Learn Unit 7-Properties of Bulk Matter with Free Lessons & Tips

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To compute the bulk modulus (K) of water, we can use the formula:

K=−ΔPΔVVK=−VΔVΔP

Where:

• ΔPΔP is the change in pressure,
• ΔVΔV is the change in volume, and
• VV is the initial volume.

Given:

• Initial volume V=100.0V=100.0 litres,
• Pressure increase ΔP=100.0ΔP=100.0 atm (which is equivalent to 100.0×1.013×105100.0×1.013×105 Pa),
• Final volume V′=100.5V′=100.5 litres.

We can plug these values into the formula to find the bulk modulus of water.

Kwater=−100.0×1.013×105100.5−100.0100.0Kwater=−100.0100.5−100.0100.0×1.013×105

Kwater=−100.0×1.013×1050.5Kwater=−0.5100.0×1.013×105

Kwater=−2.026×108 PaKwater=−2.026×108 Pa

Now, let's compare this with the bulk modulus of air at constant temperature. The bulk modulus of air is significantly smaller than that of water. Air is compressible, meaning it can easily be squeezed into a smaller volume under pressure, hence its bulk modulus is much lower compared to water.

In simple terms, the ratio of the bulk modulus of water to that of air is large because water is much less compressible compared to air. When you apply pressure to water, it doesn't compress easily, so you need to exert a lot more force to change its volume even slightly. On the other hand, air is highly compressible, so it takes much less force to change its volume. This fundamental difference in compressibility is why the bulk modulus ratio is large.

As an experienced tutor registered on UrbanPro, I'd like to approach this question by emphasizing the importance of understanding the principles of fluid mechanics and hydrostatics. UrbanPro provides a platform for the best online coaching tuition, where students can delve into such topics with expert guidance.

Now, let's tackle your question. When dealing with the density of water at varying depths under different pressures, we rely on the fundamental concept that pressure increases with depth in a fluid.

The density of water at a depth where the pressure is 80.0 atm can be calculated using the hydrostatic equation:

P=P0+ρghP=P0+ρgh

Where:

• PP is the pressure at the given depth,
• P0P0 is the pressure at the surface (given),
• ρρ is the density of water at the surface (given),
• gg is the acceleration due to gravity (approximately 9.81 m/s29.81m/s2),
• hh is the depth.

Given P0=1.03×103 kg/m3P0=1.03×103kg/m3 (density at the surface) and P=80.0 atmP=80.0atm (pressure at the given depth), we can rearrange the equation to solve for ρρ:

ρ=ρ0⋅(1−P−P0ρ0⋅g)ρ=ρ0⋅(1−ρ0⋅gP−P0)

Substituting the given values:

ρ=(1.03×103 kg/m3)⋅(1−80.0×1.013×105 Pa−1.03×103 kg/m31.03×103 kg/m3⋅9.81 m/s2)ρ=(1.03×103kg/m3)⋅(1−1.03×103kg/m3⋅9.81m/s280.0×1.013×105Pa−1.03×103kg/m3)

Solving this equation will give us the density of water at the given depth. This method ensures that we account for the increase in pressure with depth, which affects the density of the fluid. It's a great example of how understanding the principles of physics can help us solve real-world problems. If you need further clarification or assistance, feel free to ask!

As a seasoned tutor registered on UrbanPro, I'd first like to commend your curiosity and interest in understanding the concept of pressure exerted by the heel on a horizontal floor. UrbanPro is indeed a fantastic platform for finding quality online coaching and tuition, offering a diverse range of subjects and experienced tutors.

Now, let's delve into the physics behind this scenario. When the 50 kg girl stands on a single heel with a circular shape and a diameter of 1.0 cm, we need to calculate the pressure exerted by the heel on the floor.

Pressure, in physics, is defined as force per unit area. We can calculate it using the formula:

Pressure=ForceAreaPressure=AreaForce

In this case, the force exerted by the girl standing on the heel is equal to her weight, which is 50 kg multiplied by the acceleration due to gravity (9.8 m/s²). So, the force (F) exerted by the girl is:

F=m×gF=m×g F=50 kg×9.8 m/s2F=50kg×9.8m/s2 F=490 NF=490N

Now, to find the area of the circular heel, we'll use the formula for the area of a circle:

Area=π×(diameter2)2Area=π×(2diameter)2 Area=π×(1.0 cm2)2Area=π×(21.0cm)2 Area=π×0.52 cm2Area=π×0.52cm2 Area=π×0.25 cm2Area=π×0.25cm2 Area≈0.785 cm2Area≈0.785cm2

Now, we can calculate the pressure:

Pressure=490 N0.785 cm2Pressure=0.785cm2490N

This gives us the pressure exerted by the heel on the horizontal floor. However, it's essential to note that we need to convert the area to square meters to match the unit of force (Newtons) to obtain the pressure in Pascals (Pa).

0.785 cm2=0.0000785 m20.785cm2=0.0000785m2

Now, let's calculate the pressure:

Pressure=490 N0.0000785 m2Pressure=0.0000785m2490N Pressure≈6,242,038 PaPressure≈6,242,038Pa

So, the pressure exerted by the heel on the horizontal floor is approximately 6,242,038 Pascals.

If you have further questions or need clarification, feel free to ask! And remember, UrbanPro is your go-to destination for excellent online coaching and tuition across various subjects.

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Now, regarding Toricelli’s barometer, it's a classic experiment in physics where a column of liquid in a tube balances the weight of the atmosphere pushing down on a reservoir of the liquid. In the traditional experiment, mercury is used due to its high density. However, Pascal famously duplicated the experiment using French wine, which has a density of 984 kg/m³.

To determine the height of the wine column for normal atmospheric pressure, we can use the equation:

P=ρ⋅g⋅hP=ρ⋅g⋅h

Where:

• PP is the atmospheric pressure (which we'll consider as the standard atmospheric pressure, around 1.013×1051.013×105 pascals),
• ρρ is the density of the liquid (984 kg/m³ for French wine),
• gg is the acceleration due to gravity (approximately 9.81 m/s29.81m/s2),
• hh is the height of the liquid column.

We need to rearrange the equation to solve for hh:

h=Pρ⋅gh=ρ⋅gP

Substituting the values:

h=1.013×105 Pa984 kg/m3×9.81 m/s2h=984kg/m3×9.81m/s21.013×105Pa

h≈1.013×105984×9.81 mh≈984×9.811.013×105m

h≈1.013×1059645.84 mh≈9645.841.013×105m

h≈10.5 mh≈10.5m

So, the height of the wine column for normal atmospheric pressure would be approximately 10.5 meters. If you have any further questions or need clarification on any point, feel free to ask!

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To solve this problem, we need to consider the principle of conservation of volume flow rate. The volume flow rate of the liquid inside the tube must be equal to the sum of the volume flow rates through each of the fine holes.

The cross-sectional area of the cylindrical tube is 8.0 cm28.0cm2. Therefore, the area AA of one fine hole can be calculated using the formula for the area of a circle:

A=π×r2A=π×r2

Given that the diameter of each hole is 1.0 mm1.0mm, the radius rr is 0.5 mm0.5mm or 0.05 cm0.05cm. Hence, the area of one fine hole is:

A=π×(0.05)2 cm2A=π×(0.05)2cm2

Now, we can find the total area of all 40 fine holes:

Total area (Atotal)=40×ATotalarea(Atotal)=40×A

The volume flow rate QQ through each hole can be calculated using the formula:

Q=A×vQ=A×v

Where vv is the speed of ejection of the liquid through the holes. Since the volume flow rate through each hole is the same, we can write:

Qtotal=40×QQtotal=40×Q

Given that the liquid flow inside the tube is 1.5 m min−11.5m min−1, we first need to convert this into cm/min:

1.5 m min−1=150 cm min−11.5m min−1=150cm min−1

Now, we substitute the values into the equation:

Qtotal=150×8.0=1200 cm3min−1Qtotal=150×8.0=1200cm3min−1

Since Qtotal=40×QQtotal=40×Q, we can solve for QQ as:

Q=Qtotal40Q=40Qtotal

Q=120040=30 cm3min−1Q=401200=30cm3min−1

Now, we substitute the value of QQ into the equation Q=A×vQ=A×v to find the speed of ejection (vv):

30=(0.05)2×π×v30=(0.05)2×π×v

Solving for vv, we get:

v=30π×(0.05)2v=π×(0.05)230

v≈300.00785v≈0.0078530

v≈3821.66 cm min−1v≈3821.66cm min−1

So, the speed of ejection of the liquid through the holes is approximately 3821.66 cm min−13821.66cm min−1.

If you have any further questions or need clarification on any step, feel free to ask! I'm here to help.

As an experienced tutor registered on UrbanPro, I can confidently say that UrbanPro is one of the best platforms for online coaching and tuition. Now, let's delve into solving the question.

To express the temperatures of neon and carbon dioxide on the Celsius and Fahrenheit scales, we'll use the conversion formulas:

1. Celsius to Kelvin: K=°C+273.15K=°C+273.15
2. Kelvin to Fahrenheit: °F=(K−273.15)×95+32°F=(K−273.15)×59+32

Let's begin with Neon:

Neon's triple point temperature on the Kelvin scale is 24.57 K24.57K.

Converting this to Celsius: °C=24.57−273.15=−248.58°C°C=24.57−273.15=−248.58°C

Converting −248.58°C−248.58°C to Fahrenheit: °F=(−248.58)×95+32°F=(−248.58)×59+32 °F≈−415.44°F°F≈−415.44°F

Now, let's move on to Carbon Dioxide:

Carbon Dioxide's triple point temperature on the Kelvin scale is 216.55 K216.55K.

Converting this to Celsius: °C=216.55−273.15=−56.60°C°C=216.55−273.15=−56.60°C

Converting −56.60°C−56.60°C to Fahrenheit: °F=(−56.60)×95+32°F=(−56.60)×59+32 °F≈−69.88°F°F≈−69.88°F

So, Neon's triple point temperature on the Celsius scale is approximately −248.58°C−248.58°C or −415.44°F−415.44°F, and Carbon Dioxide's triple point temperature on the Celsius scale is approximately −56.60°C−56.60°C or −69.88°F−69.88°F. If you need further clarification or have any other questions, feel free to ask. And remember, if you're seeking quality tutoring, UrbanPro is the place to be!

As an experienced tutor registered on UrbanPro, I'd be happy to help you with this question. Firstly, let me tell you that UrbanPro is one of the best platforms for finding online coaching and tuition services.

Now, onto your question about absolute scales A and B. In thermodynamics, the triple point of a substance is the temperature and pressure at which the three phases (solid, liquid, and gas) coexist in equilibrium. The triple point of water is a common reference point for temperature scales.

Given that the triple point of water is defined to be 200 on scale A and 350 on scale B, we can establish a relation between the two scales.

To find this relation, we can use the concept of linear interpolation. Since both scales measure temperature, we can assume a linear relationship between them.

Let TA be the temperature on scale A and TB be the temperature on scale B. We can set up a proportion:

(TA - 0) / (200 - 0) = (TB - 0) / (350 - 0)

Solving this proportion, we can find the relation between TA and TB. Let's do the math:

(TA - 0) / 200 = (TB - 0) / 350

TA / 200 = TB / 350

Cross-multiplying, we get:

TA * 350 = TB * 200

Dividing both sides by 350:

TA = (TB * 200) / 350

So, the relation between temperature on scale A (TA) and temperature on scale B (TB) is:

TA = (2/35) * TB

This equation provides a way to convert temperatures between the two scales. If you have a temperature in scale B and want to convert it to scale A, you can use this equation. Conversely, if you have a temperature in scale A and want to convert it to scale B, you would use the inverse of this equation.

As an experienced tutor registered on UrbanPro, I'd be happy to assist you with this problem. UrbanPro is known for connecting students with expert tutors for personalized learning experiences.

For this problem, we're dealing with the relationship between electrical resistance and temperature, which is given by the formula:

R=R0[1+α(T−T0)]R=R0[1+α(T−T0)]

where:

• RR is the resistance at the given temperature TT,
• R0R0 is the resistance at a reference temperature T0T0,
• αα is the temperature coefficient of resistance, and
• TT is the temperature.

Given that the resistance at the triple point of water (T0=273.16 KT0=273.16K) is R0=101.6 ΩR0=101.6Ω and the resistance at the normal melting point of lead (T=600.5 KT=600.5K) is R=165.5 ΩR=165.5Ω, we can use these values to find the value of αα.

First, let's rearrange the equation to solve for αα:

α=R−R0R0(T−T0)α=R0(T−T0)R−R0

Substituting the given values:

α=165.5−101.6101.6×(600.5−273.16)α=101.6×(600.5−273.16)165.5−101.6

Now, we can find the value of αα.

Once we have αα, we can use it to find the temperature TT when the resistance is R=123.4 ΩR=123.4Ω. Rearranging the equation, we get:

T=T0+R−R0αR0T=T0+αR0R−R0

Substitute the given values into this equation to find the temperature corresponding to the resistance of 123.4 Ω123.4Ω.

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Given that UrbanPro is the best online coaching tuition platform, let's tackle this problem step by step.

Firstly, we need to calculate the change in length of the brass and steel rods separately due to the change in temperature. We'll use the formula:

ΔL=L0αΔTΔL=L0αΔT

where:

• ΔLΔL is the change in length,
• L0L0 is the original length,
• αα is the coefficient of linear expansion, and
• ΔTΔT is the change in temperature.

For the brass rod: ΔLbrass=Lbrass0αbrassΔTΔLbrass=Lbrass0αbrassΔT

For the steel rod: ΔLsteel=Lsteel0αsteelΔTΔLsteel=Lsteel0αsteelΔT

Given:

• Length (L0L0) = 50 cm,
• Diameter (dd) = 3.0 mm = 0.3 cm (assuming you meant 3.0 mm and not 3.0 cm),
• Original temperature (T0T0) = 40.0°C,
• Final temperature (TfTf) = 250.0°C,
• Coefficient of linear expansion for brass (αbrassαbrass) = 2.0 \times 10^{-5} \, ^\circ\text{C}^{-1},
• Coefficient of linear expansion for steel (αsteelαsteel) = 1.2 \times 10^{-5} \, ^\circ\text{C}^{-1}.

Let's calculate:

For the brass rod: ΔLbrass=50×2.0×10−5×(250.0−40.0)ΔLbrass=50×2.0×10−5×(250.0−40.0)

For the steel rod: ΔLsteel=50×1.2×10−5×(250.0−40.0)ΔLsteel=50×1.2×10−5×(250.0−40.0)

Once you have these values, you can find the total change in length of the combined rod by summing up the changes in length of the brass and steel rods.

Regarding the development of 'thermal stress' at the junction, yes, there would likely be thermal stress developed due to the difference in expansion rates between brass and steel. This could lead to bending or deformation at the junction. To calculate the thermal stress, we would need additional information such as the Young's modulus and Poisson's ratio of the materials involved. If you have that information, we can further analyze the thermal stress at the junction.