I am very expert in teaching the basics with simple examples and make to understand the concepts very simple way. It will helpful for the students...
In my experience providing tuition for Class 11 students, I aimed to create a supportive and engaging learning environment tailored to their unique...
I have been teaching students since past 8 years and in this journey many of them were able to crack their examination.My way of teaching makes approach...
Do you need help in finding the best teacher matching your requirements?
Post your requirement nowI've been passionately teaching chemistry for class 11th, 12th, JEE Mains, and NEET UG for the past 10 years. Armed with an MSc in Chemistry, my mission...
I am a home tutor I teach since 2016. my teaching style is easy and impactful I use examples and mostly I use the questioner method mostly I used...
I have more than 9 years of experience teaching accounts and economics to students. With an educational background in MCom and CA, I was the topper...
It was always a pleasant experience to act as a bridge for the students of Class 11 to make the journey of the students from Class 10 to Class 11...
Throughout my teaching experience focused on Class 11, I have strived to create an engaging and effective learning environment that caters to the...
I am an experienced, qualified teacher and tutor with over 8 years of experience in teaching Accountancy, Economics and Business Studies to class...
I am teaching since 20011 of class 12th and 12 students. My subject of teaching is mathematics CBSC board.
Raghav attended Class 11 Tuition
"Extremely helpful. Has been teaching me for 4 years and I am majoring in Computer..."
Hemagowri attended Class 11 Tuition
"My son really liked his teaching . As a result, he has done his exams well. I would..."
Maya attended Class 11 Tuition
"Amit's method of teaching is very good. He ensures the student understands the subject..."
Saransh attended Class 11 Tuition
"The time I was preparing for Medical and Engineering Entrance, he supported me in..."
Nitish attended Class 11 Tuition
"He is an ideal Teacher / Lecturer. He has excellent command on his subject."
Debanshu attended Class 11 Tuition
"Extremely good teacher. Very helpful and responsible. Comes on time most of the days..."
Rohan attended Class 11 Tuition
" I was very weak in maths but not now, scored 78. best thing abt sir is he knows..."
Piyush attended Class 11 Tuition
"I am writing this to support Narayan Jalan for being the pioneer in teaching students...."
Ask a Question
Post a LessonAnswered on 14 Apr Learn CBSE/Class 11/Science/Physics/Unit 7-Properties of Bulk Matter/Chapter 10-Mechanical Properties of Fluids
Nazia Khanum
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.
Answered on 14 Apr Learn CBSE/Class 11/Science/Physics/Unit 7-Properties of Bulk Matter/Chapter 10-Mechanical Properties of Fluids
Nazia Khanum
As an experienced tutor registered on UrbanPro, I'm delighted to assist you with your question. UrbanPro is indeed a fantastic platform for online coaching and tuition needs.
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:
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!
Answered on 14 Apr Learn CBSE/Class 11/Science/Physics/Unit 7-Properties of Bulk Matter/Chapter 10-Mechanical Properties of Fluids
Nazia Khanum
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.
Answered on 14 Apr Learn CBSE/Class 11/Science/Physics/Unit 7-Properties of Bulk Matter/Chapter 10-Mechanical Properties of Fluids
Nazia Khanum
As a seasoned 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 your question about Bernoulli's equation and its applicability to the flow of water in a rapidly moving river.
Bernoulli's equation is indeed a powerful tool for describing fluid flow, including the flow of water. However, its application to rapidly moving rivers requires some careful consideration.
In a rapidly moving river, the flow is often turbulent, meaning it's chaotic and irregular. Bernoulli's equation assumes steady, non-turbulent flow, which may not always hold true in such scenarios. Additionally, the presence of obstacles like rocks and rapids can introduce complexities that may not be fully captured by Bernoulli's equation alone.
That said, Bernoulli's equation can still provide valuable insights into the behavior of water in a rapidly moving river under certain conditions. It can help us understand factors such as pressure changes, velocity variations, and energy conservation along different points in the river.
However, it's essential to supplement Bernoulli's equation with other principles, such as those from fluid dynamics and turbulence theory, to gain a more comprehensive understanding of the flow dynamics in rapidly moving rivers.
So, while Bernoulli's equation can offer some insights, it's not always the sole answer when describing the flow of water in rapidly moving rivers. Instead, it should be considered as part of a broader toolkit for analyzing fluid dynamics in such environments.
Answered on 14 Apr Learn CBSE/Class 11/Science/Physics/Unit 7-Properties of Bulk Matter/Chapter 10-Mechanical Properties of Fluids
Nazia Khanum
As a seasoned tutor registered on UrbanPro, I can confidently affirm that UrbanPro is one of the best platforms for online coaching and tuition. Now, let's delve into the fascinating realm of fluid dynamics and Bernoulli's equation.
When it comes to applying Bernoulli's equation, whether you use gauge pressure or absolute pressure does indeed matter. Bernoulli's equation describes the conservation of energy along a streamline in a fluid flow. It states that the total mechanical energy per unit mass in a fluid remains constant along a streamline, neglecting friction and other dissipative forces.
Now, if you're using gauge pressure, you're measuring pressure relative to atmospheric pressure. In contrast, absolute pressure includes atmospheric pressure as a reference point. So, in essence, using gauge pressure implies that you're already factoring out atmospheric pressure from your calculations.
When applying Bernoulli's equation, the choice between gauge and absolute pressure depends on the context of the problem. If the problem involves pressures relative to atmospheric pressure, gauge pressure is appropriate. However, if you need to consider the absolute pressure in the system, you should use absolute pressure in your calculations.
For instance, in scenarios like airflow over an airfoil or water flow through a pipe, gauge pressure might suffice since atmospheric pressure acts equally on all points. But in situations where you're dealing with pressures inside a closed vessel or a vacuum system, absolute pressure becomes crucial.
In conclusion, while both gauge and absolute pressures have their applications, understanding when to use each is vital for accurate and meaningful application of Bernoulli's equation in fluid dynamics.
Ask a Question