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CaHeK987 [17]
3 years ago
12

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 _mT = 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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A jet engine gets its thrust by taking in air, heating and compressing it, and
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Answer:

F=ma=20\ Kg\ 400\ m/s^2=8,000\ Nw

Explanation:

Thrust is known as a reaction force which appears when a system expels or accelerates mass in one specific direction. If we know the acceleration and the mass of the air expelled by the jet engine, we can compute the thrust .

The acceleration is calculated by using the dynamics formula

\displaystyle a=\frac{v_f-v_o}{t}

The values are  

v_f=500\ m/s,\ v_o=100\ m/s,\ t=1\ sec

\displaystyle a=\frac{500-100}{1}=400\ m/s^2

The thrust is

F=ma=20\ Kg\ 400\ m/s^2=8,000\ Nw

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Which of the following correctly describe electric field lines?
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-- Electric field lines DO never cross.  <em>(A) </em>

-- Electric field lines that are close together DO indicate a stronger electric field. <em>(B) </em>

-- Electric field lines DO not affect the charge that created them.  <em>(C)</em>

-- Electric field lines DON'T begin on north poles and end on south poles.  North and South "poles" are the way we talk about magnets, not electric charges.

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A conveyer belt moves a 40 kg box at a velocity of 2 m/s. What is the kinetic energy of the box while it is on the conveyor belt
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What is the potential difference across a parallel-plate capacitor whose plates are separated by a distance of 4.0 mm where each
suter [353]

The potential difference across the parallel plate capacitor is 2.26 millivolts

<h3>Capacitance of a parallel plate capacitor</h3>

The capacitance of the parallel plate capacitor is given by C = ε₀A/d where

  • ε₀ = permittivity of free space = 8.854 × 10⁻¹² F/m,
  • A = area of plates and
  • d = distance between plates = 4.0 mm = 4.0 × 10⁻³ m.

<h3>Charge on plates</h3>

Also, the surface charge on the capacitor Q = σA where

  • σ = charge density = 5.0 pC/m² = 5.0 × 10⁻¹² C/m² and
  • a = area of plates.

<h3>The potential difference across the parallel plate capacitor</h3>

The potential difference across the parallel plate capacitor is V = Q/C

= σA ÷ ε₀A/d

= σd/ε₀

Substituting the values of the variables into the equation, we have

V = σd/ε₀

V = 5.0 × 10⁻¹² C/m² × 4.0 × 10⁻³ m/8.854 × 10⁻¹² F/m

V = 20.0 C/m × 10⁻³/8.854 F/m

V = 2.26 × 10⁻³ Volts

V = 2.26 millivolts

So, the potential difference across the parallel plate capacitor is 2.26 millivolts

Learn more about potential difference across parallel plate capacitor here:

brainly.com/question/12993474

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