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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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2 years ago

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3 years ago

At a certain instant a particle is moving in the +x direction with momentum +8 kg m/s. During the next 0.13 seconds a constant f

jeka94

Answer:

The momentum of the particle at the end of the 0.13 s time interval is 7.12 kg m/s

Explanation:

The momentum of the particle is related to force by the following equation:

Δp = F · Δt

Where:

Δp = change in momentum = final momentum - initial momentum

F = constant force.

Δt = time interval.

Let´s calculate the x-component of the momentum after the 0.13 s:

final momentum - 8 kg m/s = -7 N · 0.13 s

final momentum = -7 kg m/s² · 0.13 s + 8 kg m/s

final momentum = 7.09 kg m/s

Now let´s calculate the y-component of the momentum vector after the 0.13 s. Since the particle wasn´t moving in the y-direction, the initial momentum in this direction is zero:

final momentum = 5 kg m/s² · 0.13 s

final momentum = 0.65 kg m/s

Then, the mometum vector will be as follows:

p = (7.09 kg m/s, 0.65 kg m/s)

The magnitude of this vector is calculated as follows:

$|p| = \sqrt{(7.09 kg m/s)^{2} + (0.65 kg m/s)^{2}} = 7.12 kg m/s$

The momentum of the particle at the end of the 0.13 s time interval is 7.12 kg m/s

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3 years ago