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where B is the bulk modulus (see [link] ), V 0 size 12{V rSub { size 8{0} } } {} is the original volume, and F A size 12{ { {F} over {A} } } {} is the force per unit area applied uniformly inward on all surfaces. Note that no bulk moduli are given for gases.

What are some examples of bulk compression of solids and liquids? One practical example is the manufacture of industrial-grade diamonds by compressing carbon with an extremely large force per unit area. The carbon atoms rearrange their crystalline structure into the more tightly packed pattern of diamonds. In nature, a similar process occurs deep underground, where extremely large forces result from the weight of overlying material. Another natural source of large compressive forces is the pressure created by the weight of water, especially in deep parts of the oceans. Water exerts an inward force on all surfaces of a submerged object, and even on the water itself. At great depths, water is measurably compressed, as the following example illustrates.

Calculating change in volume with deformation: how much is water compressed at great ocean depths?

Calculate the fractional decrease in volume ( Δ V V 0 size 12{ { {ΔV} over {V rSub { size 8{0} } } } } {} ) for seawater at 5.00 km depth, where the force per unit area is 5 . 00 × 10 7 N / m 2 size 12{5 "." "00" times "10" rSup { size 8{7} } N/m rSup { size 8{2} } } {} .

Strategy

Equation Δ V = 1 B F A V 0 is the correct physical relationship. All quantities in the equation except Δ V V 0 are known.

Solution

Solving for the unknown Δ V V 0 gives

Δ V V 0 = 1 B F A . size 12{ { {ΔV} over {V rSub { size 8{0} } } } = { {1} over {B} } { {F} over {A} } } {}

Substituting known values with the value for the bulk modulus B from [link] ,

Δ V V 0 = 5.00 × 10 7 N/m 2 2 . 2 × 10 9 N/m 2 = 0.023 = 2.3%.

Discussion

Although measurable, this is not a significant decrease in volume considering that the force per unit area is about 500 atmospheres (1 million pounds per square foot). Liquids and solids are extraordinarily difficult to compress.

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Conversely, very large forces are created by liquids and solids when they try to expand but are constrained from doing so—which is equivalent to compressing them to less than their normal volume. This often occurs when a contained material warms up, since most materials expand when their temperature increases. If the materials are tightly constrained, they deform or break their container. Another very common example occurs when water freezes. Water, unlike most materials, expands when it freezes, and it can easily fracture a boulder, rupture a biological cell, or crack an engine block that gets in its way.

Other types of deformations, such as torsion or twisting, behave analogously to the tension, shear, and bulk deformations considered here.

Section summary

  • Hooke's law is given by
    F = k Δ L , size 12{F=kΔL} {}

    where Δ L size 12{ΔL} {} is the amount of deformation (the change in length), F size 12{F} {} is the applied force, and k size 12{k} {} is a proportionality constant that depends on the shape and composition of the object and the direction of the force. The relationship between the deformation and the applied force can also be written as

    Δ L = 1 Y F A L 0 , size 12{ΔL= { {1} over {Y} } { {F} over {A} } L rSub { size 8{0} } } {}

    where Y size 12{Y} {} is Young's modulus , which depends on the substance, A size 12{A} {} is the cross-sectional area, and L 0 size 12{L rSub { size 8{0} } } {} is the original length.

  • The ratio of force to area, F A size 12{ { {F} over {A} } } {} , is defined as stress , measured in N/m 2 .
  • The ratio of the change in length to length, Δ L L 0 size 12{ { {ΔL} over {L rSub { size 8{0} } } } } {} , is defined as strain (a unitless quantity). In other words,
    stress = Y × strain . size 12{"stress"=Y times "strain"} {}
  • The expression for shear deformation is
    Δ x = 1 S F A L 0 , size 12{Δx= { {1} over {S} } { {F} over {A} } L rSub { size 8{0} } } {}

    where S is the shear modulus and F is the force applied perpendicular to L 0 and parallel to the cross-sectional area A .

  • The relationship of the change in volume to other physical quantities is given by
    Δ V = 1 B F A V 0 , size 12{ΔV= { {1} over {B} } { {F} over {A} } V rSub { size 8{0} } } {}

    where B is the bulk modulus, V 0 is the original volume, and F A size 12{ { {F} over {A} } } {} is the force per unit area applied uniformly inward on all surfaces.

Questions & Answers

Three charges q_{1}=+3\mu C, q_{2}=+6\mu C and q_{3}=+8\mu C are located at (2,0)m (0,0)m and (0,3) coordinates respectively. Find the magnitude and direction acted upon q_{2} by the two other charges.Draw the correct graphical illustration of the problem above showing the direction of all forces.
Kate Reply
To solve this problem, we need to first find the net force acting on charge q_{2}. The magnitude of the force exerted by q_{1} on q_{2} is given by F=\frac{kq_{1}q_{2}}{r^{2}} where k is the Coulomb constant, q_{1} and q_{2} are the charges of the particles, and r is the distance between them.
Muhammed
What is the direction and net electric force on q_{1}= 5µC located at (0,4)r due to charges q_{2}=7mu located at (0,0)m and q_{3}=3\mu C located at (4,0)m?
Kate Reply
what is the change in momentum of a body?
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what is a capacitor?
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Capacitor is a separation of opposite charges using an insulator of very small dimension between them. Capacitor is used for allowing an AC (alternating current) to pass while a DC (direct current) is blocked.
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A motor travelling at 72km/m on sighting a stop sign applying the breaks such that under constant deaccelerate in the meters of 50 metres what is the magnitude of the accelerate
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please solve
Sharon
8m/s²
Aishat
What is Thermodynamics
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velocity can be 72 km/h in question. 72 km/h=20 m/s, v^2=2.a.x , 20^2=2.a.50, a=4 m/s^2.
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A boat travels due east at a speed of 40meter per seconds across a river flowing due south at 30meter per seconds. what is the resultant speed of the boat
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50 m/s due south east
Someone
which has a higher temperature, 1cup of boiling water or 1teapot of boiling water which can transfer more heat 1cup of boiling water or 1 teapot of boiling water explain your . answer
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I believe temperature being an intensive property does not change for any amount of boiling water whereas heat being an extensive property changes with amount/size of the system.
Someone
Scratch that
Someone
temperature for any amount of water to boil at ntp is 100⁰C (it is a state function and and intensive property) and it depends both will give same amount of heat because the surface available for heat transfer is greater in case of the kettle as well as the heat stored in it but if you talk.....
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about the amount of heat stored in the system then in that case since the mass of water in the kettle is greater so more energy is required to raise the temperature b/c more molecules of water are present in the kettle
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definitely of physics
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what is field
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physics, biology and chemistry this is my Field
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field is a region of space under the influence of some physical properties
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Another formula for Acceleration
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a=v/t. a=f/m a
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pratica A on solution of hydro chloric acid,B is a solution containing 0.5000 mole ofsodium chlorid per dm³,put A in the burret and titrate 20.00 or 25.00cm³ portion of B using melting orange as the indicator. record the deside of your burret tabulate the burret reading and calculate the average volume of acid used?
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Two bodies attract each other electrically. Do they both have to be charged? Answer the same question if the bodies repel one another.
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No. According to Isac Newtons law. this two bodies maybe you and the wall beside you. Attracting depends on the mass och each body and distance between them.
Dlovan
Are you really asking if two bodies have to be charged to be influenced by Coulombs Law?
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like charges repel while unlike charges atttact
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What is specific heat capacity
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Specific heat capacity is a measure of the amount of energy required to raise the temperature of a substance by one degree Celsius (or Kelvin). It is measured in Joules per kilogram per degree Celsius (J/kg°C).
AI-Robot
specific heat capacity is the amount of energy needed to raise the temperature of a substance by one degree Celsius or kelvin
ROKEEB
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Source:  OpenStax, College physics for ap® courses. OpenStax CNX. Nov 04, 2016 Download for free at https://legacy.cnx.org/content/col11844/1.14
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