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2 years ago

9

An architect used this diagram to design a curved balcony. She drew a circle with a radius of 40 feet and a central angle of 70°

to determine the length of railing needed for the balcony.

1 answer:

Phantasy [73]2 years ago

3 0

49FT. LENGTH=ANGLE OVER 360 x 2pi

= 70 over 360 x 2pi (40)

= 48.9 = 49FT.

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The increase in length of an aluminum rod is twice the increase in length of an Invar rod with only a third of the temperature i

cluponka [151]

Answer:

the ratio of lengths of the two rods, Aluminum to Invar is 11.27

Step-by-step explanation:

coefficient of linear expansion of aluminum, $\alpha _{Al} = 23 \times 10^{-6} /K$

Coefficient of linear expansion of Invar, $\alpha _{Iv} = 1.2 \times 10^{-6}/K$

Linear thermal expansion is given as;

$\Delta L = L_0 \times \alpha\times \Delta T\\\\where;\\\\L_0 \ is \ the \ original \ length \ of \ the \ metal\\\\\Delta L \ is \ the \ increase \ in \ length$

The increase in length of Invar is given as;

$\Delta L_{Iv} = L_0_{Iv} \times \alpha _{Iv}\times \Delta T_{Iv}$

The increase in length of the Aluminum;

$\Delta L_{ Al} = L_0_{Al} \times \alpha _{Al} \times \Delta T_{Al}\\\\from \ the\ given \ question, \ the \ relationship \ between \ the \ rods \ is \ given \ as\\\\ L_0_{Al} \times \alpha _{Al} \times \frac{1}{3} \Delta T_{Iv}= 2( L_0_{Iv} \times \alpha _{Iv} \times \Delta T_{Iv})\\\\ L_0_{Al} \times \alpha _{Al} \times \Delta T_{Iv}= 6( L_0_{Iv} \times \alpha _{Iv} \times \Delta T_{Iv})\\\\ L_0_{Al} \times \alpha _{Al} \times \Delta T_{Iv} = 6L_0_{Iv} \times 6\alpha _{Iv} \times 6 \Delta T_{Iv}\\\\$

$\frac{L_0_{Al}}{6L_0_{Iv} } = \frac{6\alpha _{Iv} \ \times \ 6 \Delta T_{Iv}}{\alpha _{Al} \ \times \ \Delta T_{Iv}} \\\\\frac{L_0_{Al}}{6L_0_{Iv} } = \frac{6\alpha _{Iv} \ \times \ 6}{\alpha _{Al} \ } \\\\\frac{L_0_{Al}}{L_0_{Iv} } = 6^3(\frac{\alpha _{Iv} }{\alpha _{Al} } )\\\\\frac{L_0_{Al}}{L_0_{Iv} } = 6^3(\frac{1.2 \times 10^{-6} }{23\times 10^{-6} } )\\\\\frac{L_0_{Al}}{L_0_{Iv} } = 6^3(\frac{1.2}{23} )\\\\\frac{L_0_{Al}}{L_0_{Iv} } = \frac{259.2}{23} \\\\\frac{L_0_{Al}}{L_0_{Iv} } = 11.27$

Therefore, the ratio of lengths of the two rods, Aluminum to Invar is 11.27

5 0

3 years ago

Divide the following polynomial, then place the ans

Montano1993 [528]

Answer:

Step-by-step explanation:

5 0

3 years ago

What is the area of a sector with radius 6" and measure of arc equal to 120°?

fenix001 [56]

The area of a sector = 0.5 * radius^2 * central angle measure (in radians)

So that:

$Area_{sector}= \frac{1}{2} * 6^2* \frac{120}{180}*\pi=12\pi=37.69911
\\
Area_{sector}=37.7$ sq in

7 0

2 years ago

A 35 foot wire is secured from the top of a flagpole to a stake in the ground. If the stake is 14 feet from the base of the flag

cupoosta [38]

the answer is 21ft because the total is 35feet so 35 - 14 = 21

6 0

3 years ago

Read 2 more answers

R is the midpoint of QS, if QR=2x and RS=x+3, what is RS?

lbvjy [14]

Q -------------R------------S

suppose this is your line,

R is the mid point

and it's given that

QR = 2x
and
RS = x+3

as R is the mid point of QS
so,
QR = RS
then,
2x = x + 3
2x - x =3
x = 3

as,
it's given that RS = x + 3
then,
RS = 3 + 3
= 6 units.

5 0

3 years ago