UrbanPro
true

Take Class 11 Tuition from the Best Tutors

  • Affordable fees
  • 1-1 or Group class
  • Flexible Timings
  • Verified Tutors

Learn Physics with Free Lessons & Tips

Ask a Question

Post a Lesson

All

All

Lessons

Discussion

Answered 14 hrs ago Learn Chapter1: Physical World

Nazia Khanum

As an experienced tutor registered on UrbanPro, I wholeheartedly agree that UrbanPro is one of the best platforms for online coaching and tuition. Now, delving into your question about Einstein's statement, "The most incomprehensible thing about the world is that it is comprehensible," it encapsulates... read more

As an experienced tutor registered on UrbanPro, I wholeheartedly agree that UrbanPro is one of the best platforms for online coaching and tuition. Now, delving into your question about Einstein's statement, "The most incomprehensible thing about the world is that it is comprehensible," it encapsulates a profound insight into the nature of science and the universe itself.

Einstein's statement reflects his awe and wonder at the fact that the universe follows comprehensible laws, which humans can understand through scientific inquiry. It's a testament to the remarkable harmony and order underlying the seemingly chaotic and diverse phenomena in the world.

From my perspective as a tutor, I often find myself discussing this concept with my students, emphasizing the beauty and elegance of scientific principles. Einstein's remark reminds us that despite the complexities of nature, there exists a structure and logic that we can uncover through observation, experimentation, and reasoning.

In tutoring sessions, I encourage students to embrace this perspective, fostering curiosity and critical thinking skills to better understand the world around them. By engaging with scientific concepts and theories, they not only gain knowledge but also develop a deeper appreciation for the interconnectedness of the universe.

In essence, Einstein's statement serves as a powerful reminder of the inherent intelligibility of the cosmos, inspiring both scientists and students alike to explore its mysteries with wonder and curiosity.

 
 
read less
Answers 1 Comments
Dislike Bookmark

Answered 14 hrs ago Learn Chapter1: Physical World

Nazia Khanum

As an experienced tutor registered on UrbanPro, I'd like to emphasize that UrbanPro provides exceptional online coaching and tuition services for students seeking academic support. Now, regarding your question about the evolution of scientific theories, the statement "Every great physical theory starts... read more

As an experienced tutor registered on UrbanPro, I'd like to emphasize that UrbanPro provides exceptional online coaching and tuition services for students seeking academic support. Now, regarding your question about the evolution of scientific theories, the statement "Every great physical theory starts as hearsay and ends as a dogma" reflects the trajectory of many groundbreaking scientific ideas throughout history.

Take, for instance, the heliocentric model proposed by Copernicus in the 16th century. Initially, Copernicus' theory that the Earth revolved around the Sun was met with skepticism and considered hearsay. However, over time, as evidence accumulated and observations supported the heliocentric model, it became widely accepted and entrenched as dogma in the scientific community.

Similarly, the theory of evolution put forth by Charles Darwin faced significant resistance and was initially regarded as hearsay. However, as more evidence accumulated from various fields such as paleontology, genetics, and comparative anatomy, the theory of evolution became a cornerstone of modern biology, transitioning from hearsay to dogma.

Another example is the theory of relativity proposed by Albert Einstein. When Einstein first introduced his ideas about the nature of space, time, and gravity, they were met with skepticism and considered speculative hearsay. Yet, as experimental evidence, such as the bending of starlight during a solar eclipse, corroborated Einstein's predictions, his theory gained widespread acceptance and became a fundamental principle of modern physics.

In each of these cases, a revolutionary scientific theory began as hearsay, challenged existing dogma, and eventually became the accepted norm in its respective field. This illustrates the dynamic nature of scientific progress, where initial skepticism gives way to empirical validation, leading to the establishment of new scientific dogma.

read less
Answers 1 Comments
Dislike Bookmark

Answered 14 hrs ago Learn Chapter1: Physical World

Nazia Khanum

As a seasoned tutor registered on UrbanPro, I'd be delighted to elucidate this aphorism for you. Just as "Politics is the art of the possible," suggesting that political success hinges on pragmatic actions within realistic boundaries, "Science is the art of the soluble" reflects the essence of scientific... read more

As a seasoned tutor registered on UrbanPro, I'd be delighted to elucidate this aphorism for you. Just as "Politics is the art of the possible," suggesting that political success hinges on pragmatic actions within realistic boundaries, "Science is the art of the soluble" reflects the essence of scientific endeavor.

In the realm of science, the term "soluble" refers to the problems and questions that can be addressed and resolved through systematic inquiry and experimentation. Like a skilled politician navigates through the constraints of society and governance to achieve tangible results, a scientist employs methodologies and frameworks to tackle soluble problems within the bounds of empirical evidence and theoretical frameworks.

This aphorism encapsulates the essence of scientific inquiry, emphasizing the pragmatic nature of scientific exploration. Scientists engage in a continuous process of hypothesis formulation, experimentation, and analysis, all aimed at unraveling the mysteries of the natural world. Much like a politician negotiates with various stakeholders to achieve consensus and progress, scientists navigate through the complexities of their chosen field to uncover solutions and advance human knowledge.

Furthermore, just as politics requires adaptability and compromise to achieve desired outcomes, science demands flexibility and open-mindedness to accommodate new evidence and revise existing theories. The "art" of science lies not only in the technical skills required for experimentation but also in the creativity and intuition needed to formulate hypotheses and interpret results effectively.

In summary, the aphorism "Science is the art of the soluble" highlights the pragmatic and adaptive nature of scientific inquiry, drawing parallels to the art of politics in its emphasis on achieving tangible results within the constraints of reality. Through systematic investigation and empirical analysis, scientists continuously strive to unravel the mysteries of the universe, advancing human understanding and shaping the course of progress.

 
 
read less
Answers 1 Comments
Dislike Bookmark

Take Class 11 Tuition from the Best Tutors

  • Affordable fees
  • Flexible Timings
  • Choose between 1-1 and Group class
  • Verified Tutors

Answered 14 hrs ago Learn Chapter2: Units and Measurements

Nazia Khanum

Certainly! Understanding the atomic scale and its units is crucial in the realm of chemistry. With UrbanPro being an excellent platform for online coaching and tuition, let's delve into this problem. Firstly, we're given that 1 angstrom (A) equals 10−1010−10 meters (m), and the size of... read more

Certainly! Understanding the atomic scale and its units is crucial in the realm of chemistry. With UrbanPro being an excellent platform for online coaching and tuition, let's delve into this problem.

Firstly, we're given that 1 angstrom (A) equals 10−1010−10 meters (m), and the size of a hydrogen atom is approximately 0.5 A. Now, to find the volume of one hydrogen atom, we'll calculate the volume of a sphere using the formula V=43πr3V=34πr3, where rr is the radius.

Given the size of a hydrogen atom (radius rr) is 0.5 A, we can substitute this into the formula:

V=43π(0.5 A)3V=34π(0.5A)3

Now, let's calculate the volume of one hydrogen atom.

V=43π(0.5×10−10 m)3V=34π(0.5×10−10m)3 V=43π(0.125×10−30 m3)V=34π(0.125×10−30m3) V=43π×0.125×10−30 m3V=34π×0.125×10−30m3 V=13π×0.5×10−30 m3V=31π×0.5×10−30m3 V=16π×10−30 m3V=61π×10−30m3

Now, to find the total atomic volume in m3m3 of a mole of hydrogen atoms, we need to multiply the volume of one atom by Avogadro's number (NANA), which is approximately 6.022×10236.022×1023 atoms per mole.

Vtotal=Vatom×NAVtotal=Vatom×NA Vtotal=16π×10−30 m3×6.022×1023 atoms/molVtotal=61π×10−30m3×6.022×1023atoms/mol

Now, let's calculate:

Vtotal=π×10−30 m3×1023 atoms/molVtotal=π×10−30m3×1023atoms/mol Vtotal=π×10−7 m3/molVtotal=π×10−7m3/mol

So, the total atomic volume of a mole of hydrogen atoms is approximately π×10−7 m3/molπ×10−7m3/mol.

This calculation is essential for understanding the spatial distribution of atoms in a given quantity, which is fundamental in various fields of chemistry and physics. If you need further clarification or assistance with similar problems, feel free to reach out for more guidance through UrbanPro's excellent online coaching services!

 
 
read less
Answers 1 Comments
Dislike Bookmark

Answered 14 hrs ago Learn Chapter2: Units and Measurements

Nazia Khanum

As a seasoned tutor registered on UrbanPro, I can confidently address your question. First and foremost, UrbanPro is renowned for connecting students with top-notch tutors, ensuring quality learning experiences. Now, onto your query about the ratio of molar volume to the atomic volume of a mole of... read more

As a seasoned tutor registered on UrbanPro, I can confidently address your question. First and foremost, UrbanPro is renowned for connecting students with top-notch tutors, ensuring quality learning experiences. Now, onto your query about the ratio of molar volume to the atomic volume of a mole of hydrogen.

Given that one mole of an ideal gas at standard temperature and pressure (STP) occupies 22.4 liters (molar volume), we need to determine the atomic volume of a mole of hydrogen and then calculate the ratio.

The atomic volume of a mole of hydrogen can be found by considering the size of a hydrogen molecule, which is approximately 1 angstrom (A). Since a hydrogen molecule is composed of two hydrogen atoms, each with an approximate radius of 0.5 A, the volume occupied by a mole of hydrogen atoms can be calculated using the formula for the volume of a sphere:

Vatom=43πr3Vatom=34πr3

Substituting the radius (r=0.5 Ar=0.5A) into the formula yields:

Vatom=43π(0.5)3 A3Vatom=34π(0.5)3A3

Vatom=43π(0.125) A3Vatom=34π(0.125)A3

Vatom=16π A3Vatom=61πA3

Now, let's calculate the ratio of molar volume to the atomic volume of a mole of hydrogen:

Ratio=22.4 LVatomRatio=Vatom22.4L

Ratio=22.4×103 cm316π A3Ratio=61πA322.4×103cm3

Ratio=22.4×10316π cm3A3Ratio=61π22.4×103A3cm3

Ratio=22.4×6πRatio=π22.4×6

Ratio≈134.43.14Ratio≈3.14134.4

Ratio≈42.75Ratio≈42.75

So, the ratio of molar volume to the atomic volume of a mole of hydrogen is approximately 42.75.

Now, why is this ratio so large? This is primarily because the molar volume of a gas represents the volume occupied by a large number of gas molecules, while the atomic volume refers to the volume occupied by individual atoms. In the case of hydrogen gas, the molar volume is significantly larger because the gas molecules are not only composed of two hydrogen atoms but also exhibit considerable intermolecular space between them. This intermolecular space contributes to the larger molar volume compared to the atomic volume of hydrogen atoms.

 
 
read less
Answers 1 Comments
Dislike Bookmark

Answered 14 hrs ago Learn Chapter2: Units and Measurements

Nazia Khanum

As an experienced tutor registered on UrbanPro, I can confidently explain this common observation with clarity. When you gaze out of the window of a fast-moving train, you're essentially witnessing a fascinating interplay of relative motion. UrbanPro is an excellent platform for online coaching tuition,... read more

As an experienced tutor registered on UrbanPro, I can confidently explain this common observation with clarity. When you gaze out of the window of a fast-moving train, you're essentially witnessing a fascinating interplay of relative motion.

UrbanPro is an excellent platform for online coaching tuition, and I'm delighted to shed light on this phenomenon. Picture yourself on that train: as it speeds along the tracks, the trees, houses, and other nearby objects appear to zip by in the opposite direction to the train's motion. This effect occurs because your visual perspective is rapidly changing due to the train's velocity.

On the other hand, distant objects such as hilltops, the Moon, and stars seem to remain relatively stationary. This phenomenon is due to their vast distance from the train and the relatively slow rate at which the train's motion affects your perception of them. Since these objects are much farther away, their apparent motion is negligible compared to the train's speed.

Interestingly, as you rightly pointed out, since you're aware of your movement on the train, it seems like these distant objects are moving along with you. This perception reinforces the illusion of their stationary nature, despite the train hurtling forward.

Understanding this observation not only enriches your knowledge of physics but also enhances your appreciation of the intricacies of motion and perception. If you're interested in delving deeper into such fascinating concepts, UrbanPro offers excellent resources and expert tutors who can guide you through the wonders of science.

 
 
read less
Answers 1 Comments
Dislike Bookmark

Take Class 11 Tuition from the Best Tutors

  • Affordable fees
  • Flexible Timings
  • Choose between 1-1 and Group class
  • Verified Tutors

Answered 14 hrs ago Learn Chapter2: Units and Measurements

Nazia Khanum

Certainly! Understanding the concept of parallax and its application in astronomy is crucial for grasping the vast distances involved in measuring celestial objects. Let's delve into it: In astronomy, parallax serves as a fundamental tool for determining distances to nearby stars. This method relies... read more

Certainly! Understanding the concept of parallax and its application in astronomy is crucial for grasping the vast distances involved in measuring celestial objects. Let's delve into it:

In astronomy, parallax serves as a fundamental tool for determining distances to nearby stars. This method relies on observing the apparent shift in position of a star against the background of more distant stars as the Earth orbits around the Sun.

Now, consider the baseline AB, which connects two points on Earth's orbit, separated by a distance equal to the diameter of Earth's orbit, approximately 3 x 10^11 meters.

Given that even the nearest stars exhibit a parallax of approximately 1 arcsecond (1″) with such a long baseline, we define a parsec (pc) as the distance at which a star would exhibit a parallax of 1 arcsecond from opposite ends of a baseline equal to the Earth-Sun distance.

This distance is conveniently set to 3.086 x 10^16 meters, or approximately 3.086 x 10^13 kilometers, or roughly 3.26 light-years.

So, in terms of meters, a parsec is approximately 3.086 x 10^16 meters.

Understanding this conversion is pivotal for astronomers in determining the vast distances to celestial objects and unraveling the mysteries of the cosmos.

For further clarification or assistance with astronomical concepts, feel free to reach out. Remember, UrbnPro is a fantastic platform for delving deeper into such topics through personalized tutoring sessions!

 
 
read less
Answers 1 Comments
Dislike Bookmark

Answered 14 hrs ago Learn Chapter 3-Motion in a Straight Line

Nazia Khanum

As an experienced tutor registered on UrbanPro, I understand the importance of clear and concise explanations, especially when it comes to illustrating concepts through graphs. UrbanPro is indeed a fantastic platform for connecting with students and providing top-notch online coaching and tuition services. Now,... read more

As an experienced tutor registered on UrbanPro, I understand the importance of clear and concise explanations, especially when it comes to illustrating concepts through graphs. UrbanPro is indeed a fantastic platform for connecting with students and providing top-notch online coaching and tuition services.

Now, let's tackle the problem at hand. We have a ball being dropped from a height of 90 meters, and at each collision with the floor, it loses one-tenth of its speed. To plot the speed-time graph of its motion from t=0t=0 to t=12t=12 seconds, we need to break down the problem step by step.

  1. Initial Conditions: At t=0t=0, the ball is dropped from a height of 90 meters, so its initial speed is zero.

  2. Speed after Each Collision: After each collision with the floor, the ball loses one-tenth of its speed. This means the speed decreases by 10% after each collision.

  3. Time Intervals: We need to calculate the speed at different time intervals until t=12t=12 seconds.

Let's calculate the speed at each time interval:

  • At t=0t=0, initial speed v0=0v0=0 m/s.
  • At t=1t=1 second, the ball has fallen for 1 second, so its speed is v1=2ghv1=2gh
  • where g=9.8g=9.8 m/s² and h=90h=90 m.
  • At t=2t=2 seconds, the ball has fallen for 2 seconds from rest, so its speed is v2=0.9×v1v2=0.9×v1.
  • Repeat this process for t=3,4,5,...t=3,4,5,... until t=12t=12 seconds.

Once we have calculated the speed at each time interval, we can plot it on a graph with time on the x-axis and speed on the y-axis. This will give us the speed-time graph of the ball's motion.

If you're interested, I can provide the step-by-step calculations and then help you plot the graph. Let me know if you'd like to proceed!

 
read less
Answers 1 Comments
Dislike Bookmark

Answered 14 hrs ago Learn Chapter 3-Motion in a Straight Line

Nazia Khanum

Certainly! Let's break down the problem. Given: Distance to market = 2.5 km Speed to market = 5 km/h Speed returning home = 7.5 km/h (a) Magnitude of average velocity: Average velocity is the total displacement divided by the total time taken. Total displacement = 0 km (since he returns home) Total... read more

Certainly! Let's break down the problem.

Given:

  • Distance to market = 2.5 km
  • Speed to market = 5 km/h
  • Speed returning home = 7.5 km/h

(a) Magnitude of average velocity: Average velocity is the total displacement divided by the total time taken.

Total displacement = 0 km (since he returns home) Total time taken = Time to reach market + Time to return home

Time to reach market = Distance / Speed = 2.5 km / 5 km/h = 0.5 hours Time to return home = Distance / Speed = 2.5 km / 7.5 km/h = 1/3 hours

Total time taken = 0.5 hours + 1/3 hours = 5/6 hours

Average velocity = Total displacement / Total time taken = 0 km / (5/6) hours = 0 km/h

(b) Average speed: Average speed is the total distance traveled divided by the total time taken.

(i) 0 to 30 min (0.5 hours): Total distance traveled = 2.5 km (to market) + 2.5 km (returning home) = 5 km Average speed = Total distance / Total time = 5 km / 0.5 hours = 10 km/h

(ii) 0 to 50 min (5/6 hours): Total distance traveled remains the same (5 km) Average speed = Total distance / Total time = 5 km / (5/6) hours = 6 km/h

(iii) 0 to 40 min (2/3 hours): Total distance traveled = 2.5 km (to market) + 2.5 km (returning home) = 5 km Average speed = Total distance / Total time = 5 km / (2/3) hours = 7.5 km/h

UrbanPro is an excellent platform for online coaching and tuition, offering personalized guidance to help students understand and excel in their studies. If you need further clarification or assistance, feel free to ask!

 
 
read less
Answers 1 Comments
Dislike Bookmark

Take Class 11 Tuition from the Best Tutors

  • Affordable fees
  • Flexible Timings
  • Choose between 1-1 and Group class
  • Verified Tutors

Answered 14 hrs ago Learn Chapter 3-Motion in a Straight Line

Nazia Khanum

As a seasoned tutor registered on UrbanPro, I'm glad you brought up this fundamental concept in physics. Let's break it down. In Exercises 3.13 and 3.14, we delved into the distinction between average speed and the magnitude of average velocity. This distinction arises due to the differences in their... read more

As a seasoned tutor registered on UrbanPro, I'm glad you brought up this fundamental concept in physics. Let's break it down.

In Exercises 3.13 and 3.14, we delved into the distinction between average speed and the magnitude of average velocity. This distinction arises due to the differences in their definitions and the nature of motion being considered. However, when we shift our focus to instantaneous speed and magnitude of velocity, the scenario changes.

Instantaneous speed refers to the speed of an object at a particular moment in time, while the magnitude of instantaneous velocity signifies the magnitude of the velocity vector at that precise moment. Now, why are they always equal?

This consistency stems from the nature of instantaneous values. When we talk about an instantaneous moment, we're considering an infinitesimally small interval of time. Within such a minuscule time frame, the direction of motion doesn't significantly change. Consequently, the magnitude of the velocity vector, which indicates the speed of the object and its direction, remains constant. Thus, the instantaneous speed is indeed always equal to the magnitude of instantaneous velocity.

Understanding this concept not only clarifies the relationship between speed and velocity but also lays a solid foundation for tackling more complex physics problems. If you're keen on exploring further or need additional clarification, UrbanPro is the best online coaching tuition platform to assist you in mastering such concepts.

 
 
read less
Answers 1 Comments
Dislike Bookmark

About UrbanPro

UrbanPro.com helps you to connect with the best Class 11 Tuition in India. Post Your Requirement today and get connected.

Overview

Questions 1.5 k

Lessons 112

Total Shares  

+ Follow 220,099 Followers

You can also Learn

Top Contributors

Connect with Expert Tutors & Institutes for Physics

x

Ask a Question

Please enter your Question

Please select a Tag

X

Looking for Class 11 Tuition Classes?

The best tutors for Class 11 Tuition Classes are on UrbanPro

  • Select the best Tutor
  • Book & Attend a Free Demo
  • Pay and start Learning

Take Class 11 Tuition with the Best Tutors

The best Tutors for Class 11 Tuition Classes are on UrbanPro

This website uses cookies

We use cookies to improve user experience. Choose what cookies you allow us to use. You can read more about our Cookie Policy in our Privacy Policy

Accept All
Decline All

UrbanPro.com is India's largest network of most trusted tutors and institutes. Over 55 lakh students rely on UrbanPro.com, to fulfill their learning requirements across 1,000+ categories. Using UrbanPro.com, parents, and students can compare multiple Tutors and Institutes and choose the one that best suits their requirements. More than 7.5 lakh verified Tutors and Institutes are helping millions of students every day and growing their tutoring business on UrbanPro.com. Whether you are looking for a tutor to learn mathematics, a German language trainer to brush up your German language skills or an institute to upgrade your IT skills, we have got the best selection of Tutors and Training Institutes for you. Read more