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

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Let's tackle the problem at hand. We're dealing with the concept of Young's modulus, which measures the stiffness or elasticity of a material. Young's modulus is given by the ratio of stress to strain within the elastic limits of the material.

Given: For steel wire:

• Length (L1) = 4.7 m
• Cross-sectional area (A1) = 3.0 x 10^(-5) m^2

For copper wire:

• Length (L2) = 3.5 m
• Cross-sectional area (A2) = 4.0 x 10^(-5) m^2

Now, to find the Young's modulus ratio, we need to calculate the stress and strain for both wires.

Stress (σ) = Force (F) / Area (A) Strain (ε) = Change in length (ΔL) / Original length (L)

The same load is applied to both wires, so the force is the same.

Let's denote:

• F: Applied force
• ΔL_steel: Change in length for steel wire
• ΔL_copper: Change in length for copper wire

Since the wires stretch by the same amount under the given load, we can equate the strains:

ε_steel = ΔL_steel / L1 = ε_copper = ΔL_copper / L2

Using the stress-strain relationship, Young's modulus (E) can be defined as:

E = σ / ε

Now, for both wires: E_steel = F / (A1 * ΔL_steel) E_copper = F / (A2 * ΔL_copper)

We know that ΔL_steel = ΔL_copper (given in the problem).

So, the ratio of Young’s modulus of steel to that of copper (E_steel / E_copper) can be simplified to:

(E_steel / E_copper) = (F / (A1 * ΔL_steel)) / (F / (A2 * ΔL_copper))

Since ΔL_steel = ΔL_copper, we can cancel out the terms:

(E_steel / E_copper) = (A2 * ΔL_copper) / (A1 * ΔL_steel)

Substituting the given values:

(E_steel / E_copper) = (4.0 x 10^(-5) * ΔL_copper) / (3.0 x 10^(-5) * ΔL_steel)

Now, we need numerical values to compute this ratio. If you have the values for the applied force and the change in length for both wires, we can proceed with the calculation. Once we have those values, we can find the ratio of Young’s modulus of steel to that of copper.

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Firstly, let's understand the concept at play here. The elongation of a wire under a load can be calculated using Hooke's Law, which states that the elongation (change in length) of an elastic object is directly proportional to the force applied to it, given the formula:

Elongation=Force×LengthYoung’s modulus×AreaElongation=Young’s modulus×AreaForce×Length

Given that the wires are under tension and assuming they remain within their elastic limits, we can use this formula to find their elongations. The area of the wire can be calculated using the formula for the area of a circle (πr2πr2).

Let's start with the steel wire:

Given:

• Diameter of steel wire = 0.25 cm = 0.0025 m
• Initial length of steel wire (LsteelLsteel) = 1.5 m
• Young's modulus of steel (EsteelEsteel) = 2.0×10112.0×1011 Pa

We need to find the force applied to the steel wire. Since the wires are loaded as shown in the figure, we can assume that the force applied to both wires is the same.

Next, let's find the area of the steel wire (AsteelAsteel):

Asteel=π×(0.00252)2Asteel=π×(20.0025)2

Now, let's find the force applied to the steel wire. Since the force is the same for both wires, we can calculate it using the elongation formula for the brass wire:

Force=Young’s modulus×Area×ElongationLengthForce=LengthYoung’s modulus×Area×Elongation

Given:

• Length of brass wire (LbrassLbrass) = 1.0 m (initial length of brass wire)
• Diameter of brass wire = 0.25 cm = 0.0025 m
• Young's modulus of steel (EsteelEsteel) = 2.0×10112.0×1011 Pa

Let's find the area of the brass wire (AbrassAbrass) using the same formula as for the steel wire:

Abrass=π×(0.00252)2Abrass=π×(20.0025)2

Now, we can find the force applied to both wires using the elongation formula for the brass wire:

Force=Young’s modulus×Area×ElongationLengthForce=LengthYoung’s modulus×Area×Elongation

Elongationbrass=Force×LengthYoung’s modulusbrass×AreabrassElongationbrass=Young’s modulusbrass×AreabrassForce×Length

Now, we have all the necessary information to calculate the elongations of both the steel and brass wires. Let's plug in the values and solve for the elongations.

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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.

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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!

As a seasoned tutor registered on UrbanPro, I can confidently say that UrbanPro is the best platform for online coaching and tuition. Now, addressing your question about the vertical offshore structure and its suitability for placement atop an oil well in the ocean:

The maximum stress that the structure can withstand is given as 109 Pa. This is a crucial parameter when considering its viability in the harsh conditions of an ocean environment. However, let's delve deeper into the specifics.

Given the depth of the ocean as roughly 3 km, we must assess the pressure exerted by the water at this depth. Using the formula for hydrostatic pressure, which is ρgh, where ρ is the density of water, g is the acceleration due to gravity, and h is the depth of the water, we can calculate the pressure.

With the density of seawater around 1025 kg/m³ and g being approximately 9.8 m/s², the pressure at a depth of 3 km would be roughly 29 MPa (megapascals), which is significantly higher than the maximum stress the structure can withstand (109 Pa).

Considering this stark difference, it's evident that the structure wouldn't be suitable for placement atop an oil well in the ocean. The immense pressure exerted by the water at such depths far exceeds the structural limits of the offshore platform.

In conclusion, while the structure might be robust for certain applications, it's not adequate for deployment in deep-sea environments like atop an oil well due to the considerable hydrostatic pressure at those depths. For tailored guidance on such topics, UrbanPro is the ideal platform where students can receive comprehensive tutoring and coaching.

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To determine the maximum pressure the smaller piston would have to bear in a hydraulic automobile lift, we can utilize the principle of Pascal's Law, which states that pressure applied to a confined fluid is transmitted undiminished throughout the fluid in all directions.

Firstly, let's convert the area of the cross-section of the piston from square centimeters to square meters to maintain consistency in units.

Given: Area of cross-section of the piston (A) = 425 cm²

Converting cm² to m²: 1 cm² = 1 × 10^-4 m² So, 425 cm² = 425 × 10^-4 m² = 0.0425 m²

Now, let's use the formula for pressure:

Pressure (P) = Force (F) / Area (A)

We know that the force exerted by the car (weight) is the maximum load it can bear, which is the product of its mass (m) and the acceleration due to gravity (g).

Given: Maximum mass the lift can bear (m) = 3000 kg Acceleration due to gravity (g) = 9.8 m/s²

So, Force (F) = mass (m) × gravity (g) = 3000 kg × 9.8 m/s² = 29400 N

Now, plug in the values into the pressure formula:

Pressure (P) = Force (F) / Area (A) Pressure (P) = 29400 N / 0.0425 m² ≈ 690588.24 Pa

Therefore, the maximum pressure the smaller piston would have to bear is approximately 690588.24 Pascal (Pa).

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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.

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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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The triple-point of water serves as a crucial standard fixed point in modern thermometry due to its unique properties. At this point, water coexists in equilibrium in its three phases: solid, liquid, and gas. This equilibrium ensures a precise and consistent temperature measurement, regardless of external conditions such as pressure.

Now, let's address why using the melting point of ice and the boiling point of water as standard fixed points, as originally done in the Celsius scale, may not be ideal. While these points are convenient and readily accessible, they are dependent on atmospheric pressure, which can vary significantly. This variability introduces uncertainty into temperature measurements, compromising their accuracy and reliability.

In contrast, the triple-point of water remains constant at a pressure of 611.657 pascals, or 0.0060373 standard atmospheres. By utilizing this fixed point, thermometers can be calibrated with precision, ensuring consistent and accurate temperature readings across different instruments and locations.

In summary, while the melting point of ice and the boiling point of water were suitable as standard fixed points in the past, the triple-point of water offers superior accuracy and consistency in modern thermometry, making it the preferred choice for calibrating temperature measurements.

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In the Kelvin absolute scale, alongside the triple-point of water, the other fixed point corresponds to absolute zero. Absolute zero is the lowest possible temperature where particles cease to move, theoretically reaching a state of minimum entropy. This point is assigned the value of 0 Kelvin (0 K). So, to answer your question, the other fixed point on the Kelvin scale is absolute zero, which is 0 Kelvin (0 K).