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alexira [117]
3 years ago
14

An old wheat-grinding wheel in a museum actually works. The sign on the wall says that the wheel has a rotational acceleration o

f 220 rad/s2 as its spinning rotational speed increases from zero to 1600 rpm.How long does it take the wheel to attain this rotational speed?
Physics
1 answer:
SIZIF [17.4K]3 years ago
5 0

Answer:

0.76 s

Explanation:

We use the formula for rotational speed, ω given its rotational acceleration α . ω = ω₀ + α(t - t₀) where ω₀ = initial rotational speed, ω = final rotational speed, ω₀ = initial rotational speed, α = rotational acceleration and t = initial time and t₀ = final time.

Since the wheel starts from rest, ω₀ = 0 and t₀ = 0. Given that α = 220 rad/s² and ω = 1600 rpm = 1600 × 2π/60 rad/s = 167.55 rad/s. Substituting these values into the equation ω = ω₀ + α(t - t₀)

167.55 = 0 + 220(t - 0) = 220t

So, t = 167.55/220 = 0.762 s ≅ 0.76 s

So, it takes the wheel 0.76 s to attain this rotational speed

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Answer:

the energy difference between adjacent levels decreases as the quantum number increases

Explanation:

The energy levels of the hydrogen atom are given by the following formula:

E=-E_0 \frac{1}{n^2}

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n is the level number

We can write therefore the energy difference between adjacent levels as

\Delta E=-13.6 eV (\frac{1}{n^2}-\frac{1}{(n+1)^2})

We see that this difference decreases as the level number (n) increases. For example, the difference between the levels n=1 and n=2 is

\Delta E=-13.6 eV(\frac{1}{1^2}-\frac{1}{2^2})=-13.6 eV(1-\frac{1}{4})=-13.6 eV(\frac{3}{4})=-10.2 eV

While the difference between the levels n=2 and n=3 is

\Delta E=-13.6 eV(\frac{1}{2^2}-\frac{1}{3^2})=-13.6 eV(\frac{1}{4}-\frac{1}{9})=-13.6 eV(\frac{5}{36})=-1.9 eV

And so on.

So, the energy difference between adjacent levels decreases as the quantum number increases.

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The magnitude of the Poynting vector of a planar electromagnetic wave has an average value of 0.724 W/m2. What is the maximum va
Hitman42 [59]

Answer:

7.78x10^-8T

Explanation:

The Pointing Vector S is

S = (1/μ0) E × B

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S = (1/μ0) E B

where S, E and B are magnitudes. The average value of the Pointing Vector is

<S> = [1/(2 μ0)] E0 B0

where E0 and B0 are amplitudes. (This can be derived by finding the rms value of a sinusoidal wave over an integer number of wavelengths.)

Also at any instant,

E = c B

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E0 = c B0

Substituting for E0,

<S> = [1/(2 μ0)] (c B0) B0 = [c/(2 μ0)] (B0)²

Solve for B0.

Bo = √ (0.724x2x4πx10^-7/ 3 x10^8)

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