Respuesta :
Answer:
Explanation: Bulk Modulus (K) = [tex]\frac{Volumetric Stress}{Volumetric Strain}[/tex]
This is the same as saying it equal to the change in pressure with respect to the change in volume divided by initial volume.
Bulk modulus elasticity may also be expressed in terms of pressure and density:
Remember: Strain is a unit-less quantity.
[tex]K=\frac{(p_1 - p_0)}{\frac{(\rho_1- \rho_0)}{\rho_0}}[/tex]
Here,
- ρ0 and ρ1 are the initial and final density values.
- p₀ & p₁ are the initial and final pressure values.
Given data:
bulk modulus, [tex]k= 1.8 \times10^{9} N.m^{-2}[/tex]
volumetric strain, [tex]\epsilon _{v}=2\times10^{-3}[/tex]
To find: Force exerted by the juice per square centimeter, [tex]\sigma _{v}[/tex]
Solution: [tex]\sigma _{v}=\epsilon _{v}\times K [/tex]
⇒ [tex]\sigma _{v}= 2\times10^{-3}\times1.8 \times10^{9} [/tex]
[tex]\sigma _{v}= 3.6\times10^{6} N.m^{-2} =3.6\times10^{2}N.cm^{-2}[/tex]
Answer:
The magnitude of force per [tex]cm^{2}[/tex] is [tex]3.6\times 10^{6}\ N/m^{2}[/tex]
Solution:
As per the question:
Increase in the volume, [tex]\frac{\Delta V}{V_{o}} = 2\times 10^{- 3}[/tex]
Bulk Modulus, B = [tex]1.8\times 10^{9}\ N/m^{2}[/tex]
Now, to calculate the Normal Force's magnitude, F exerted by the juice:
Volume change of an elastic substance on the application of a force is given by:
[tex]\frac{\Delta V}{V_{o}} = \frac{1}{B}\times(\frac{F}{Area,\ A})[/tex]
[tex]\frac{F}{Area,\ A} = B(\frac{\Delta V}{V_{o}})[/tex]
Now, putting suitable values in the above eqn:
[tex]\frac{F}{Area,\ A} = 1.8\times 10^{9}\times 2\times 10^{- 3} = 3.6\times 10^{6}\ N/m^{2}[/tex]
Here, F is the force exerted on the juice by the container per [tex]cm^{2}[/tex], there an equal reaction force per [tex]cm^{2}[/tex] will be exerted by the juice on the container.
