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Studentka2010 [4]
2 years ago
7

Reposting this because no body gave me the equation. An equation would be appreciated :)

Mathematics
2 answers:
Margaret [11]2 years ago
6 0

Answer:

\frac{6q}{5}  - 3 + 3 \leqslant 28 -  \frac{4q}{5}  + 3 \\  \\  \frac{6q}{5} \leqslant  -  \frac{4q}{5}   + 31 \\  \\  \frac{6q}{5}  +  \frac{4q}{5}  \leqslant  -  \frac{4q}{5}  +  \frac{4q}{5}  + 31 \\  \\  \frac{10q}{5}  \leqslant 31 \\  \\  2q \leqslant 31 \\  \\ 2q \div 2 \leqslant 31 \div 2 \\  \\ q \leqslant  \frac{31}{2}

I hope I helped you^_^

o-na [289]2 years ago
3 0

bring the like terms together,

\frac{6q}{5}  +  \frac{4q}{5}  \leqslant 28 + 3 + 3 - 3

\frac{10q}{5}  \leqslant 31

2q \leqslant 31

q \leqslant  \frac{31}{2}

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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}

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}

so

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

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so correct option is C)  2.8

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Answer:

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