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A single-turn circular loop of wire of radius 5.0 cm lies in a plane perpendicular to a spatially uniform magnetic field. During a 0.02500.0250-\text{s}s time interval, the magnitude of the field increases uniformly from 200 to 300 mT. Determine the magnitude of the emf induced in the loop

Answers

Answer 1

Answer:

Given Information:

time = Δt = 0.0250 seconds

Radius = r = 5 cm = 0.05 m

Change in Magnetic field = ΔB = (0.300 - 0.200) T

Number of turns = N = 1

Required Information:

Magnitude of induced emf = ξ = ?

Answer:

Magnitude of induced emf = ξ = 3.141x10⁻² V

Explanation:

The EMF induced in a circular loop of wire in a changing magnetic field is given by

ξ = -NΔΦ/Δt

Where change in flux ΔΦ is given by

ΔΦ = ΔBA

ΔΦ = ΔBπr²

ΔΦ = (0.300 - 0.200)*π*(0.05)²

ΔΦ = 7.854x10⁻⁴ T.m²

ξ = -NΔΦ/Δt

ξ = -(1*7.854x10⁻⁴)/0.0250

ξ = -3.141x10⁻² V

The negative sign is due to Lenz law.

Answer 2

Answer:

-0.0314 V

Explanation:

Parameters given:

Initial magnetic field, Bini = 200 mT = 0.2T

Final magnetic field, Bfin = 300mT = 0.3 T

Number of turns, N = 1

Radius, r = 5 cm = 0.05 m

Time, t = 0.025 secs

Induced EMF is given as:

EMF = [-(Bfin - Bini) * N * pi * r²] / t

EMF = [-(0.3 - 0.2) * 1 * 3.142 * 0.05²] / 0.025

EMF = (-0.1 * 3.142 * 0.0025) / 0.025

EMF = -0.0314 V

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Given a double slit apparatus with slit distance 1 mm, what is the theoretical maximum number of bright spots that I would see when I shine light with a wavelength 400 nm on the slits

Answers

Answer:

The maximum number of bright spot is $n_(max) =5001$

Explanation:

From the question we are told that

The slit distance is $d = 1 \ mm = 0.001 \ m$

The wavelength is $\lambda = 400 \ nm = 400*10^(-9 ) \ m$

Generally the condition for interference is

$n * \lambda = d * sin \theta$

Where n is the number of fringe(bright spots) for the number of bright spots to be maximum $\theta = 90$

=> $sin( 90 )= 1$

$n = (d )/(\lambda )$

substituting values

$n = ( 1 *10^(-3) )/( 400 *10^(-9) )$

$n = 2500$

given there are two sides when it comes to the double slit apparatus which implies that the fringe would appear on two sides so the maximum number of bright spots is mathematically evaluated as

$n_(max) = 2 * n + 1$

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You are an electrician installing the wiring in a new home. The homeowner desires that a ceiling fan with light kits be installed in five different rooms. Each fan contains a light kit that can accommodate four 60-watt lamps. Each fan motor draws a current of 1.8 amperes when operated on high speed. It is assumed that each fan can operate more than three hours at a time and therefore must be considered a continuous-duty device. The fans are to be connected to a 15-ampere circuit. Because the devices are continuous-duty, the circuit current must be limited to 80% of the continuous connected load. How many fans can be connected to a single 15-ampere circuit

Answers

Answer:

3 fans per 15 A circuit

Explanation:

From the question and the data given, the light load let fan would have been

(60 * 4)/120 = 240/120 = 2 A.

Next, we add the current of the fan motor to it, so,

2 A + 1.8 A = 3.8 A.

Since the devices are continuos duty and the circuit current must be limited to 80%, then the Breaker load max would be

0.8 * 15 A = 12 A.

Now, we can get the number if fans, which will be

12 A/ 3.8 A = 3.16 fans, or approximately, 3 fans per 15 A circuit.

Final answer:

The total power draw of each fan is 3.8 amperes. Thus, considering a limit of 80% usage of 15 amperes, only 3 fans can be connected to a single circuit to keep the total power draw below 12 amperes.

Explanation:

The question is asking how many ceiling fans, each with a certain power draw, can be connected on a single 15-ampere circuit, considering that each fan is a continuous-duty device. The power draw of each fan when the motor is operated at high speed and the light kit is fully loaded is the sum of the power draw of the motor and the light kit. As the power draw of each motor is 1.8 amperes and the light kit is 240 watts or 2 amperes (calculated using the formula Power = Voltage x Current; assuming a voltage of 120 volts), the total power draw of each fan is 3.8 amperes. Considering the limit of 80% of the continuous load, only 12 amperes (80% of 15) can be used. Thus, 3 fans can be connected to the circuit as it reaches 11.4 amperes, close enough to the 12 amperes limit.

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Enunciado: Una bola se lanza verticalmente de la parte superior de un edificio con una velocidad inicial de 25 m/s. La bola impacta al suelo en la base del edificio 7 segundos después de ser lanzada. (Marque la respuesta correcta) ¿Qué altura subió la bola (medida desde la parte superior del edificio)? a) 19.6 m b) 12.75 m c) 31.88 m d) 40 m e) 20 m

Answers

La altura vertical máxima alcanzada es de 31,88 m.

Tenemos la siguiente información de la pregunta;

Velocidad inicial = 25 m/s

Velocidad final = 0 m/s (a la altura máxima)

tiempo empleado = 3,5 minutos (el tiempo empleado para subir y bajar es igual).

Usando la ecuación;

v^2 = u^2 - 2gh

Dado que v = 0

u^2 = 2gh

h = tu^2/2g

h = (25)^2/2 *9.8

h = 31,88 m

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The subject of this question is kinematics. The ball reached a height of 65.1 meters.

To determine the height that the ball reached, we can use the kinematic equation for vertical motion:

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The record for the world’s loudest burp is 109.9 dB, measured at a distance of 2.5 m from the burper. Assuming that this sound was emitted as a spherical wave, what was the power emitted by the burper during his record burp?