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gladu [14]
3 years ago
10

Match the following ERA’s Question 4

Physics
1 answer:
inna [77]3 years ago
8 0
1. Cenozoic ERA
2. Mesozoic ERA
3. Paleozoic ERA
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In a double-slit experiment, the third-order maximum for light of wavelength 510 nm is located 17 mm from the central bright spo
Marysya12 [62]

Answer:

14.9 mm

Explanation:

We know dsinθ = mλ where d = separating of slit, m = order of maximum = 3 and λ = wavelength = 510nm = 510 × 10⁻⁹ m

Also tanθ = L/D where L = distance of m order fringe from central bright spot = 17 mm = 0.017 m and D = distance of screen from slit = 1.6 m

So, sinθ = mλ/d

Since θ is small, sinθ ≅ tanθ

So,

mλ/d = L/D

d = mλD/L

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

d = 3 × 510 × 10⁻⁹ m × 1.6 m/0.017 m

d = 2448 × 10⁻⁹ m²/0.017 m

d = 144000 × 10⁻⁹ m

d = 1.44 × 10⁻⁴ m

d = 0.144 × 10⁻³ m

d = 0.144 mm

Now, for the second-order maximum, m' of the 670 nm wavelength of light,

m'λ'/d = L'/D where m' = order of maximum = 2, λ' = wavelength of light = 670 nm = 670 × 10⁻⁹ m, d = slit separation = 0.144 mm = 0.144 × 10⁻³ m, L' = distance of second order maximum from central bright spot and D = distance of screen from slit = 1.6m

So, L' = m'λ'D/d

So, substituting the values of the variables into the equation, we have

L' = 2 × 670 × 10⁻⁹ m × 1.6 m/0.144 × 10⁻³ m

L' = 2144  × 10⁻⁹ m²/0.144 × 10⁻³ m

L' = 14888.89 × 10⁻⁶ m

L' = 0.01488 m

L' ≅ 0.0149 m

L' = 14.9 mm

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3 years ago
How efficient is an incandescent light bulb?
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As heaters, they are somewhere around 80% - 85% efficient.
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The magnetic field of a solenoid depends upon n, which represents select one:
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The number of loops
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A simple accelerometer is constructed inside a car by suspending an object of mass m from a string of length L that is tied to t
Fittoniya [83]

Answer:

Part a)

\theta = tan^{-1}\frac{a}{g}

so here the angle made by the string is independent of the mass

Part b)

a = 4.16 m/s^2

Explanation:

Part a)

Let the string makes some angle with the vertical so we have force equation given as

Tcos\theta = mg

T sin\theta = ma

so we will have

tan\theta = \frac{ma}{mg}

\theta = tan^{-1}\frac{a}{g}

so here the angle made by the string is independent of the mass

Part b)

Now from above equation if we know that angle made by the string is

\theta = 23 degree

so we will have

tan23 = \frac{a}{g}

a = g tan23

a = 9.81(tan23)

a = 4.16 m/s^2

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