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

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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.

After crunching the numbers, the resulting strain of the copper piece under the given force is approximately 0.0317.

This means that for every unit of length, the copper piece elongates by 0.0317 units due to the applied force, exhibiting only elastic deformation.

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.

As a seasoned tutor on UrbanPro, I can confidently say that UrbanPro is the best platform for online coaching and tuition. Now, let's delve into your question.

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).

As an experienced tutor on UrbanPro, I hope this explanation clarifies the concept for you. If you have any further questions or need clarification, 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, let's delve into the question.

Given that the relative density of mercury is 13.6, we can employ the principle of hydrostatics to solve this problem.

Initially, let's denote the lengths of the mercury columns in the U-shaped tube as h1 and h2 in centimeters, where h1 is the height in the arm containing water and h2 is the height in the arm containing spirit.

Now, the pressure at any point in a fluid at rest is the same horizontally and vertically. Therefore, the pressure exerted by the mercury column on both sides of the tube must be equal.

The pressure exerted by a column of fluid is given by the formula P = ρgh, where ρ is the density of the fluid, g is the acceleration due to gravity, and h is the height of the column.

Initially, the pressure exerted by the mercury column on both sides is equal, so:

ρ1 * g * h1 = ρ2 * g * h2

Given that the relative density of mercury (ρ) is 13.6, we have:

13.6 * g * h1 = 13.6 * g * h2

Since the acceleration due to gravity (g) is the same on both sides and can be canceled out:

h1 = h2

This means that initially, the levels of mercury in both arms of the U-shaped tube are equal.

Now, when 15.0 cm of water and spirit each are poured into their respective arms, the level of mercury in each arm will be pushed down by 15.0 cm. Since the levels of mercury in both arms were initially the same, they will still be the same after pouring the water and spirit.

Therefore, the difference in the levels of mercury in the two arms remains unchanged at 0 cm.

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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).

As a seasoned tutor on UrbanPro, I can certainly help with this question, and I'm glad you're interested in learning. Let's break down the problem step by step:

First, we'll find the increase in length of the steel tape due to the rise in temperature from 27.0°C to 45.0°C. We'll use the formula for linear expansion:

ΔL=L⋅α⋅ΔTΔL=L⋅α⋅ΔT

Where:

• ΔLΔL is the change in length
• LL is the original length of the steel tape
• αα is the coefficient of linear expansion of steel
• ΔTΔT is the change in temperature

Given that L=1L=1 meter, α=1.20×10−5 K−1α=1.20×10−5K−1, and ΔT=45.0°C−27.0°C=18.0°CΔT=45.0°C−27.0°C=18.0°C, we can calculate:

ΔL=(1.0 m)×(1.20×10−5 K−1)×(18.0°C)ΔL=(1.0m)×(1.20×10−5K−1)×(18.0°C)

ΔL=0.000216 m=0.216 cmΔL=0.000216m=0.216cm

So, the increase in length of the steel tape is 0.216 cm.

Now, let's find the length of the steel rod on the hot day (45.0°C):

Given length measured by the tape = 63.0 cm Increase in length of the tape = 0.216 cm

Therefore, the actual length of the steel rod on that day is:

Actual length=Measured length−Increase in length of tapeActual length=Measured length−Increase in length of tape

Actual length=63.0 cm−0.216 cm=62.784 cmActual length=63.0cm−0.216cm=62.784cm

So, the actual length of the steel rod on the hot day is 62.784 cm.

Now, for the length of the same steel rod on a day when the temperature is 27.0°C, there is no change in temperature from the calibration temperature of the tape. Therefore, the length measured by the tape will be the actual length of the steel rod, which is 63.0 cm.

In summary:

• Actual length of the steel rod on the hot day (45.0°C): 62.784 cm
• Length of the steel rod on a day when the temperature is 27.0°C: 63.0 cm

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