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3 years ago

What is calculated by speed of light/wavelength?

Physics

1 answer:

zhenek [66]3 years ago

6 0

Frequency of oscillation = speed of light/wave length

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What is the formula for calculating the efficiency of a heat engine?

Anton [14]

Answer:

Efficiency = StartFraction T Subscript h Baseline minus T Subscript C Baseline over T Subscript h Baseline EndFraction times 100. Efficiency equals T Subscript c Baseline minus T Subscript h Baseline over T Subscript h Baseline all times 100.

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3 years ago

Read 2 more answers

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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2 years ago

A proton is released from rest at the positive plate of a parallelplatecapacitor. It crosses the capacitor and reaches the negat

Triss [41]

Answer:

2.1406 × $10^6$ m/sec

Explanation:

we know that energy is always conserved

so from the law of energy conservation

$qV=\frac{1}{2}mv^2$

here V is the potential difference

we know that mass of proton = 1.67× $10^{-27}$ kg

we have given speed =50000m/sec

so potential difference $V=\frac{\frac{1}{2}\times 1.67\times 10^{-27}50000^2}{1.6\times 10^{-19}}=13.045$

now mass of electron =9.11× $10^{-31}$

so for electron

$\frac{1}{2}\times 9.11\times 10^{-31}v^2=1.6\times 10^{-19}\times 13.045=2.1406\times 10^6 m/sec$

so the velocity of electron will be 2.1406× $10^6$ m/sec

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3 years ago

In an application, Germanium is

Zigmanuir [339]

Answer:

produce electronics

Explanation:

The uses of Germanium are recorded beneath: Germanium's principle use is to deliver strong state hardware, semiconductors and fiber optic frameworks. As a phosphor in fluorescent lights.

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3 years ago

Observe the given figure and find the the gravitational force between m1 and m2.

Leno4ka [110]

Answer:

The gravitational force between m₁ and m₂, is approximately 1.06789 × 10⁻⁶ N

Explanation:

The details of the given masses having gravitational attractive force between them are;

m₁ = 20 kg, r₁ = 10 cm = 0.1 m, m₂ = 50 kg, and r₂ = 15 cm = 0.15 m

The gravitational force between m₁ and m₂ is given by Newton's Law of gravitation as follows;

$F =G \cdot \dfrac{m_{1} \cdot m_{2}}{r^{2}}$

Where;

F = The gravitational force between m₁ and m₂

G = The universal gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²

r₂ = 0.1 m + 0.15 m = 0.25 m

Therefore, we have;

$F = 6.67430 \times 10^{-11} \ N \cdot m^2/kg \times \dfrac{20 \ kg\times 50 \ kg}{(0.1 \ m+ 0.15 \ m)^{2}} \approx 1.06789 \times 10^{-6} \ N$

The gravitational force between m₁ and m₂, F ≈ 1.06789 × 10⁻⁶ N

8 0

2 years ago