Respuesta :
Given that,
Outer diameter = 14 mm
Inner diameter = 11 mm
Length = 21 cm
Young's modulus = 68 GPa
Tensile strength = 7 Mpa
(a). We need to calculate the effective spring constant of the straw with respect to elongation
Using formula of effective spring constant
[tex]\dfrac{Y}{\Delta l}=\dfrac{YA}{l}[/tex]
[tex]k=\dfrac{YA}{l}[/tex]
Where, k = effective spring constant
Y= Young's modulus
A = area
l = length
Put the value into the formula
[tex]k=\dfrac{68\times10^{9}\times\dfrac{\pi}{4}(0.014^2-0.011^2)}{21\times10^{-2}}[/tex]
[tex]k=1.90\times10^{7}\ N/m[/tex]
(b). Force = 90 N
We need to calculate the stretch the straw
Using formula of stretch
[tex]\Delta l=\dfrac{F}{k}[/tex]
Where, F = force
k = effective spring constant
Put the value into the formula
[tex]\Delta l=\dfrac{90}{1.90\times10^{7}}[/tex]
[tex]\Delta l=0.0000047\ m[/tex]
[tex]\Delta l=4.7\times10^{-6}\ m[/tex]
(c). We need to calculate the extend the length of the straw before it breaks
Using formula of extend length
[tex]E_{max}=\dfrac{Y\Delta l}{l}[/tex]
[tex]\Delta l=\dfrac{E_{max}l}{Y}[/tex]
Put the value into the formula
[tex]\Delta l=\dfrac{7\times10^6\times21\times10^{-2}}{68\times10^{9}}[/tex]
[tex]\Delta l=0.0000216\ m[/tex]
[tex]\Delta l=0.0216\times10^{-3}\ m[/tex]
[tex]\Delta l=0.0216\ mm[/tex]
Hence, (a). The effective spring constant of the straw with respect to elongation is [tex]1.90\times10^{7}\ N/m[/tex]
(b). The stretch the straw is [tex]4.7\times10^{-6}\ m[/tex]
(c). The extend length of the straw before it breaks is 0.0216 mm.
