Mathematics, 31.07.2019 21:20 danielbamf1414

Acertain wire stretches 0.90 cm when outward forces with magnitude f are applied to each end. the same forces are applied to a wire of the same material but with three times the diameter and three times the length. the second wire stretches
a. 3 cm
b. 1 cm
c. 2.7 cm
d. 9 cm

Answers: 3

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Answer from: LuluMathLover101

Option A

Step-by-step explanation:

Given -

Length to which a wire stretches when outward forces with magnitude F are applied to each end = 0.90 cm

Let the length of first wire = l_1

Let the diameter of first wire = d_1

The dimension of the second wire is equal to

$l_2= 3l_1\\d_2= 3d_1$

We know that

strain $= \frac{dl}{l} \\$

and strain $= Y\frac{F}{A}$

Putting the above two equations together, we get -

$\frac{dl}{l} = Y\frac{F}{A}$

For wire 1,

$\frac{dl_1}{l_1} = Y\frac{F}{A_1}$ ---------Eq(1)

For wire 2,

$\frac{dl_2}{l_2} = Y\frac{F}{A_2}$ ---------Eq(1)

Dividing equation 1 by equation 2, we get

$\frac{dl_1}{dl_2} = \frac{l_2}{l_1} \frac{A_2}{A_1}$ $\frac{dl_1}{dl_2}= 3* \frac{(3d_1)^2}{d_1^2} \\dl_2= \frac{0.90}{3*9}\\dl_2=0.30$ cm

Hence, option A is correct

Answer from: Karinaccccc

When the same forces are applied to a wire of the same material but with three times the diameter and three times the length of the first wire then the second wire stretches 3 cm.

Given :

A certain wire stretches 0.90 cm when outward forces with magnitude F are applied to each end.The same forces are applied to a wire of the same material but with three times the diameter and three times the length.

Let the length of the first wire be 'L' and the diameter be 'D'. Then the length and diameter of the second wire will be:

L' = 3L

D' = 3D

The formula of strain is given by:

$\rm strain = \dfrac{dl}{l} = \gamma \dfrac{F}{A}$

So, for the first wire :

$\rm \dfrac{dL}{L}= \gamma \dfrac{F}{A}$ ---- (1)

For the second wire:

$\rm \dfrac{dL'}{L'}= \gamma \dfrac{F}{A'}$ ----- (2)

Now, divide equation (1) by equation (2).

$\rm \dfrac{\dfrac{dL}{L}}{\dfrac{dL'}{L'}}=\dfrac{\dfrac{1}{A}}{\dfrac{1}{A'}}$

Simplify the above equation.

$\rm \dfrac{dL}{dL'}=\dfrac{L}{L'}\times\dfrac{A'}{A}$

$\rm \dfrac{dL}{dL'}=\dfrac{L}{3L}\times\dfrac{(3D)^2}{D^2}$

$\rm \dfrac{dL}{dL'}=\dfrac{1}{3}\times {9}$

$\rm \dfrac{dL}{dL'}=3$

Therefore, the correct option is A) 3 cm.

For more information, refer to the link given below:

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Answer from: Quest

step-by-step explanation:

given that there are two triangles j'k'l' and jkl>

the former is got by dilating the second one.

when drawn on a grid, (graph paper) coordinates of vertices of first triangle is

(6,12)j, (9,6)k and (12,9)l

side jk has length = distance between j and k

= $\sqrt{(9-6)^{2} +(12-6)^{2} } \\=\sqrt(45)=3\sqrt(5)$

side j'k'=distance between j' and k'

= $\sqrt{(3-2)^{2} +(2-4)^{2} } \\=sqrt(5)$

ie j'k'l' has sides 1/3 of that of corresponding sides of jkl.

hence scale factor of 1/3 is used to get j'k'l'from jkl

Answer from: Quest

pls refer ro the top for answer

Offering 20 pts ! i need with my homework ! (must include steps)

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