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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 1

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.

Answer 2

Answer:

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.

Learn more about Circuit Capacity here:

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A capacitor consists of two closely spaced metal conductors of large area, separated by a thin insulating foil. It has an electrical capacity of 3000.0 μF and is charged to a potential difference of 60.0 V. Calculate the amount of energy stored in the capacitor. Tries 0/20 Calculate the charge on this capacitor when the electrical energy stored in the capacitor is 6.53 J. Tries 0/20 If the two plates of the capacitor have their separation increased by a factor of 5 while the charge on the plates remains constant, by what factor is the energy stored in the capacitor increased?

Answers

Answer:

1 = 5.4 J

2 = 0.1979 C

3 = 5

Explanation:

Energy in a capacitor, E is

E = 1/2 * C * V²

E = 1/2 * 3000*10^-6 * 60²

E = 1/2 * 3000*10^-6 * 3600

E = 1/2 * 10.8

E = 5.4 J

E = Q²/2C = 6.53 J

E * 2C = Q²

Q² = 6.53 * 2 * 3000*10^-6

Q² = 13.06 * 3000*10^-6

Q² = 0.03918

Q = √0.03918

Q = 0.1979 C

The Capacitor, C is inversely proportional to the distance of separation, D. Thus, if D is increased by 5 to be 5D, then C would be C/5. And therefore, our energy stored in the capacitor is increased by a factor of 5.

Light has wavelength 600 nm in a vacuum. it passes into glass, which has an index of refraction of 1.5. what is the frequency of the light inside the glass

Answers

Light has wavelength 600 nm in a vacuum ,the frequency of the light is 2 × $10^(-13)$ Hz.

What is wavelength?

The separation between such a wave motion's crests and troughs would be known as the wavelength of photons.

What is frequency ?

The total number of waves that pass a specific location in a predetermined amount of time is known as frequency.

Calculation of frequency

Given data:

wavelength = 600 nm = 600 × $10^(-9)$ m

index of refraction = 1.5.

Frequency can be calculated by using the formula:

v = f × wavelength

f = wavelength / v

Where, f = Frequency , v is velocity.

put the given data in above equation.

f = wavelength / v

f = 600 × $10^(-9)$ m / 3 × $10^(8)$

f = 200 × $10^(-15)$ .

f = 2 × $10^(-13)$

Therefore, the frequency of the light is 2 × $10^(-13)$ Hz.

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v = f lambda

in vac ... 3X10^8 = 600x10^-9xf

in glass speed slower, poss 2/3 that of vacuum

Air contains 78.08% nitrogen, 20.095% oxygen, and 0.93% argon. calculate the partial pressure of oxygen if the total pressure of the air sample is 1.7 atm.a.

Answers

partial pressure in a mixture of two or more gases will be given by formula

$P_(partial)$ = mole fraction of gas * total pressure

now here mole fraction is same as percentage of gas in the mixture

Now mole fraction of oxygen is 0.20095 (20.095%)

now here pressure of oxygen in the mixture is given as

$P_(o_2) = 0.20095 * P_(total)$

$P_(o_2) = 0.20095 * 1.7$

$P_(o_2) = 0.342 atm$

so pressure due to oxygen in the mixture will be 0.342 atm

Answer:

20.095

Explanation:

Recall your experimental setup from Lab 05A: a constant force was applied to a disc by attaching a mass to a light string wrapped around a mass-less pulley and hanging the mass over the edge of the apparatus. In the lab, you used energy conservation arguments to derive an expression for the angular velocity of the disc after the mass had fallen a distance x . Your goal now is to use kinematics and dynamics to confirm your expression. Use the following symbols throughout this question: m is the mass of the hanging mass, M is the mass of the disc, r is the radius of the pulley, R is the radius of the disc, x is the distance the mass has fallen, and g is the acceleration due to gravity. What is the linear acceleration of the mass after it has fallen a distance x

Answers

Answer:

w = $\sqrt{(2gy)/(r^2 + (1)/(2) R^2 ) }$

Explanation:

For this exercise let's start by applying Newton's second law to the mass with the string

W - T = m a

In this case, as the system is going down, we will assume the vertical directional down as positive.

T = W - m a

Now we apply Newton's second law for rotational motion to the pulley of radius r. We will assume the positive counterclockwise rotations

∑ τ = I α

T r = I α

the moment of inertia of the disk is

I = ½ M R²

angular and linear acceleration are related

a = α r

we substitute

T r = (½ m R²) (a / r)

T = ½ m ( $(R)/(r)$ )² a

we write our two equations

T = W - m a

T = ½ m ( $(R)/(r)$ )² a

we solve the system of equations

W - m a = ½ m (\frac{R}{r} )² a

m g = m a [ 1 + ½ (\frac{R}{r} )² ]

a = $(g)/( 1 + (1)/(2) ((R)/(r))^2 )$

this acceleration is constant throughout the trajectory, so with the angular and lineal kinematics relations

w² = w₀² + 2 α θ

v² = v₀² + 2 a y

as the system is released its initial angular velocity is zero

w² = 0 + 2 α θ

v² = 0 + 2 a y

we look for the angular acceleration

a =α r

α = a / r

α = $(g)/(r (1 + (1)/(2) ((R)/(r))^2 )$

we look for the angle, remember that they must be measured in radians

θ = s / r

in this case we approximate the arc to the distance

s = y

θ = y / r

we substitute

w = $\sqrt{2 (g)/( r( (1)/(2) ((R)/(r))^2 ) (y)/(r) }$

w = $\sqrt{(2gy)/(r^2 + (1)/(2) R^2 ) }$

for the simple case where r = R

w = $\sqrt{ (2gy)/( (3)/(2) R^2 ) }$

w = $\sqrt{ (4)/(3) (gy)/(R^2) }$

Instantaneous speed is...a) A speed of 1000 km/h
b) The speed attained at a particular instant in time.
c) The speed that can be reached in a particular amount of time.

PLEASE HURRY

Answers

Answer:

The speed attained at a particular instant in time.

Explanation:

Instantaneous speed is the speed attained at a particular instant in time.

It is given by :

$v=(dx)/(dt)$

It is equal to the rate of change of speed.

It can be also defined as when the speed of an object is constantly changing, the instantaneous speed is the speed of an object at a particular moment (instant) in time.

Hence, the correct option is (b).

In the standing broad jump, one squats and then pushes off with the legs to see how far one can jump. Suppose the extension of the legs from the crouch position is 0.55 m and the acceleration achieved during the time the jumper is extending their legs is 1.2 times the acceleration due to gravity, g .How far can they jump? State your assumptions. (Increased range can be achieved by swinging the arms in the direction of the jump.)

Answers

Answer:

1.32 m.

Explanation:

Below is an attachment containing the solution.

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