Chapter 11 Flashcards by Josh Stirrup

Density

Density is the mass per unit volume of an object. Objects made from low density materials typically have a lower mass

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How can we find which has more density

To find how much more dense one substance is compared with another.
We can measure the mass of equal volume of the two substances.
The substance with the greater mass in the smae volume is more dense.
For example a lead sphere of volume 1cm³ has a mass of 11.3g whereas an aluminium sphere of the same volume has a mass of 2.7g.

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Desnity of a substance

Is defined as its mass per unit volume.

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Density = what?

M/V
Mass = m
Volume = v
Density = p

Density = kgm-³

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Cubiod

M / width x base x height

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Cylinder

M / π x r² x height

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Sphere

H / 4/3 x π x r³

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Ireegular solid

M / volume pf water displaced

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Alloy

Is a mixture of two or more metals

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Desnity of alloys calculation

Mass of metal A = Pa Va,
Mass of metal B = Pb Vb
Total mass = Pa Va + Pb Vb

P = Pa Va + Pb Vb / V
= PaVa / v + PbVb/v

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Centimetres³ into metres³

100³ = 1,000,000 = 10⁶

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How to work out desnity of a regular solid

Masure its mass using a top pan balance

Meausre its dimensions by using a venier calipers then calculate its volume using an appropriate equation
Then calculate the density from mass / volume

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How to work out the density of a liquid

Masure the mass of an empty measuring cylinder. Pour some of the liquid into the measuring cylinder and measure the volume of the liquid directly.
Measure the mas of the cylinder and the liquid to enbale the mass of the liquid to be calculated
Then do mass / volume

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How to measure the density of an irregular solid

Measure the mass of an object
Immerse the object on a thread in liquid in a measuring cylinder observe the increase in the liquid level. This is the volume of the object
Then calculate the density by using mass / volume

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Example question 1

Volume = 0.05m x 0.08m x 0.2 = 8x10-⁴
Desnity = p=m/v
= 2.5 / 8x10-4
= 3125

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alloy tube of volume 1.8x 10-4 m° consists of 60% aluminium
and 40% magnesium by volume. Calculate a the mass of Ai aluminium,
Aii magnesium in the tube, b the density of the alloy. The density of
aluminium = 2700 kg m-3. The density of magnesium = 1700 kg m-3

Ai) 
Aluminium = 2700kgm-3 
2700 x (1.8x10-4 x0.6)
Mass = 0.29kg

Aii)
Magnesium = 1700kgm-3
1700 x (1.8x10-4 x 0.4)
Mass = 0.12kg

B)0.29 + 0.12 / 1.8x10-4 = 2277
=2300kgm-3

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Hookes law 1660

The force needed to stretch a spring is directly proportional to the extension of the spring from its natural length, provided the elastic limit is not exceeded.
Up to the limit of proportionality

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Force = what? Hookes law equation

F = k x ΔL
F  = force 
K = spring constant (Nm-1)
ΔL = change in length (m)

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What is spring constant

This is a simple measure of stiffness but doesn’t take into account the objects physical dimensions.
Y axis = force
X axis = extension
Line = elastic and inelastic lastic
The bottom half of the line is elastic
The top half of the line is inelastic lastic
Half way is the elastic limit of proportional
Gradient = K

Spirngs in parallel

Srpings in a parallel with extend less for a given force

K total = k1 +k2 …

Springs in series

1 / ktotal = 1/k1 + 1/k2

Energy stored in a spring is the area under the …

F = k x ΔL
½ x (KxΔL)xΔL

Ep = ½ x FxΔL = ½ xKxΔL²

Ep elastic potential energy stored in a stretched spring

Deformation of a spring

Stress is the force per unit cross-sectonal area applied to a sample or material: 
σ = F/A 
σ = stress (nm-2 or Pa) 
F = force (n) 
A = cross sectional area (m2)

Strain

Is the extenson per unit length undergone by a sample of material: 
ε = ΔL / L
ε = strain 
ΔL = extension 
L v length (m) 
Tensile = under tensile

Young modulas

Is a comprehensive measurment of a materials stiffness is given by: E = σ÷ε = F/A ÷ ΔL/L = FxL / AxΔL E = young modulas (Pa or Nm-2)

Graph

Stress y axis Strain x axis Look at (photo)

Stuff we can read off:

Stiffness from E, gradient of straight line section Strength from UTS Elasticity from strain in straight line secton Ductility from strain under plastic deformation Necking occurs between UTS and B

Tensile stress equation

``` σ = F/A F = force A = area σ = stess ```

Brittel

Material snaps without any noticable yeild. Glass breaks without any give.

Ductile

Material can be drawn into a wire copper is more ductile than steel Stretches easily without breaking

Stiffness

the stiffness of different materials can be compared using the gradient of the stress-strain line, which is equal to the Young modulus of the material. Thus steel is stiffer than copper.

Strength

The strength of a material is its ultimate tensile stress (UTS), which is its maximum tensile stress. Steel is stronger than copper because its maximum tensile stress is greater.

Metal wire

Metal wire stretches elastically until elastic limit is reached Further deformation is platsic and permanent

Rubber

Deformation does not obey Hookes law , as limit of proportionality is very low However elastic limit is high so returns to its orginal

Polythene

Polythene immediately exceeds both the elastic limit and the limit of proportionality. Deformation is permanent and non proportional

Area

Area under a force vs extention graph is the enegry stored Area under loading curve for rubber is greater than unloading curve This tells us the difference in energy into the internal enrgy of the rubber.

Describe how you would use this arrangement to measure the mass of the rock sample. State the measurements you would make and explain how you would use the measurements to find the mass of the rock sample.

``` Measure intital length Put a known masss on it (mg) Measure new length Le - li = extension K = E/e alternatively put more known masses on and measure several extensions Plot Force vs extension and calculate gradient Change in Y / chnage in X = F/e = spring constant = K Put on rock Measure extension Use graoh to find the force Force Y axis Extension X axis Divide force by G = mass ```

Mass = what ?

Mass = density x volume

What is meant by the elastic limit of a wire

The maxium amount that a material can be stretched and still to its orginial length when the force is removed

Young modulus ratio

Ratio of tensile stress to tensile strain

Young modulus equation

``` E = FxL / A x ΔL A = area E = young modulus F = force ΔL = change in length ```

Young modulus | Cross sectional area

The cross sectional area chnages as the rod is compressed or stretched

Tesnile strain

``` ε = e/L E = extension L = length ε = strain ```

Tesnile stress

Teh stretching force applied per unit cross sectional

Tensile strain

Extesnion produced per unit length

Must obey hookes law from A to B where B is the limit of proportionality beyond at B elastic limit reached undergoes plastic deformation

Unloading

At C the load is removed linear relation between stress and strain which does not return to its original length