THERMAL PROPERTIES OF MATTER

1. A steel bar (200m) long 20 . If the extremes of temperature to which it might be exposed are -30 to +40 how much will it constract and expand? Given the coefficient of linear expansion of steel is 12 .α ˚

Lo=200m T=20 TF = 40 coefficient= 12x10 K
For the expansion

L = Lo T
=(12x10 K )(200m)(40-20)
=4.8x10 m

For the contraction

L = Lo T
=(12x10 K )(200m)(-30-20)
=12x10 m
Or the decreace in length is 12.0cm



2. A cooper plate (α=1.70 Xˉ⁵Kˉ¹) has dimension of 80cmx15cm at 15˚C. what is the decrease in area when it heated to 35˚C.

α=1.70 Xˉ⁵Kˉ¹ Ao=80cmx15cm T=15˚C TF =35˚C
A= Ao (1+2 α ΔT)
= 0.12(1+2(1.70 Xˉ⁵Kˉ¹)(35˚C-15˚C)
= 0.12m²




3. A circular cooper rng at 20.0˚C has a hole with an area of 9.98cm². what
Minimum temperature must it have so that it can be slipped onto a steel metal rod having a cross sectional area of 10.0cm²? Given α=1.70 Xˉ⁵Kˉ¹

ΔA=2 αAoΔT
10.0cm²-9.98cm²= 2(1.70 Xˉ⁵Kˉ¹)( 9.98cm²)(ΔT)
0.2 = 3.39x10ˉ⁴ΔT

ΔT = (0.2 / 3.39 x10ˉ⁴)
= 58.9 ˚C





4. Two metal rods, one aluminium and one brass, are each comped at one. At 0.0˚C the rods are each 50.0cm long are separated by 0.024cm their unfastened ends. At what temperature will the rods just come into contract.

Lo =50.0cm To = 0.0˚C
αbr = 19x10ˉ⁶Kˉ¹
αAl = 2.28x10ˉ⁶Kˉ¹
αbr+αA l =19x10ˉ⁶Kˉ¹+2.28x10ˉ⁶Kˉ¹
= 0.024cm
ΔT = 0.024cm / (19X10ˉ⁶ + 2.25 X 10ˉ⁶)(50.0cm)

= 0.024cm / (2.075 X 10ˉ⁵)

ΔT = 11.56˚C

TF = To +ΔT
= 0.0˚C + 11.56˚C
= 12˚C





5. A quartz sphere is 8.75cm in diameter. What will be its change in volume if it is heated from 30˚Cto 200˚C. The coefficient of linear expansion of the quartz is 0.4x10ˉ⁶Kˉ¹.
Vol of sphere = 4/3 ∏r³
= 4/3∏r³ x (8.75x10ˉ²/2)
= 3.51 x 10ˉ⁴m³
V = Vo (1+3αΔT)

= 3.51 x 10ˉ⁴m³ (1+3(0.4x10ˉ⁶Kˉ¹.)(200-30)
= 3.511 x 10ˉ⁴m³
ΔV = Vf-Vo
= 3.511 x 10ˉ⁴m³ -3.51 x 10ˉ⁴m³
= 0.1 x 10ˉ⁶m³





6. A linear rod 40cm long at 40˚C is heated to 60˚C. The length of the rod is then measured to be 40.0105cm. what is the coefficient of linear expansion of the metal.

ΔV = βVo ΔT
ΔV = V-Vo
= 0.40105 – 0.4
= 1.05x10 ˉ⁴
β = 2α
β= (ΔV/ VoΔT)
= (1.05 x 10ˉ⁴)/(0.4 (60-20))
= 6.56 x 10ˉ⁶X2
= 13 x10ˉ⁶Kˉ¹





7. An iron ring is to fit snugly on a cylinder iron rod. At 20˚C, the diameter of the rod is 6.420cm. to slip over the rod diameter by about 0.008cm. to what temperature must the ring be brought if its hole is to be large enough so it will slip over the rod? Given α iron = 12 x 10ˉ⁶Kˉ¹


To = 20˚C
Lo = 6.420cm
ΔL= 0.008cm
αiron = 12 x 10ˉ⁶Kˉ¹

ΔT = ΔL / αiron Lo
= (0.008 x 10ˉ²)/ (12 x 10ˉ⁶(6.420 x 10ˉ²))
= 103.8˚C

ΔTF = T + To
= 103.8 + 20
= 123.8˚C





8. The confficient of linear expansion of aluminium is 23 x 10ˉ⁶Kˉ¹. a circular hole in an aluminium plate is 2.725cm in diameter at 0˚C. What is the diameter of the hole if the temperature of the plate is raised to 100˚C.

α= 23 x 10ˉ⁶Kˉ¹.
Vo = 2.725cm
To = 0˚C
TF = 100˚C
ΔV = Vo (1+3αΔT)
=2.725x 10ˉ² (1+3(23x10ˉ⁶Kˉ¹)(100-0)
= 2.7cm





9. The coefficient of linear expansion of steel is 12 x 10ˉ⁶Kˉ¹. A railroad track is made of invalidual rail of steel 1.0km in length . by what length wouldthese rails change between a cold day when the temperature is -10˚C and hot day at 30˚C.

αsteel= 12 x 10ˉ⁶Kˉ¹ Lo = 1000m To = -10˚C TF = 30˚C

ΔL = αsteel LoΔT
= 12 x 10ˉ⁶Kˉ¹(1000m)(30-(-10))
= 48x 10 ˉ²
= 48cm

10. A cooper plate a length of 0.12m and a width of 0.10m at 25˚C. The plate is uniformly heated to 175˚C. If the linear expansion coefficient for cooper is 1.70 x 10ˉ⁵Kˉ¹ . what is the change in the area plate as a result of the increase in temperature?


Ao = 0.12cm
To = 25˚C
TF = 175˚C
α= 1.70 x 10ˉ⁵Kˉ¹

ΔA = 2αAoΔT
= 2(1.70 x 10ˉ⁵Kˉ¹)(0.12)(175-25)
=6.12 x 10ˉ⁵m²

THERMAL PROPERTIES OF MATTER

1) A sheet of copper has an area of 500cm² at 0˚C. find the area of this sheet at 80˚C.
The coefficient of area expansion is to good approximation 2α.thus

Answe:

∆A=2αA∆t
=2(1.67x10⁻⁵)(500)(80˚-0˚)
=1.34cm²

The new area is
500 cm²+1.34 cm²
=501.3cm²


2) If an anisotropic solid has coefficients of liner expansion αx, αy. Αz for three mutually perpendicular direction in the solid, what is the coefficient of volume expansion for the solid?

Answer:

Consider a cube,with edges parallel to X,Y,Z, of dimension L∘ at T=0. After a change in temperature ∆T=(T-0), the dimensions change to
Lx=L∘(1+αxT) Ly=L∘(1+αyT) Lz=L∘(1+αzT)
And the volume of the parallelepiped is
V=V∘(1+αxT)(1+αyT)(1+αzT)≈ V∘[1+(αx+αy+αz)T]
Where V∘=L∘³. Therefore, the coefficient of volume expansion is given by β≈αx+αy+αz. (note : the symbols β and γ are often used for the coefficient of value expansion.)


3) Calculate the increase in volume of 100cm³ of mercury when its temperature change from 10 to 35˚C.

Answer:

∆V=βV∘∆T. Then volume coefficient of expansion of mercury β=1.8x10⁻⁴˚C⁻¹. then ∆V=(1.8x10⁻⁴˚C⁻¹)(100cm³)(25˚C)=0.45cm³


4) The coefficient of liner expansion of glass. If a bottle holds 50.000cm³ at 15˚C, what is its capacity at 25˚C?

Answer:

The glass expands uniformly in all dimensions, so the capacity expands in proportion to the glass. The volume coefficient of expansion is related to the linear coefficient byβ=3α. Then V=V∘(1+3α∆T)=(50.000cm³)[1+3(8.3x10⁻⁶˚C⁻1)(10˚c)]. V=50.012cm³


5) Determine the change in volume of a block of iron 5cm x 10cm x 6cm, when the temperature change from 15˚C to 47˚C.

Answer:

V∘=300cm³; β=3α.then
∆V=3αV∘∆T=3(1.2 x 10⁻⁵˚C⁻¹)(300cm³)(32˚C)=0.35cm³



6) A glass vessel is filled with exactly 1L of turpentine at 20˚C. What volume of the liquid will overflow if the temperature is raised to 86˚C?

Answer:

As the temperature raises, both the capacity of the glass vessel and the volume of the turpentine increase. The amount of the overflow is the difference in these increases : ∆VT-∆VG=(βT-βG)V∆T. noting that βG=3αG, and using the data , we have
∆VT-∆VG=(9.4x10⁻⁴˚C⁻¹)-(0.25 x 10⁻⁴˚C⁻¹)(1L)(66˚C)=60mL


7) The density of gold is 19.30g/cm³ at 20˚C. compute the density of gold at 90˚C.

Answer:

d=M/V d1/d2=V2/V1
since mass constant. The indices 1 and 2 refer to 90˚C and 20˚C, respectively
d1/d2=V2/[V2(1+3α∆T)]
=1/[(1+3α∆T)]≈ 1-3α∆T (since 3α∆T<<1)
d1=d2(1-3α∆T)=(19.3g/cm³)[1-3(14.2 x 10⁻⁶˚C⁻¹)(70˚C)]
d1=19.24g/cm³


8) The density of mercury at 0˚C is 13600kg/m³.calculate the density of mercury at 50˚C.

Answer:

Let ρ1=ρ∘[V∘/V1]=ρ∘[V∘/(V∘+∆V)]=ρ∘[1/(1+(∆V/V∘)]
But ∆V/V∘=β∆T=(1.82 x 10⁻⁴˚C⁻¹)(50˚C)=0.0091
Substitution gives ρ1=(13600kg/m³)[1/(1+0.0091)]=13477kg/m³


9) the mercury in a thermometer has a volume of 210mm³ at 0˚C, at which temperature the diameter of the bore is 0.2mm. How far apart are the degree marks on the stem?

Answer:

We assume that the temperature rises from 0˚C to t. so ∆t=t. Then
∆VHg =β Hg V Hg ∆t, ∆VG = 3αGVG ∆t, where here VHg is the volume occupied by mercury at 0˚C and VG is the volume of the glass container occupied by the mercury at 0˚C. Indeed hAt = ∆VHg-∆VG, where h is the height that the mercury rises and At is the cross section of the bore at temperature t. At = ∏r(1+2αG∆t) with r=0.1mm, the radius at 0˚C, then
Sh=[(βHg-3αG)VHgt]/[ ∏r²(1+2αG t)]
=[ [(182-25) x 10⁻⁶˚C⁻1](210mm³)t]/[0.0314[1+(16.6 x 10⁻⁶˚C⁻¹)t]]


10) A glass beaker hold exactly 1L at 0˚C. (a) what is its volume at 50˚C? (b) If the beaker is filled with mercury at 0˚C, what volume of mercury overflow when the temperature is raised to 50˚C?

Answer:

a) The volume of the beaker after the temperature change is.
Vbeaker =V∘(1+3αG∆T)=(1)[1+3(8.3 x 10⁻⁶)(50)]=1.001L.
b)for the mercury expansion,
Vmercury=V∘(1+β∆T)=(1)[1+(1.82 x 10⁻⁴)(50)]=1.009L
The overflow is thus 1.009-1.001=0.008L, or 8mL

THERMAL PROPERTIES OF MATTER

1. A pie plate is filled up to the brim with pumpkin pie filling. The pie plate is made of Pyrex and its expansion can be neglected. It is a cylinder with an inside depth of 2.10 cm and an inside diameter of 30.0cm. It is prepared at a room temperature of 68 degrees Fahrenheit and placed in an oven at 400 degrees Fahrenheit. When it’s taken out, 151cc of the pie filling has flowed out and over the rim. Determine the coefficient of volume expansion of the pie filling, assuming it is a fluid.

Answer:

The volume expanded is 151 cc. the volume of reference, that is, the V of cylinder is 0.25*3.1415*2.10*(30.0) ^2the coefficient of volume expansion= 151/ VThe expansion of the Pyrex plate (container) is neglected.So V (final) =V (Initial)=V of the cylinder.

2. At the bottom of an old mercury-in-glass thermometer is a 46.0-mm3 resevoir filled with mercury. When the thermometer was placed under your tongue, the warmed mercury would expand into a very narrow cylindrical channel, called a capillary, whose radius was 2.50 x 10-2 mm. Marks were placed along the capillary that indicated the temperature. Ignore the thermal expansion of the glass and determine how far (in mm) the mercury would expand into the capillary when the temperature changed by 1.0 C°

Answer:

The change in the mercury's volume is equal to beta*V*delta T, where beta is the coefficient given above, V is the original volume, and delta T is the temperature change. Since the diameter of the capillary is unchanged, the change in volume there will be pi*r-squared*delta L, where r is the radius, and deltaL is the change in the length of the mercury column. Setting the volume changes equal to each other, and doing a little algebra,beta*V*deltaT = pi*r-squared*deltaLdeltaL = beta*V*deltaT / pi*r-squareddeltaL = 1.8x10(-4)*46*1.0 / 3.14*[2.5x10(-2)]*[2.5x10(-2)]deltaL = 4.21 mm

3. The coefficient of volume expansion for ethyl alcohol is 1.12x10^-3 1/K. Copper has a coefficient of volume expansion of 1.67x10^-3 1/K. If a copper cylinder is filled to the brim with ethyl alcohol at 5 C, what percentage of the alcohol will spill if the temperature is raised by 25 C.

Answer:

The temperature change is 20 C. So, the volume expansion of the copper will be20x1.67x10^-3 or 3.34x10^-2 liters and the volume expansion of the alcohol will be 20x1.12x10^-3=2.24x10^-2 l

4. The density of gasoline is 730 kg/m3 at 0°C. Its average coefficient of volume expansion is 9.60 x 10-4/C°. If 1.00 gallon of gasoline occupies 0.00380 m3, how many extra kilograms of gasoline would you get if you bought 8.0 gal of gasoline at 0°C rather than at 19.0°C from a pump that is not temperature compensated?

Answer:

Density of gasoline at 0°C = 730 kg/m3Density of gasoline at 19°C = 730 (1 + 19(9.60 x 10-4)) kg/m3 = 743.3152 kg/mm=dVV = 8 gal = 0.0304 m3difference in m = 743.3152(0.0304) - 730(0.0304) kg= 0.40478208 kg

5. A pair of eyeglass frames are made of an epoxy plastic (coefficient of linear expansion = 1.30 10-4°C-1). At room temperature (20.0°C), the frames have circular lens holes 2.10 cm in radius. To what temperature must the frames be heated if lenses 2.11 cm in radius are to be inserted into them?

ANSWER:

ΔL = αLoΔT(.0211 - .0210) = (1.3e-4)(.021)(T - 20).0001 = (2.73e-6)(T - 20)36.63003663 = T - 20T = 56.63003663T = 56.6°C

The frames must be heated to 56.6°C for the frames to increase from a 2.1cm radius to a 2.11cm radius.

6. The aptly-named Steel Bridge over the Willamette River in Portland, Oregon, has a central span steel truss that is 64.1 m long. What is its change in length over the year from the average minimum annual temperature of 0.90°C to the average maximum of 26.3°C? Use a coefficient of linear expansion of 1.17×10^−5 (1/C°).

Answer:

delta L = length chg of central truss = (L)(Cx) (delta T) = (64.1)(1.17e-5)(26.3 - 0.9)delta L = 75e-5(25.4) = 1905e-5 = 0.0191 m = 1.91 cm

7. We cut a circular piece from the rectangular plate. Which ones of the processes given below can help us in passing through the circular piece from the hole?

Answer:

I. Increasing the temperatures of rectangular plate and circular piece
II. Decreasing the temperature of the circular piece
III. Decreasing the temperatures of the rectangular plate and circular piece

I. If we increase the temperatures of the plate and circular piece, expansion of the hole and the circular piece will be the same. Thus, this option can help us.
II. If we decrease the temperature of the circular piece, it contracts and hole becomes larger than the piece. This option can also help us.
III. If we decrease the temperatures of the plate and circular piece, hole and circular piece contract in same size. This process can also help us.

8. There are three same metal rods having same length and thickness. If the temperatures of them are given like; T, 2T and 3T find the relations of final lengths of the rods.

Answer:

We find the final temperatures of the system by the formula;
Tfinal=T1+T2+T3/3=6T/3=2T
Since the temperature of the first rod increase, its final length also increases. Temperature of the second rod stays same, thus there won’t be change in the length of this rod. Finally, temperature of the third rod decreases, thus its contract and final length of it decreases with respect to initial length. As a result relation of the final lengths of the rods;
L1>L2>L3

week 7:electomagnetic wave

(week 6) wave and sound

Wave and sound

(1-0) types of wave

Water wave

Sound waves travel in air with a speed of 343m/s. The lowest frequancy sound we can hear is 20.0 Hz; the highest frequency is 20.0kHz. Find the wave length of sound for frequencies of 20.0Hz and 20.0kHz

Solution

Solve Equation for λ

λ = v/f = (343 m/s)/(20.0s⁻¹) = 17.2 m

λ = v/f = (343 m/s)/(20,000s⁻¹) = 1.72 m

(2.0) wave and string

The speed of a wave on a string

A 5.0-m length of rope, with a mass of 520g, is pulled taut with a tension of 46N. Find the speed of waves on the rope.

Solution

First, calculate the mass per length, μ:

520g = 0.52kg

μ = m/ L = (0.52kg)/(5.0m) = 0.10kg/m

Now, substitute μ and F into Equation (2.0):

ν = √(F/ μ) = √[46N/(0.10kg/m)]

(3.0) harmonic wave functions

A loudspeaker puts out 0.15 W of sound through a square area 2.0 m on each side. What is the intensity of this sound?

Solution

Applying Equation 14-5, with A = (2.0m)², we find

I = intensity = ?

P = power = 0.15W

A = area = (2.0m)²

I = P/A = (0.15W)/(2.0m)² = 0.038W/m²

(4.0) harmonic wave functions

The power of song

Two people relaxing on a deck listen to a songbird sing. One person, only 1.00m from the bird, hears the sound with an intensity of 2.80x10⁻⁶W/m². (a) what intensity is heard by the second person , who is 4.25m from the bird? Assume that no reflected sound is heard by either person. (b) What is the power output of the bird’s song?

Our problem, the two observers, one at a distance of r1 = 1.00m from the bird, the other at a distance of r2 = 4.25m. The sound emitted by the bird is assumed to spread out spherically, with no reflections

Strategy

a. the two intensities are related by equation 4.0 ,with r1 = 1.00m and r2 = 4.25m.

b. the power output can be obtained from the definition of intensity, I =( P/A). we can calculate P for either observer, noting that A = 4πr²

solution

part (a)

1. Substitute numerical values into equation (4.0)

I2 = (r1/r2)²I1 = (1.00m/4.25m)²(2.80 X 10⁻⁶ W/m²)

= 1.55 X 10⁻⁷ W/m²

Part (b)

2. Solve I =( P/A) for the power ,p, using data for observer 1:

I1 =( P/A1)

P = I1 A1 = (2.80 X 10⁻⁶W/m²)[4π(1.00m)²]

= 3.52 X 10⁻⁵W

3. As a check, report the calculation for observer 2:

I2 =( P/A2)

P = I2A2 = (1.55 X 10⁻⁷W/m²)[4π(4.25m)²]

= 3.52 X 10⁻⁵W

(5.0)

At a distance of 5m from a source the sound level is 90dB. How far away has the level dropped to 50dB

Solution

I1 = (P/4πr1²) and I2 = (P/4πr2²), so (I2 /I1) = (r1²/r2²)

Β1 = 10 log (I1 /I0 ) = 90dB , so (I1 /I0 ) = 10⁹

similarly (I2 /I0 ) = 10⁵

thus (I2 /I1) = (10⁵/10⁹) = 10⁻⁴ = (r1²/r2²), so r2 = 10²r1 = 500m

TUTORIAL

ELASTICITY


1.An iron rod 4.00m long and 0.500cm²in cross section stretches 1.00mm when amass of 225 Kg is hung from its lower end.compute young’s modulus for the iron.

σ = F/A = (225 kg) (9.8m/s²) / (0.500 X 10⁻⁴ m²)
=4.41×10⁷ Pa
ε = ∆s/S∘
= 1.00 x 10 ⁻³m / 4.00 m
=2.5×10⁻⁴
Y= stress/strain
= σ/ε
=4.41x10⁷ / 2.5x10⁻⁴
=1.764×10¹¹
=176 Gpa


2. A load of 50 kg is applied to the lower end of steel rod 80 cm long and 0.60 cm in diameter, How much will the rod strectch?

m=50 kg×9.81 j²=3×10⁻³m ℓ∘ = 80cm
d=0.6cm÷2 A=πj² = 80×10⁻²
=0.3 =πj²(3×10⁻³)² = 0.8
=2.83×10⁻⁵m y=190kPa
Aℓo=Fℓo / Y∆
= (4.90 x 0.8)/(190x 10⁹)(2.83x 10⁵)
=392.4 / 5.377 x 10⁸
=72.97×10⁻⁷
=73μm


3. A platform is suspended by four wires at its corner. The wires are 3.0m long and lave diameter of 2.0mm. Young’s modulus for material is 180GPa.How far will the platform drop (due to elongation of the wires)if a50kg load is placed at the center of the platform?

Y=180GPa
=180×10⁹ Pa
D = 2mm÷2
=1×10×⁻³
Y = Fℓo / YA
ℓo=3m A=πj²
F=50×9.81 = (3.14×10⁻⁵)m
=490.5N
Aℓo=490.5N (3m²) / (180)(3.14 x 10 ˉ⁶)
=1471.5Nm/5.652x 10⁻⁴
=2.6×10⁶m
=2.6/4
=0.65mm

4. Two parallel and opposite forces,each 4000N,aree aplied tangentialy to the upper and lower faces of a cubical metal block 25cm on a side.Find the angel of shear and the displament of the upper surface relative to the lower surface.The shear modulus for the metl is 80GPa.

tanθ=2 x 10ˉ⁷ / 25 x 10⁻²
= 8.0×10⁻⁷ rad
Displament=4000 / (2.5 x 10ˉ²)
80GPa = 64000/∑
∑=64000/80G
=8×10⁻⁷(25×10⁻²)
=2×10⁻⁷

5.The bolk moduls of water is 2.1 Gpa. Compute the volume comtraction of 100ml of water when subjected to a pressure of 1.5 Mpa.

B= 2.1GPa
Vo= 100mL / 1000
=0.1 mL
∆p = 1.5Mpa
∆v = (∆p)(Vo) / B
=(1.5 Mpa)(0.1mL) / 2.1 GPa)
=150000 / 2.1GPa
= -71.43×10³
= -0.0714mL

6. The compressibility of water is 5.0×10⁻¹⁰m²/N,find the decrease in volume of 100ml of water subjected to a pressure of 15mPa.

Bulk strains = ∆Vo / Vo
AVo=B×VoA
=15×10⁶Pa(5.0×10⁻¹Nm⁻²)
=7.5×10⁻³
=0.75mL

TUTORIAL

HYDROSTATIC


1. Find the pressure due to the fluid at a depth of 76cm in still
Water (pw=1.00g/cm³)

P=pgh
=(0.001kg/m³)(9.8m/s²)(0.76m)
=7.448kpa = 7.5 kPa
(b)mecury.(p=13.6 g/cm³)
P=pgh
=(0.0136kg/m³)(9.8m/s²)(0.76m)
=1.01×10⁵N/m²=101.3×10³ N/m²

2. A weight piston confines afluid density p in a closed container. The combined weight of piston and weight is 200N, and the cross-sentional area of piston is A=8.0cm².Find the total pressure at point B if the mercury and h=25cm.What would an ordinary pressure gouge read at B?

P = F/A + Pgh (2000/8 x 10⁻⁴)+13600(9.8m/s²)(25×10⁻²)
=2500000+33354
= 2.8×10⁵Pa
P=2.8×10⁵Pa+Pm
=2.8×10⁵Pa+1.01×10³Nm
=3.8×10⁵Pa
Pm=(13.6g/cm³)(9.81)90.76)
=0. 136kg/m³)(9.8m/s²)(0.76m)
=1.01×10⁵Nm

3. A vertical test tube has 2.0 cm of oil (p=0.8g/cm³)flooting 8.0cm of water. What is the pressure at botton of the tube due to the fluid in it ?

P₁=Pgh
=800kg/m (9.81m/s²)(2×10⁻²)
=156.96
P₂=Pgh
=1000kg/m (9.81m/s²)(8×10⁻²)
=784.8
P₁+P₂ = 156.96+784.8
=0.94 kPa

4. The u-tube devuce connected to the tank. Wahat is the pressure in the tank if atmosphera pressure is 76cm of mercury? The density of mercury is 13.6g/cm³.

P= 76cm – 13.6g/cm³(9.81ms²)(5×10⁻²)
= 76cm – 13600kg/cm³(9.81ms²)(5×10⁻²)
= 101.325 kPa -66708
=94.65×10³
=95 kPa

5. The mass of a block of aluminiumis 25.0g. (a) what is its volume?(b0what will be the tension in a string that suspends the block when the block is totally submerged in water?the density of aluminium is 2700kg/m²?

V=m/p ft=vg(e-pal)
=0.025/2700 =(9.26×10⁻⁶×9.81)(1000-2700)
= 9.26 =0.154N

6. A solid aluminium cylinder with p= 2700kg/m³ has a measured of 67g in air and 45g when immersed in turpentine. Determine the density of turpentine.

FB=V₁P₁а
PT = FB / V1a
FB+Fr=mg
FB=mg-Fr
=(0.067×9.81)-(0.045×9.81)
=0.65727-0.4415
=0.21582
V1 = m/p = 0.067 / 2700
=2.481×10⁻⁵
P₁=0.21582
=(2.481×10⁻⁵)(9.81m/s²)
=8.9×10²kg/m³