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ANTONII [103]
3 years ago
8

Find the area of a Semicircle whose radius is 3.5 cm (take pi as 22/7​

Mathematics
2 answers:
Vera_Pavlovna [14]3 years ago
4 0

Answer:

TSA=115.5cm²

Step-by-step explanation:

T.S.A=3πr²

Given r=3.5cm

eq.

3×22/7×3.5×3.5

=3×22×0.5×3.5

=115.5cm²

Roman55 [17]3 years ago
4 0

Answer:

19.25cm^{2} \\

Step-by-step explanation:

<u>Formula</u>

<u />\frac{\pi r^{2} }{2}<u />

<u />

<u>Solve</u>

<u />\frac{(\frac{22}{7} )(3.5cm)^{2}}{2}= \frac{(\frac{22}{7} )(3.5cm)(3.5cm)}{2}=\frac{((\frac{22}{7} )(3.5cm))(3.5cm)}{2}=\frac{(\frac{77}{7}cm)(3.5cm) }{2}=

\frac{(11cm)(3.5cm)}{2}=\frac{38.5cm^{2}}{2}=19.25cm^{2}

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Answer:

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Step-by-step explanation:

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A string passing over a smooth pulley carries a stone at one end. While its other end is attached to a vibrating tuning fork and
nasty-shy [4]

Answer:

correct option is C)  2.8

Step-by-step explanation:

given data

string vibrates form =  8 loops

in water loop formed =  10 loops

solution

we consider  mass of stone = m

string length = l

frequency of tuning = f

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

\sqrt{\frac{Tension}{mass\ per\ unit \length }}   =   f × \lambda_1

here tension = mg

so

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

and

case (2)  

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

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when block is immersed

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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

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\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}

so

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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Answer:

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