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3 years ago

Assuming that the tungsten filament of a lightbulb is a blackbody, determine its peak wavelength if its temperature is 3 200 K.

Physics

1 answer:

rosijanka [135]3 years ago

6 0

Answer:

the peak wavelength when the temperature is 3200 K = $9.05625*10^{-7} \ m$

Explanation:

Given that:

the temperature = 3200 K

By applying Wien's displacement law ,we have

$\lambda _m$ T = 0.2898×10⁻² m.K

The peak wavelength of the emitted radiation at this temperature is given by

$\lambda _m$ = $\frac{0.2898*10^{-2} m.K}{3200 K}$

$\lambda _m$ = $9.05625*10^{-7} \ m$

Hence, the peak wavelength when the temperature is 3200 K = $9.05625*10^{-7} \ m$

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Currently, the Sun is thought to be located in the galactic _____________ about ____________ the center of the galaxy.

Ede4ka [16]

Answer:

a. Disk, 28 thousand light-years from

Explanation:

Since, a galactic disc is a component of disc galaxies, for instance spiral galaxies and lenticular galaxies. It consists of a stellar component and a gaseous component.

Also, the Sun lies within the galactic disk or in other words it is thought to be located in the galactic disk,

The Sun is located about 26,000 light-years away from the centre of the galaxy.

∵ From the given options 28,000 is nearest to 26,000

Hence, the sun is about 28 thousand light-years from the centre of the galaxy.

i.e. OPTION 'a' would be correct.

8 0

3 years ago

A wheel with moment of inertia 25 kg. m2 and angular velocity 10 rad/s begins to speed up, with angular acceleration 15 rad/sec2

Pani-rosa [81]

Answer:

(A) Angular speed 40 rad/sec

Rotation = 50 rad

(b) 37812.5 J

Explanation:

We have given moment of inertia of the wheel $I=25kgm^2$

Initial angular velocity of the wheel $\omega _0=10rad/sec$

Angular acceleration $\alpha =15rad/sec^2$

(a) We know that $\omega =\omega _0+\alpha t$

We have given t = 2 sec

So $\omega =10+15\times 2=40rad/sec$

Now $\Theta =\omega _0t+\frac{1}{2}\alpha t^2=10\times 2+\frac{1}{2}\times 15\times 2^2=50rad$

(b) After 3 sec $\omega =10+15\times 3=55rad/sec$

We know that kinetic energy is given by $Ke=\frac{1}{2}I\omega ^2=\frac{1}{2}\times 25\times 55^2=37812.5J$

7 0

3 years ago

the average american receives 2.28 mSv dose equivalent from radon each year. Assuming you receive this dose, and it all comes fr

Dmitry [639]

Answer:

The approximate number of decays this represent is $N= 23*10^{10}$

Explanation:

From the question we are told that

The amount of Radiation received by an average american is $I_a = 2.28 \ mSv$

The source of the radiation is $S = 5.49 MeV \ alpha \ particle$

Generally

$1 \ J/kg = 1000 mSv$

Therefore $2.28 \ mSv = \frac{2.28}{1000} = 2.28 *10^{-3} J/kg$

Also $1eV = 1.602 *10^{-19}J$

Therefore $2.28*10^{-3} \frac{J}{kg} = 2.28*10^{-3} \frac{J}{kg} * \frac{1ev}{1.602*10^{-19} J} = 1.43*10^{16} ev/kg$

An Average american weighs 88.7 kg

The total energy received is mathematically evaluated as

$1 kg ------> 1.423*10^{16}ev \\88.7kg --------> x$

Cross-multiplying and making x the subject

$x = 88.7 * 1.423*10^{16} eV$

$x = 126.2*10^{16}eV$

Therefore the total energy deposited is $x = 126.2*10^{16}eV$

The approximate number of decays this represent is mathematically evaluated as

N = $\frac{x}{S}$

Where n is the approximate number of decay

Substituting values

$N = \frac{126 .2*10^{16}}{5.49*10^6}$

$N= 23*10^{10}$

7 0

3 years ago

The ratio of output power to input power, in percent, is called?

Harrizon [31]

That's efficiency. There's no law that it must be stated in percent.

4 0

2 years ago

Read 2 more answers

An average hole drift velocity of 103 cm/sec results when 2 V is applied across a 1 cm long semiconductor bar. What is the hole

Helga [31]

Answer:

ε = 2 V/cm

Explanation:

To calculate the mobility inside this bar, we just need to apply the expression that let us determine the mobility. This expression is the following:

ε = ΔV / L

Where:

ε: Hole mobility inside the bar

ΔV: voltage applied in the bar

L: Length of the bar

We already have the voltage and the length so replacing in the above expression we have:

ε = 2 V / 1 cm

The data of the speed can be used for further calculations, but in this part its not necessary.

Hope this helps

8 0

2 years ago