A 500 W heating coil designed to operate from 110 V is made of Nichrome 0.500 mm in diametera.Assuming the resistivity of the nichrome remains constant at is 20.0 degrees C value find the length of wire used.b. Now consider the variation of resistivity with temperature. What power is delivered to the coil of part (a) when it is warmed to 1200 degrees C.?

Answers

Answer 1
Answer:

(a) Length of the wire is 3.162 m

(b)Power delivered to the coil is 339.7 W

Electrical Power:

The electrical power is given by

P = V² / R

R = V² / P

Resistance of the heating coil, R

R = (110² / 500)

R = 12100 / 500

R = 24.2 Ω

Now the resistivity of a wire is given by

ρ= RA/L

here ρ = 1.50×10⁻⁶ Ωm

so after rearranging we get:

L = RA / ρ

Now, the radius of wirer = 0.5 / 2 mm = 0.25 mm = 2.5×10⁻⁴ m

So the cross sectional area can be calculated as follows

A = \pi r^2\n\nA = \pi * (2.5*10^(-4))^2\n\nA = 1.96*10^(-7) m^2

hence,

L = (24.2 *1.96*10^(-7) / 1.50*10^(-6)) \n\nL = 3.162\; m

(b)The dependency of resistance with temperature is as follows:

R = R₀[1 +  αΔT]

α = 4*10^(-4)^\;oC^(-1) for Nichrome

R' = R [1 + \alpha (1200 - 20) ]\n\nR' = R[1 + \alpha (1180) ]\n\nR' = 24.2[ 1 + 4*10^(-4) * 1180 ]\n\nR' = 24.2[1 + 0.472]\n\nR' = 24.2 * 1.472\n\nR' = 35.62 \;\Omega

So the power generated is :

P = V² / R

P = (110² / 35.62)

P = 12100/ 35.62

P = 339.70 watts

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Answer 2
Answer:

Answer:

a) 3.162 m

b) 339.7 W

Explanation:

Assume ρ = 1.50*10^-6 Ωm, and

α = 4.000 10-4(°C)−1 for Nichrome

To solve this, we would use the formula

P = V² / R

So when we rearrange and make R subject of formula, we have

R = V² / P

Resistance of the heating coil, R

R = (110² / 500)

R = 12100 / 500

R = 24.2 ohms

Recall the formula for resistivity of a wire

R = ρ.L/A

Again, in rearranging and making L subject of formula, we have

L = R.A / ρ

To make it uniform, we convert our radius from mm to m.

Diameter, D = 0.5 mm

Radius of wire = 0.5 / 2 mm = 0.25 mm = 0.00025 m

We then use this radius to find our area

A = πr²

A = π * 0.00025²

A = 1.96*10^-7 m²

And finally, we solve for L

L = (24.2 * 1.96*10^-7 / 1.50*10^-6) =

L = 3.162 m

(b)

Temperature coefficient of resistance.

R₁₂₀₀ = R₂₀[1 + α(1200 - 20.0) ]

R₁₂₀₀ = R₂₀[1 + α(1180) ]

R₁₂₀₀ = 24.2[ 1 + 4.*10^-4 * 1180 ]

R₁₂₀₀ = 24.2[1 + 0.472]

R₁₂₀₀ = 24.2 * 1.472

R₁₂₀₀ = 35.62 ohms

Putting this value of R in the first formula from part a, we have

P = V² / R

P = (110² / 35.62)

P = 12100/ 35.62

P = 339.70 watts


Related Questions

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A golf pro swings a golf club, striking a golf ball that has a mass of 55.0 g. The club is in contact with the ball for only 0.00340 s. After the collision, the ball leaves the club at a speed of 46.0 m/s. What is the magnitude of the average force (in N) exerted on the ball by the club?

Answers

To solve this problem it is necessary to apply the concepts related to Newton's second law and the equations of motion description for acceleration.

From the perspective of acceleration we have to describe it as

a = (\Delta v)/(\Delta t)

Where,

\Delta v = Velocity

\Delta t= time

At the same time by the Newton's second law we have that

F = ma

Where,

m = mass

a = Acceleration

Replacing the value of acceleration we have

F = m ((\Delta v)/(\Delta t))

Our values are given as,

m = 55*10^(-3)Kg

v = 46m/s

t = 0.00340s

Replacing we have,

F = m ((\Delta v)/(\Delta t))

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F = 744.11N

Therefore the magnitude of the average force exerted on the ball by the club is 744.11N

____ can be calculated if you know the distance that an object travels in one unit of time. A.motion
B.meter
C.Rate
D.Speed
E.velocity
F.slope
G.refrence point

PLS HELP NOW !!!

Answers

Speed can be calculated if you know the distance that an object travels in one unit of time, therefore the correct answer is option D.

What is speed?

The total distance covered by any object per unit of time is known as speed. It depends only on the magnitude of the moving object.

The unit of speed is a meter/second. The generally considered unit for speed is a meter per second.

Thus, Speed can be calculated if you know the distance that an object travels in one unit of time, therefore the correct answer is option D.

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Answer:

D.Speed

Explanation:

The speed of an object is the distance the object travels in one unit of time.

A force of 40 N is applied in a direction perpendicular to the end of a 9 m long bar that pivots about its other end. Find the torque that this force produces about the pivot point. magnitude

Answers

Answer:

360 Nm

Explanation:

Torque: This is the force that tend to cause a body to rotate or twist. The S.I unit of torque is Newton- meter (Nm).

From the question,

The expression of torque is given as

τ = F×d......................... Equation 1

Where, τ = Torque, F = force, d = distance of the bar perpendicular to the force.

Given: F = 40 N, d = 9 m

Substitute into equation 1

τ  = 40(9)

τ  = 360 Nm.

Answer:

360Nm

Explanation:

Torque is defined as the rotational effect of a force. The magnitude of a torque τ, is given by;

τ = r F sin θ

Where;

r = distance from the pivot point to the point where the force is applied

F = magnitude of the force applied

θ = the angle between the force and the vector directed from the point of application to the pivot point.

From the question;

r = 9m

F = 40N

θ = 90° (since the force is applied perpendicular to the end of the bar)

Substitute these values into equation (i) as follows;

τ = 9 x 40 sin 90°

τ = 360Nm

Therefore the torque is 360Nm

Energy is the capacity to do work, but not to produce heat

Answers

Energy and Work Energy is the capacity to do work or to produce heat. Internal energy is the sum of kinetic energy and the potential energy. ... The KE would increase because heating something causes an increase in temperature.

Final answer:

Energy is the capacity to do work but not to produce heat. In physics, energy can exist in various forms, including mechanical and thermal energy.

Explanation:

Energy is the capacity to do work and is an important concept in physics. In the context of this question, it is stated that energy is the capacity to do work but not to produce heat. This highlights the distinction between the two forms of energy. For example, mechanical energy can be used to perform work on an object and cause it to move, while thermal energy is associated with heat and not directly related to work. However, it's important to note that energy can be converted from one form to another, such as converting mechanical energy to thermal energy in a friction process.

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A 30 kg child on a 2 m long swing is released from rest when the swing supports make an angle of 34 ◦ with the vertical. The acceleration of gravity is 9.8 m/s 2 . If the speed of the child at the lowest position is 2.31547 m/s, what is the mechanical energy dissipated by the various resistive

Answers

Answer:

Energy dissipated = 13.453 Joules

Explanation:

In order to solve this problem, we first compute the gravitational potential energy the child has, and then find the kinetic energy at the lowest position.

The gravitational potential energy (relative to lowest position) is found as follows:

G.P.E = mass * gravity * height

Where,  Height = 2 - 2 * Cos(34°)

Height = 0.3193 meters

G.P.E = 30 * 9.8 * 0.3193

G.P.E = 93.874 J

Kinetic energy:

K.E = 0.5 * mass * velocity^2

K.E = 0.5 * 30 * 2.31547^2

K.E = 80.421 J

Energy dissipated = G.P.E - K.E

Energy dissipated = 93.874 - 80.421

Energy dissipated = 13.453 J

A 2.13-kg object on a frictionless horizontal track is attached to the end of a horizontal spring whose force constant is 5.00 N/m. The object is displaced 3.54 m to the right from its equilibrium position and then released, initiating simple harmonic motion. (a) What is the force (magnitude and direction) acting on the object 3.50 s after it is released

Answers

Answer:

17.54N in -x direction.

Explanation:

Amplitude (A) = 3.54m

Force constant (k) = 5N/m

Mass (m) = 2.13kg

Angular frequency ω = √(k/m)

ω = √(5/2.13)

ω = 1.53 rad/s

The force acting on the object F(t) = ?

F(t) = -mAω²cos(ωt)

F(t) = -2.13 * 3.54 * (1.53)² * cos (1.53 * 3.50)

F(t) = -17.65 * cos (5.355)

F(t) = -17.57N

The force is 17.57 in -x direction