All categories

3 years ago

Why does the chord of an electric heater not glow while the heating element does

Physics

1 answer:

8_murik_8 [283]3 years ago

4 0

Answer:

The heating element of the heater is made up of alloy which has very high resistance so when current flows through the heating element, it becomes too hot and glows red. But the resistance of cord which is usually of copper or aluminum is very low so it does not glow.

You might be interested in

You are designing a flywheel. It is to start from rest and then rotate with a constant angular acceleration of 0.200 rev/s^2. Th

Rama09 [41]

Answer:

I = 8.75 kg m

Explanation:

This is a rotational movement exercise, let's start with kinetic energy

K = ½ I w²

They tell us that K = 330 J, let's find the angular velocity with kinematics

w² = w₀² + 2 α θ

as part of rest w₀ = 0

w = √ 2α θ

let's reduce the revolutions to the SI system

θ = 30.0 rev (2π rad / 1 rev) = 60π rad

let's calculate the angular velocity

w = √(2 0.200 60π)

w = 8.683 rad / s

we clear from the first equation

I = 2K / w²

let's calculate

I = 2 330 / 8,683²

I = 8.75 kg m

4 0

3 years ago

The normal eye, myopic eye and old age

yanalaym [24]

Answer:

1) f’₀ / f = 1.10, the relationship between the focal length (f'₀) and the distance to the retina (image) is given by the constructor's equation

2) the two diameters have the same order of magnitude and are very close to each other

Explanation:

You have some problems in the writing of your exercise, we will try to answer.

1) The equation to be used in geometric optics is the constructor equation

$\frac{1}{f} = \frac{1}{p} + \frac{1}{q}$

where p and q are the distance to the object and the image, respectively, f is the focal length

* For the normal eye and with presbyopia

the object is at infinity (p = inf) and the image is on the retina (q = 15 mm = 1.5 cm)

$\frac{1}{f'_o} = 1/ inf + \frac{1}{1.5}$

f'₀ = 1.5 cm

this is the focal length for this type of eye

* Eye with myopia

the distance to the object is p = 15 cm the distance to the image that is on the retina is q = 1.5 cm

1 / f = 1/15 + 1 / 1.5

1 / f = 0.733

f = 1.36 cm

this is the focal length for the myopic eye.

In general, the two focal lengths are related

f’₀ / f = 1.5 / 1.36

f’₀ / f = 1.10

The question of the relationship between the focal length (f'₀) and the distance to the retina (image) is given by the constructor's equation

2) For this second part we have a diffraction problem, the point diameter corresponds to the first zero of the diffraction pattern that is given by the expression for a linear slit

a sin θ= m λ

the first zero occurs for m = 1, as the angles are very small

tan θ = y / f = sin θ / cos θ

for some very small the cosine is 1

sin θ = y / f

where f is the distance of the lens (eye)

y / f = lam / a

in the case of the eye we have a circular slit, therefore the system must be solved in polar coordinates, giving a numerical factor

y / f = 1.22 λ / D

y = 1.22 λ f / D

where D is the diameter of the eye

D = 2R₀

D = 2 0.1

D = 0.2 cm

the eye has its highest sensitivity for lam = 550 10⁻⁹ m (green light), let's use this wavelength for the calculation

* normal eye

the focal length of the normal eye can be accommodated to give a focus on the immobile retian, so let's use the constructor equation

\frac{1}{f} = \frac{1}{p} + \frac{1}{q}

sustitute

$\frac{1}{f} = \frac{1}{25} + \frac{1}{1.5}$

$\frac{1}{f}$ = 0.7066

f = 1.415 cm

therefore the diffraction is

y = 1.22 550 10⁻⁹ 1.415 / 0.2

y = 4.75 10⁻⁶ m

this is the radius, the diffraction diameter is

d = 2y

d_normal = 9.49 10⁻⁶ m

* myopic eye

In the statement they indicate that the distance to the object is p = 15 cm, the retina is at the same distance, it does not move, q = 1.5 cm

$\frac{1}{f} = \frac{1}{15} + \frac{1}{ 1.5}$

$\frac{1}{f}$ = 0.733

f = 1.36 cm

diffraction is

y = 1.22 550 10-9 1.36 10-2 / 0.2 10--2

y = 4.56 10-6 m

the diffraction diameter is

d_myope = 2y

d_myope = 9.16 10-6 m

$\frac{d_{normal}}{d_{myope}}$ = 9.49 /9.16

\frac{d_{normal}}{d_{myope}} = 1.04

we can see that the two diameters have the same order of magnitude and are very close to each other

8 0

3 years ago

The current supplied by a battery as a function of time is I(t) = (0.88 A) e^(-t*6 hr). What is the total number of electrons tr

Degger [83]

Answer:

e. 1.2 x 10²³

Explanation:

According to the problem, The current equation is given by:

$I(t)=0.88e^{-t/6\times3600s}$

Here time is in seconds.

Consider at t=0 s the current starts to flow due to battery and the current stops when the time t tends to infinite.

The relation between current and number of charge carriers is:

$q=\int\limits {I} \, dt$

Here the limits of integration is from 0 to infinite. So,

$q=\int\limits {0.88e^{-t/6\times3600s}}\, dt$

$q=0.88\times(-6\times3600)(0-1)$

q = 1.90 x 10⁴ C

Consider N be the total number of charge carriers. So,

q = N e

Here e is electronic charge and its value is 1.69 x 10⁻¹⁹ C.

N = q/e

Substitute the suitable values in the above equation.

$N= \frac{1.9\times10^{4} }{1.69\times10^{-19}}$

N = 1.2 x 10²³

8 0

3 years ago

Why is the rock cycle really a cycle?

Naddik [55]

The rock cycle is a group of changes that describes the formation, breakdown, and reformation of a rock as a result of sedimentary, igneous, and metamorphic processes. All rocks are made up of minerals. Volcanic processes form extrusive igneous rocks. Extrusive rocks cool quickly on or very near the surface of the earth.

6 0

3 years ago

A 1500-kilogram car is at rest on a level track. The driver gives the engine gas and the car begins to accelerate forward. The c

Nutka1998 [239]

1. The acceleration of the car is 3.375 m/s².

2. The force moving the car is 5062.5 N

From the question given above, the following data were obtained:

Mass of car = 1500 Kg

Initial velocity (u) = 0 m/s

Final velocity (v) = 27 m/s

Time (t) = 8 s

<h3>Acceleration (a) =? </h3><h3>Force (F) =? </h3>

<h3>1. Determination of the acceleration </h3>

Initial velocity (u) = 0 m/s

Final velocity (v) = 27 m/s

Time (t) = 8 s

<h3>Acceleration (a) =?</h3>

$a = \frac{v - u}{t} \\\\a = \frac{27 - 0}{8} \\\\a = \frac{27}{8}\\$

Therefore, the acceleration of the car is 3.375 m/s²

<h3>2. Determination of the force moving the car.</h3>

Mass of car = 1500 Kg

Acceleration (a) = 3.375 m/s²

<h3>Force (F) =?</h3>

F = 1500 × 3.375

Therefore, the force moving the car is 5062.5 N

Learn more: brainly.com/question/13001406

8 0

3 years ago

Why does the chord of an electric heater not glow while the heating element does​

Why does the chord of an electric heater not glow while the heating element does