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maria [59]
3 years ago
7

Find the mass of the wire formed by the intersection of the sphere x 2 + y 2 + z 2 = 1 and the plane x + z = 0 if the density of

the wire is 4x 2 grams per unit length.
Mathematics
1 answer:
Marat540 [252]3 years ago
6 0
<span>Answer: x²+(â’xâ’z)²+z² = 1 → 2x²+2zx+(2z²â’1) = 0 Solve as a quadratic : x = { â’2z ± âš(4z²â’8(2z²â’1) } / 4 = { â’z ± âš(2â’3z²) } / 2 A more manageable form can be derived by setting z = âš(2/3).sint to remove âš This gives x = â’âš(1/6).sint + âš(1/2).cost (can omit ± sign since â’ is covered by Ď€â’t) Using x+y+z = 0 gives y = â’xâ’z = â’âš(1/6)sint â’ âš(1/2).cost So a parameterisation in terms of t â [0,2Ď€] is x = â’âš(1/6).sint + âš(1/2).cost, y = â’âš(1/6).sint â’ âš(1/2).cost, z = âš(2/3).sint (ds/dt)² = (â’âš(1/6).costâ’âš(1/2).sint)² + (â’âš(1/6).cost+âš(1/2).sint)² + (âš(2/3).cost)² = 1 â´ â« Ďds = â« ( â’âš(1/6).sint + âš(1/2).cost )²dt, [t=0,2Ď€] = 2Ď€/3</span>
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The value of pressure at an altitude of 10000 ft = 10 \frac{lb}{in^{2} }

Step-by-step explanation:

Given data

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Pressure at 4000 ft = 12.6 \frac{lb}{in^{2} }

If temperature is constant then the atmospheric pressure is varies with the altitude according to law

P (h) = P_{0} e^{- kh} ------ (1)

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12.6 = 14.7 e^{- 4000k}

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㏑ 0.857 = - 4000 k

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Thus the atmospheric pressure at an altitude of 10,000 ft is

P_{10000} = 14.7 × e^{- kh} ----- (2)

Product of k & h is

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k h = 0.385

Put his value of k h = 0.385  in equation (2) we get

P_{10000} = 14.7 × e^{-0.385 }

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7 0
3 years ago
A small company 's net income for the first 6 months of the year was 76.500 and for the last six months it was 100,000 . What is
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Answer:The ratio of net income in the first 6 months, to the last six months is $76,500 / $100,000. This simplifies intuitively as follows:

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The denominator 200 is only divisible by the prime numbers 2 and 5, and since the numerator 153 is not divisible by either 2 or 5, this means that this is in simplest form, and the final answer is 153/200.

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2 years ago
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