All categories

3 years ago

How is the thermal energy of the radiator transferred to the surrounding air

1 answer:

6 0

Radiant energy. The heat in the radiator does not transfer from the radiator by convection because the heat in the radiator is inside the radiator and not a part of the surrounding air. There is a casing surrounding that heat source; the air cannot get to that heat source directly.

Here's what really happens. Radiant energy from the radiator honeycomb heats the surrounding air. And the heated surrounding air then carries off that heat by convection. My point is this... thermal energy is transferred from the radiator to the surrounding air as radiant energy. Only after that energy is in the air does convection play a role in the heat transfer.
oldprof

You might be interested in

A hydrogen discharge lamp emits light with two prominent wavelengths: 656 nm (red) and 486 nm (blue). The light enters a flint-g

mezya [45]

Answer:

The angle between the red and blue light is 1.7°.

Explanation:

Given that,

Wavelength of red = 656 nm

Wavelength of blue = 486 nm

Angle = 37°

Suppose we need to find the angle between the red and blue light as it leaves the prism

$n_{r}=1.572$

$n_{b}=1.587$

We need to calculate the angle for red wavelength

Using Snell's law,

$n_{r}\sin\theta_{i}=n_{a}\sin\theta_{r}$

Put the value into the formula

$1.572\sin37=1\times\sin\theta_{r}$

$\theta_{r}=\sin^{-1}(\dfrac{1.572\sin37}{1})$

$\theta_{r}=71.0^{\circ}$

We need to calculate the angle for blue wavelength

Using Snell's law,

$n_{b}\sin\theta_{i}=n_{a}\sin\theta_{b}$

Put the value into the formula

$1.587\sin37=1\times\sin\theta_{b}$

$\theta_{b}=\sin^{-1}(\dfrac{1.587\sin37}{1})$

$\theta_{b}=72.7^{\circ}$

We need to calculate the angle between the red and blue light

Using formula of angle

$\Delta \theta=\theta_{b}-\theta_{r}$

Put the value into the formula

$\Delta \theta=72.7-71.0$

$\Delta \theta=1.7^{\circ}$

Hence, The angle between the red and blue light is 1.7°.

8 0

3 years ago

The energy of a photon was found to be 3.38 x 10-19 J. Planck's constant is 6.63 x 10-34 J. S. Which color of light corresponds

GalinKa [24]

Answer:

The correct option is (c).

Explanation:

Given that,

The energy of a photon is, $E=3.38\times 10^{-19}\ J$

We need to tell the color of this light. We know that, the energy of a photon is given by :

$E=\dfrac{hc}{\lambda}$

Where

c is the speed of light

$\lambda=\dfrac{hc}{E}\\\\\lambda=\dfrac{6.63\times 10^{-34}\times 3\times 10^8}{3.38 \times 10^{-19}}\\\\\lambda=5.88\times 10^{-7}\ m\\\\\lambda=588\ nm$

The wavelength of yellow light is approx 580 nm. Hence, we can say that this photon corresponds to yellow light.

6 1

3 years ago

A uniform magnetic field passes through a horizontal circular wire loop at an angle 19.5 ∘ from the vertical. The magnitude of t

nlexa [21]

To solve this problem, we will apply the concepts related to Faraday's law that describes the behavior of the emf induced in the loop. Remember that this can be expressed as the product between the number of loops and the variation of the magnetic flux per unit of time. At the same time the magnetic flux through a loop of cross sectional area is,

$\Phi = BA Cos \theta$

Here,

$\theta$ = Angle between areal vector and magnetic field direction.

According to Faraday's law, induced emf in the loop is,

$\epsilon= -N \frac{d\Phi }{dt}$

$\epsilon = -N \frac{(BAcos\theta)}{dt}$

$\epsilon = -NAcos\theta \frac{dB}{dt}$

$\epsilon = -N\pi r^2 cos\theta \frac{d}{dt} ( ( 3.75 T ) + ( 3.05T/s ) t + ( -6.95 T/s^2 ) t^2)$

$\epsilon = -N\pi r^2 cos\theta( (3.05T/s)-(13.9T/s)t )$

At time $t = 5.71s$ , Induced emf is,

$\epsilon = -(1) \pi (0.220m)^2 cos(19.5\°)( (3.05T/s)-(13.9T/s)(5.71s))$

$\epsilon = 10.9V$

Therefore the magnitude of the induced emf is 10.9V

4 0

3 years ago

Read 2 more answers

A 13-kg sled is moving at a speed of 3.0 m/s. At which of the following speeds will the sled have twice as much kinetic energy?

AysviL [449]

K.E. = 1/2 mv²
K.E. is directly proportional to v^2
So, when K.E. increase by 2, K.E. increase by root. 2
v' = 1.41v
original v value was 3 so, final would be:
v' = 1.41*3 = 4.23
After round-off to it's tenth value, it will be:
v' = 4.2

So, option B is your answer!

Hope this helps!

7 0

3 years ago

are the properties of a specific element the same as the properties of a molecule made up of that element ? why or why not

konstantin123 [22]

Because it’s science

8 0

3 years ago