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3 years ago

Help ASAP 15 points

Mathematics

1 answer:

zhuklara [117]3 years ago

3 0

Answer:

Step-by-step explanation:

a) Let's $x$ the length of the side of the small square.

Thus,

$x^{2} + (\sqrt{40})^2 = (\sqrt{50})^2$

$x^{2} + 40 = 50$

$x^{2} = 50 - 40$

$x^{2} = 10$

$x = \sqrt{10}$

$x=3.16$

b) Now, the area of the small square is

$A= x^{2}$

$A=(3.16)^2$

$A=9.99$

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If you project S onto the (x,y)-plane, it casts a "shadow" corresponding to the trapezoidal region

T = {(x,y) : 0 ≤ x ≤ 10 - y and -4 ≤ y ≤ 4}

Let z = f(x, y) = √(16 - y²) and z = g(x, y) = -√(16 - y²), each referring to one half of the cylinder to either side of the plane z = 0.

The surface element for the "positive" half is

dS = √(1 + (∂f/∂x)² + (∂f/dy)²) dx dy

dS = √(1 + 0 + 4y²/(16 - y²)) dx dy

dS = √((16 + 3y²)/(16 - y²)) dx dy

The the surface integral along this half is

$\displaystyle \iint_T xz \,dS = \int_{-4}^4 \int_0^{10-y} x \sqrt{16-y^2} \sqrt{\frac{16+3y^2}{16-y^2}} \, dx \, dy$

$\displaystyle \iint_T xz \,dS = \int_{-4}^4 \int_0^{10-y} x \sqrt{16+3y^2}\, dx \, dy$

$\displaystyle \iint_T xz \,dS = \frac12 \int_{-4}^4 (10-y)^2 \sqrt{16+3y^2} \, dy$

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You'll find that the integral over the "negative" half has the same value, but multiplied by -1. Then the overall surface integral is 0.

8 0

3 years ago

<img src="https://tex.z-dn.net/?f=%28a%2Bb%29%5E%7B2%7D" id="TexFormula1" title="(a+b)^{2}" alt="(a+b)^{2}" align="absmiddle" cl

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Answer:

$({a + b})^{2}$

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${a}^{2} + 2ab + {b}^{2}$

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nirvana33 [79]

Answer:

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Andre45 [30]

Answer:

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Step-by-step explanation:

We know that the formula fro deflection is:

δ = PL/AE + LαΔT

As deflection is equal to zero in this case, the formula becomes

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we see that only vertical force acting is weight

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W = 981N = ∑ $F_{y}$ where 1N = 1kgm/s²

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Now by substituting all these values in equation 1, wee get

ΔT = -P/αAE = - 588.6/(20*10⁻⁶*60*10⁻⁶*105*10⁹)

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3 years ago