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taurus [48]
3 years ago
7

hat are the wavelengths of peak intensity and the corresponding spectral regions for radiating objects at (a) normal human body

temperature of 37°C, (b) the temperature of the filament in an incandescent lamp, 1500°C, and (c) the temperature of the surface of the sun, 5800 K?
Physics
1 answer:
bagirrra123 [75]3 years ago
5 0

Answer:

(a)

\lambda _{m}=9.332 \times 10^{-6}m

(b)

\lambda _{m}=1.632 \times 10^{-6}m

(c) \lambda _{m}=4.988 \times 10^{-7}m

 

Explanation:

According to the Wein's displacement law

\lambda _{m}\times T = b

Where, T be the absolute temperature and b is the Wein's displacement constant.

b = 2.898 x 10^-3 m-K

(a) T = 37°C = 37 + 273 = 310 K

\lambda _{m}=\frac{b}{T}

\lambda _{m}=\frac{2.893\times 10^{-3}}{310}

\lambda _{m}=9.332 \times 10^{-6}m

(b) T = 1500°C = 1500 + 273 = 1773 K

\lambda _{m}=\frac{b}{T}

\lambda _{m}=\frac{2.893\times 10^{-3}}{1773}

\lambda _{m}=1.632 \times 10^{-6}m

(c) T = 5800 K

\lambda _{m}=\frac{b}{T}

\lambda _{m}=\frac{2.893\times 10^{-3}}{5800}

\lambda _{m}=4.988 \times 10^{-7}m

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two children sit on a merry go round. on sits 0.5 m from the center and the other sits 3 m from the center. Which child has grea
GuDViN [60]

i would say that the child with more linear speed is the cild that is 3 meters away from the center of the merry go round. because the child that is 0.5 meters from the center of the merry go round is less linear because the steering of the merry go round is started from the outer part of the merry go round so it would make more sense that the child that is 3 meters from the center of the merry go round would be more linear in speed.

hope this helps!

8 0
3 years ago
Select to show the energy of pendulum 1. Be sure that friction is set to none. Drag the pendulum to an angle (with respect to th
iragen [17]

Answer:

it have Potential energy

Explanation:

given data

Drag the pendulum to an angle 30∘

to find out

what form of energy does it have

solution

we know that pendulum start no kinetic energy when it release from any rest position then in starting it have potential energy only so that when pendulum is angle 30∘ at some height from ground so when it start it have potential energy same as in starting.

we know that the total energy is always conserve  

so it have potential energy

3 0
3 years ago
What results when energy is transformed while juggling three bowling pins?
Nady [450]

Answer:

his is an example of the transformation of gravitational potential energy into kinetic energy

Explanation:

The game of juggling bowling is a clear example of the conservation of mechanical energy,

when the bolus is in the upper part of the path mechanical energy is potential energy; As this energy descends, it becomes kinetic energy where the lowest part of the trajectory, just before touching the hand, is totally kinetic.

At the moment of touching the hand, a relationship is applied that reverses the value of the speed, that is, now it is ascending and the cycle repeats.

Therefore this is an example of the transformation of gravitational potential energy into kinetic energy

8 0
2 years ago
Suppose an empty grocery cart rolls downhill in a parking lot. The cart has a maximum speed of 1.3 m/s when it hits the side of
Ket [755]

The cart comes to rest from 1.3 m/s in a matter of 0.30 s, so it undergoes an acceleration <em>a</em> of

<em>a</em> = (0 - 1.3 m/s) / (0.30 s)

<em>a</em> ≈ -4.33 m/s²

This acceleration is applied by a force of -65 N, i.e. a force of 65 N that opposes the cart's motion downhill. So the cart has a mass <em>m</em> such that

-65 N = <em>m</em> (-4.33 m/s²)

<em>m</em> = 15 kg

4 0
3 years ago
Star X has an apparent magnitude of 1. Star Y has an apparent magnitude of 4. Both stars are in the same star cluster. Which sta
sashaice [31]

Answer:

Explanation:

From the given information:

Since both stars are in the same cluster, the magnitude and luminosity relationship can be calculated as:

m_1 - m_2 = -2.5 log _{10} (\dfrac{L_1}{L_2})

Given that;

m_1 = 1 and

m_2 = 4

Therefore,

1 - 4 = -2.5 log _{10} ( \dfrac{L_1}{L_2})

3 = -2.5 log _{10} ( \dfrac{L_1}{L_2})

Making \dfrac{L_1}{L_2} the subject of the formula:

\implies \dfrac{L_1}{L_2}= 10^{(\dfrac{3}{2.5})}

=15.84

≅ 16

Hence, we can conclude that star X is more luminous by a factor of 16

7 0
3 years ago
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