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nordsb [41]
3 years ago
11

9. Does a blackbody radiator emit light waves? Explain.

Physics
1 answer:
Yakvenalex [24]3 years ago
4 0

Answer:

"A blackbody is an object that absorbs all of the radiation that it receives(that is,it does not reflect any light,nor does it allow any light to pass through it and out the other side).The energy that the blackbody absorbs heats it up,and then it will emit its own radiation."

Explanation:

Risk*

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When does the transfer of thermal energy end?​
Dahasolnce [82]

Answer:

I think when the object transferring the thermal energy reaches the same temp as the object absorbing it

Explanation:

6 0
3 years ago
Find the angle for the third-order maximum for 591 nm wavelength light falling on a diffraction grating having 1460 lines per ce
Marat540 [252]

Answer:

15.32°

Explanation:

We have given the wavelength \lambda =591nm=591\times 10^{-9}m

Diffraction grating is 1460 lines per cm

So  d=\frac{10^{-2}}{1460}=6.71\times 10^{-6}m (as 1 m=100 cm )

For maximum diffraction

dsin\Theta =m\lambda here m is order of diffraction

So 6.71\times 10^{-6}sin\Theta =3\times 591\times 10^{-9}

sin\Theta =0.264

\Theta =15.32^{\circ}

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3 years ago
How does frequency relate to pitch?
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The sensation of a frequency is commonly referred to as the pitch of a sound. A high pitch sound corresponds to a high frequency sound wave and a low pitch sound corresponds to a low frequency sound wave. ... That is, two sound waves sound good when played together if one sound has twice the frequency of the other.
7 0
3 years ago
In a fast-pitch softball game the pitcher is impressive to watch, as she delivers a pitch by rapidly whirling her arm around so
svetlana [45]

Answer:

(a). The magnitude of the total acceleration of the ball is 239.97 m/s².

(b). The angle of the total acceleration relative to the radial direction is 11.0°

Explanation:

Given that,

Radius of the circle = 0.681 m

Angular acceleration = 67.7 rad/s²

Angular speed =18.6 rad/s

We need to calculate the centripetal acceleration of the ball

Using formula of centripetal acceleration

a_{c}=\omega^2\times r

Put the value into the formula

a_{c}=(18.6)^2\times0.681

a_{c}=235.5\ m/s^2

We need to calculate the tangential acceleration of the ball

Using formula of tangential acceleration

a_{t}=r\alpha

Put the value into the formula

a_{t}=0.681\times67.7

a_{t}=46.104\ m/s^2

(a). We need to calculate the magnitude of the total acceleration of the ball

Using formula of total acceleration

a=\sqrt{a_{c}^2+a_{t}^2}

Put the value into the formula

a=\sqrt{(235.5)^2+(46.104)^2}

a=239.97\ m/s^2

(b). We need to calculate the angle of the total acceleration relative to the radial direction

Using formula of the direction

\theat=\tan^{-1}(\dfrac{a_{t}}{a_{c}})

Put the value into the formula

\theta=\tan^{-1}(\dfrac{46.104}{235.5})

\theta=11.0^{\circ}

Hence, (a). The magnitude of the total acceleration of the ball is 239.97 m/s².

(b). The angle of the total acceleration relative to the radial direction is 11.0°

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3 years ago
A car travels 45 km due north and 70 km west. What is the car's displacement?
Elis [28]

Answer:

A

Explanation:

7 0
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