic field that is directed into the page. The coil originally has an area of 0.325 m^2. It is squished to have no area in 0.225 s. What is the magnitude of the average induced emf in volts if the uniform magnetic field has a strength of 1.5 T?

EMF = 108.3 Volts

Explanation:

As per Faraday's law of electromagnetic induction we know that rate of change in magnetic flux will induce EMF in closed loop

So it is given as

now we know that

N = 50 turns

B = 1.5 T

A = 0.325 m^2

now we have

Consider a 26-MeV proton moving perpendicularly to a 1.35 T field in a cyclotron. Find the radius of curvature, in meters, of the path of the proton while moving through the cyclotron.

0.545 m

Explanation:

K. E = 26 MeV = 26 x 1.6 x 10^-13 J = 4.16 x 10^-12 J

B = 1.35 T

Let r be the radius of curvature

The formula for the kinetic energy of a cyclotron is given by

m = 1.67 x 10^-27 kg, q = 1.6 x 10^-19 c

r = 0.545 m

A screwdriver is being used in a 13.5 T magnetic field. what maximum emf can be induced in V along its 10.5 cm length when it moves through the field at 0.85 m/s?

EMF = 1.20 V

Explanation:

It is given that,

Magnetic field used by the screwdriver, B = 13.5 T

Length of screwdriver, l = 10.5 cm = 0.105 m

Speed with which it is moving. v = 0.85 m/s

We need to find the maximum EMF induced in the screwdriver. It is given by :

So, the maximum emf of the screwdriver is 1.20 V. Hence, this is the required solution.

An MRI technician moves his hand from a region of very low magnetic field strength into an MRI scanner's 2.00 T field with his fingers pointing in the direction of the field. His wedding ring has a diameter of 2.5 cm and it takes 0.45 s to move it into the field. What average current is induced in the ring in A if its resistance is 0.0100 Ω ?

2.62 A

Explanation:

B = 2 T, Diameter = 2.5 cm , radius, r = 0.0125 m, t = 0.45 s, r = 0.01 ohm

Induced emf, e = rate of change of magnetic flux

e = A x dB / dt = 3.14 x (0.0125)^2 x 2 / 0.45

e = 0.026 V

induced current, i = e / R = 0.026 / 0.01 = 2.62 A

The average induced current in the loop is 0.218 A.

Induced emf in the loop

The emf induced in the loop is determined by applying Faraday's law as shown below;

emf = dФ/dt

emf = BA/t

where;

A is the area

A = πr² = πd²/4

A = π x (0.025)²/4

A = 4.908 x 10⁻³ m²

emf = (2 x 4.908 x 10⁻³)/(0.45)

emf = 2.18 x 10⁻³ V

Average induced current

The average induced current in the loop is calculated as follows;

I = emf/R

I = 2.18 x 10⁻³/0.01

I = 0.218 A

The emf of a battery is equal to its terminal potential difference: A) under all conditions B) only when the battery is being charged C) only when a large cwrrent is in the battery D) only when there is no current in the battery E) under no conditions

option (C)

Explanation:

EMF stand for electro motive force. the emf of a battery is the potential between the two electrodes when it is not use in the circuit.

The terminal potential difference is the potential difference between the electrodes of a cell when it is in use.

EMF is only when the current is very large in the battery.

The emf of a battery is equal to its terminal potential difference only when a large cwrrent is in the battery.

C) only when a large cwrrent is in the battery.

A battery with an emf of 12 V and an internal resistance of 1 Ω is used to charge a battery with an emfof 10 V and an internal resistance of 1 Ω. The current in the circuit is : A) 2 A B) 1 A C) 4A D) 11A E) 22 A

Option B

Explanation:

The net emf in the circuit

E = 12 - 10 = 2 V

Total effective resistance,

r = 1 + 1 = 2 ohm

By using Ohm's law

E = I × R

I = 2 / 2 = 1 A

A wire with a length of 150 m and a radius of 0.15 mm carries a current with a uniform current density of 2.8 x 10^7A/m^2. The current is: A) 2.0 A B) 0.63 A C) 5.9A D) 300 A E) 26000 A

The current is 2.0 A.

(A) is correct option.

Explanation:

Given that,

Length = 150 m

Current density

We need to calculate the current

Using formula of current density

Where, J = current density

A = area

I = current

Put the value into the formula

Hence, The current is 2.0 A.

A 60-watt light bulb carries a current of 0.5 ampere. The total charge passing through it in one hour is: A) 3600 C B) 3000C C) 2400C D) 120 C E) 1800 C

Total charge, q = 1800 C

Explanation:

It is given that,

Power of light bulb, P = 60 watts

It carries a current, I = 0.5 A

Time, t = 1 hour = 3600 seconds

We need to find the total charge passing through it in one hour. We know that current through an electrical appliance is defined as the charge flowing per unit time i.e.

q = 1800 C

So, the total charge passing through it is 1800 C. Hence, this is the required solution.

A cosmic ray electron moves at 6.5x 10^6 m/s perpendicular to the Earth's magnetic field at an altitude where the field strength is 10x 10^-5 T. What is the radius, in meters, of the circular path the electron follows?

Explanation:

It is given that,

Speed of cosmic ray electron,

Magnetic field strength,

We need to find the radius of circular path the electron follows. It is given by :

r = 0.36 meters

So, the radius of circular path is 0.36 meters. Hence, this is the required solution.

A proton travels at a speed 0.25 x 10^7 m/s perpendicular to a magnetic field. The field causes the proton to travel in a circular path of radius 0.975 m. What is the field strength, in tesla?

The magnetic field strength of the proton is 0.026 Tesla.

Explanation:

It is given that,

Speed of the proton,

The radius of circular path, r = 0.975 m

It is moving perpendicular to a magnetic field such that the magnetic force is balancing the centripetal force.

q = charge on proton

B = 0.026 Tesla

So, the magnetic field strength of the proton is 0.026 Tesla.

A cosmic ray proton moving toward the Earth at 3.5x 10^7 ms experiences a magnetic force of 1.65x 10^-16 N. What is the strength of the magnetic field if there is a 45° angle between it and the proton's velocity?

Magnetic field,

Explanation:

It is given that,

Velocity of proton,

Magnetic force,

Charge of proton,

We need to find the strength of the magnetic field if there is a 45° angle between it and the proton's velocity. The formula for magnetic force is given by :

B = 0.0000416 T

Hence, this is the required solution.

Consider a 150 turn square loop of wire 17.5 cm on a side that carries a 42 A current in a 1.7 T. a) What is the maximum torque on the loop of wire in N.m? b) What is the magnitude of the torque in N·m when the angle between the field and the normal to the plane of the loop θ is 14°?

(a) 328 Nm

(b) 79.35 Nm

Explanation:

N = =150, side = 17.5 cm = 0.175 m, i = 42 A, B = 1.7 T

A = side^2 = 0.175^2 = 0.030625 m^2

(a) Torque = N x i x A x B x Sinθ

For maximum torque, θ = 90 degree

Torque = 150 x 42 x 0.030625 x 1.7 x Sin 90

Torque = 328 Nm

(b) θ = 14 degree

Torque = 150 x 42 x 0.030625 x 1.7 x Sin 14

Torque = 79.35 Nm

Light with a wavelength of 612 nm incident on a double slit produces a second-order maximum at an angle of 25 degree. What is the separation between the slits?

Speration between the slits

Explanation:

Given that,

Wavelength of light,

It is incident on a double slit produces a second-order maximum at an angle of 25 degree.

The condition for maxima is given as :

d = seperation between the slits

m = order, m = 2

d = 0.00000289

or

So, the seperation between the silts is . Hence, this is the required solution.

Light with a wavelength of 488 nm is incident on a single slit with a width of 6.23 x 10^-4 m. If the screen is 2.7 m away, what is the distance to the first antinodal line?

The distance to the first anti-nodal line is .

Explanation:

Given that,

Wavelength = 488 nm

Width

Distance D =2.7 m

We need to calculate the distance to the first anti-nodal line

Using formula of the distance for first anti-nodal line

.....(I)

Where, n = number of fringe

d = width

D = distance from the screen

=wavelength of light

Put the all value in the equation (I)

Hence, The distance to the first anti-nodal line is .

(d) If η = 40% and TH = 427°C, what is TC, in °C?

Explanation:

Given that,

Efficiency of heat engine,

Temperature of hot source,

We need to find the temperature of cold sink i.e. . The efficiency of heat engine is given by :

So, the temperature of the cold sink is 256.2°C. Hence, this is the required solution.

(b) If TH = 500°C, TC = 20°C, and Wcycle = 200 kJ, what are QH and QC, each in kJ?

QC = 122 KJ

QH = 2.64 x 122 = 322 KJ

Explanation:

TH = 500 Degree C = 500 + 273 = 773 K

TC = 20 degree C = 20 + 273 = 293 K

W cycle = 200 KJ

Use the formula for the work done in a cycle

Wcycle = QH - QC

200 = QH - QC ..... (1)

Usse

TH / TC = QH / QC

773 / 293 = QH / QC

QH / QC = 2.64

QH = 2.64 QC Put it in equation (1)

200 = 2.64 QC - QC

QC = 122 KJ

So, QH = 2.64 x 122 = 322 KJ

(c) If η = 60% and TC = 40°F, what is TH, in °F?

2b2t hope that helps

A 1 kg ball is hung at the end of a rod 1 m long. If the system balances at a point on the rod one third of the distance from the end holding the mass, what is the mass of the rod?

The mass of the rod on which balances the ball hung at one end is 2 kg.

The given parameters;

• mass of the ball, m = 1 kg
• length of the rod, L = 1 m

The one-third distance from the point holding the mass is calculated as follows;

p = ¹/₃ x 1 = 0.33 m

0 0.33 m 1 m

------------------------------Δ----------------------------------------------------------

↓ ↓

W 1 kg

W(0.33) = 1(1 - 0.33)

W(0.33) = 1(0.67)

Thus, the mass of the rod on which balances the ball hung at one end is 2 kg.

2 kg

Explanation:

Assuming the rod's mass is uniformly distributed, the center of mass is at half the length.

Sum of the moments at the balance point:

-(Mg)(L/3) + (mg)(L/2 − L/3) = 0

(Mg)(L/3) = (mg)(L/2 − L/3)

(Mg)(L/3) = (mg)(L/6)

2M = m

M = 1 kg, so m = 2 kg.

The mass of the rod is 2 kg.

In a certain cyclotron a proton moves in a circle of radius 0.740 m. The magnitude of the magnetic field is 0.960 T. (a) What is the oscillator frequency? (b) What is the kinetic energy of the proton?

Part a)

Part b)

Explanation:

Part a)

As we know that radius of circular path of a charge moving in constant magnetic field is given as

now we have

now the frequency of oscillator is given as

PART b)

now for kinetic energy of proton we will have

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