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madreJ [45]
3 years ago
13

Pls help with number 5

Mathematics
1 answer:
vova2212 [387]3 years ago
7 0
I believe it is asking for the same number basically written 3 different ways, so it could be 1 1/2, 6/4 and 1.5 because those are all the same number
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Write an equation perpendicular to x = 5 that passes through the point (-4,3)
Neporo4naja [7]

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2 years ago
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}

relative density = 2.78 ≈ 2.8

so correct option is C)  2.8

3 0
3 years ago
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