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bonufazy [111]
3 years ago
8

A 4.00-kg cylinder, of length 21.0 cm and diameter 12.0 cm , about the central axis of the cylinder, if the cylinder is thin-wal

led a
Physics
1 answer:
insens350 [35]3 years ago
3 0
We are given
m = 4 kg
L = 21 cm
D = 12 cm
Thin-walled cylinder

I think what is asked is the moment of inertia of the cylinder about its axis. The formula is
I = MR²
substituting the given values
I = 4 (12/2)²
I = 144 kg cm²
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Suppose you need your silicon circuit element to run continuously for 3 minutes before it shuts off long enough to cool back dow
Zigmanuir [339]

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MER=5.044 \times 10^{-4} \mathrm{~J} / \mathrm{sec}

<h3>What is the maximum rate at which energy can be added to the circuit element?</h3>

Generally, the equation for P is  mathematically given as

P=\ln s \frac{\Delta T}{\Delta t}

Therefore

Rate\ of\ Change\ of\ Temp =\frac{p}{lnS}

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3 0
1 year ago
How many diffraction maxima are contained in a region of the Fraunhofer single-slit pattern, subtending an angle of 2.12°, for a
Luba_88 [7]

Answer:

6

Explanation:

We are given that

\theta=2.12^{\circ}

Slid width,a=0.110 mm=0.11\times 10^{-3} m

1mm=10^{-3} m

Wavelength,\lambda=582 nm=582\times 10^{-9} m

1nm=10^{-9} m

We have to find the number of diffraction maxima are contained in a region of the Fraunhofer single-slit pattern.

asin\theta=\frac{2N+1}{2}\lambda

Using the formula

0.11\times 10^{-3}sin(2.12)=\frac{2N+1}{2}(582\times 10^{-9})

2N+1=\frac{0.11\times 10^{-3}sin(2.12)\times 2}{582\times 10^{-9}}

2N+1=13.98

2N=13.98-1=12.98

N=\frac{12.98}{2}\approx 6

Hence, 6 diffraction maxima are contained in a region of the Fraunhofer single-slit pattern

5 0
3 years ago
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