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4 years ago

Poiont S is the midpoint of RT. Complete the statement: ST = 38 ft, RT =

Mathematics

2 answers:

Luden [163]4 years ago

5 0

RT=76. If S is the midpoint, the line is split into two equal parts, so 38+38=76

vazorg [7]4 years ago

3 0

472.92581847840239283

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during the winter months, freshwater fish sense the water getting colder and swim to the bottom of lakes and rivers to find warm

Jlenok [28]

Answer: 28 feet.

Step-by-step explanation:

Given: During the winter months, freshwater fish sense the water getting colder and swim to the bottom of lakes and rivers to find warmer water.

The total depth of the lake = 32 feet

Since, fish swims 7/8 of the depth, then the depth of lake to which the fish swim= $\frac{7}{8}\times32\ feet$

Since, $32\div8=4$

therefore, $\frac{7}{8}\times32=7\times4=28$

Hence, the fish swim 28 feet in lake.

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3 years ago

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Alexandra [31]

I believe it neither

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3 years ago

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4 years ago

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andre [41]

I believe this is correct

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3 years ago

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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

$\sqrt{\frac{mg}{\mu}}$ = 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

3 0

4 years ago