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3 years ago

PLEASE ANSWER What is the area of minor sector AOB?

Mathematics

1 answer:

scoray [572]3 years ago

3 0

<h3>Answer: Approximately 13 square units (choice B)</h3>

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Explanation:

The given reflex angle is 215 degrees. A reflex angle is anything over 180 degrees, but less than 360. Subtract 215 from 360 to get the measure of angle AOB

angle AOB = 360 - 215 = 145

angle AOB = 145 degrees

We'll use this later.

Now find the area of the full circle. Use the formula A = pi*r^2. The radius is r = sqrt(10) which can be found through the distance formula or the pythagorean theorem. You want to find the length of either OA or OB to get the radius.

The area of the circle is

A = pi*r^2

A = pi*(sqrt(10))^2

A = 10pi

This is the exact area of the full circle, but we want just a fractional portion of it. Specifically we want the pie slice that is formed by angle AOB

area of sector AOB = [ (angle AOB)/360 ] * (area of full circle)

area of sector AOB = (145/360)*10pi

area of sector AOB = 145pi/36

area of sector AOB = 145*3.14/36

area of sector AOB = 12.647 approximately

area of sector AOB = 13 square units approximately, after rounding to the nearest whole number

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Answer:

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Step-by-step explanation:

given data

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in water loop formed = 10 loops

solution

we consider mass of stone = m

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volume = v

density of stone = $\rho$

case (1)

when 8 loop form with 2 adjacent node is $\frac{\lambda }{2}$

so here

$l = \frac{8 \lambda _1}{2}$ ..............1

$l = 4 \lambda_1\\\\\lambda_1 = \frac{l}{4}$

and we know velocity is express as

velocity = frequency × wavelength .....................2

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here tension = mg

$\sqrt{\frac{mg}{\mu}}$ = f × $\lambda_1$ ..........................3

and

case (2)

when 8 loop form with 2 adjacent node is $\frac{\lambda }{2}$

$l = \frac{10 \lambda _1}{2}$ ..............4

$l = 5 \lambda_1\\\\\lambda_1 = \frac{l}{5}$

when block is immersed

equilibrium eq will be

Tenion + force of buoyancy = mg

T + v × $\rho$ × g = mg

and

T = v × $\rho$ - v × $\rho$ × g

from equation 2

f × $\lambda_2$ = f × $\frac{1}{5}$

$\sqrt{\frac{v\rho _{stone} g - v\rho _{water} g}{\mu}} = f \times \frac{1}{5}$ .......................5

now we divide eq 5 by the eq 3

$\sqrt{\frac{vg (\rho _{stone} - \rho _{water})}{\mu vg \times \rho _{stone}}} = \frac{fl}{5} \times \frac{4}{fl}$

solve irt we get

$1 - \frac{\rho _{stone}}{\rho _{water}} = \frac{16}{25}$

relative density $\frac{\rho _{stone}}{\rho _{water}} = \frac{25}{9}$

relative density = 2.78 ≈ 2.8

so correct option is C) 2.8

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4 years ago

Help needed thanks it greatly appretiated

Ivan

c is the correct answer of this question

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3 years ago

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