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Verify that the SI unit of \(h \rho g\) is \(\mathrm{N} / \mathrm{m}^{2}\)

Short Answer

Expert verified
We verified that the SI unit of the expression "丑蚁驳" is 狈/尘虏 by identifying the SI units of the variables (h: m, 蚁: 办驳/尘鲁, g: 尘/蝉虏), multiplying them, and simplifying the result to obtain \(\frac{N}{m^2}\).

Step by step solution

01

List the SI units of the given variables

1. h: Meters (m) 2. 蚁: Kilograms per cubic meter (办驳/尘鲁) 3. g: Meters per second squared (尘/蝉虏)
02

Multiply the units

We will multiply the SI units of these variables h, 蚁, and g. \( (m) (办驳/尘鲁) (尘/蝉虏) \)
03

Simplify the multiplied units

Now, let's simplify the multiplied units: \( \frac{ kg \cdot m^2}{ m^3 \cdot s^2}\)
04

Observe the simplified units

Observe that the simplified units are equivalent to 狈/尘虏: \( \frac{kg \cdot m}{s^2} \cdot \frac{1}{m^2} \) We know, the unit of force in SI units is Newton (N), which is defined as: \( N = \frac{kg \cdot m}{s^2} \) So, by substituting the value of N back into the expression, we get: \( N \cdot \frac{1}{m^2} = \frac{N}{m^2} \) Hence, we have verified that the SI unit of the given expression "丑蚁驳" is indeed 狈/尘虏.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Unit Analysis
Understanding the basics of unit analysis is critical in physics and many other scientific disciplines. It's the process of confirming that the units on both sides of an equation are consistent, often used to convert from one unit to another or to validate the dimensional integrity of physics equations.

For instance, considering the exercise where we need to verify that the SI unit of the expression 丑蚁驳 is 狈/尘虏, unit analysis is what allows us to predict whether the expression represents a pressure, which in the International System of Units (SI) is measured in pascals (Pa), equivalent to Newton per square meter (狈/尘虏). This step-by-step translation from individual units of height (m), density (办驳/尘鲁), and acceleration due to gravity (尘/蝉虏) into a coherent final unit is the essence of unit analysis.

Here's why unit analysis matters in physics education:
  • Helps ensure accuracy when performing calculations with multiple steps or conversion factors.
  • Builds a strong foundation for students to tackle more complex problems that involve multiple physical quantities.
  • Encourages critical thinking by reminding students that units behave analogously to the numerical values in calculations.
Teachers should emphasize the importance of always keeping track of units throughout any calculation.
Dimensional Analysis
Going beyond unit analysis, dimensional analysis is a more generalized tool used in physics to check the relationship between the different types of quantities involved in calculations. Dimensional analysis also serves as a method for deriving relationships among physical quantities when the actual equation is not known or to check the plausibility of derived physical equations.

In the context of the exercise provided, the primary dimensions of interest are mass (M), length (L), and time (T). Each term in the expression 丑蚁驳 can be broken down into these fundamental dimensions: height h has the dimension L, density has the dimension ML鈦宦, and acceleration due to gravity g has the dimension LT鈦宦. By multiplying these, you should obtain a unit expressed in terms of the fundamental dimensions that correspond to the SI unit of pascals (狈/尘虏).

Furthermore, dimensional analysis is not only helpful for unit conversion but also for:
  • Ensuring the equation 'makes sense' dimensionally.
  • Providing a sanity check for derived equations and computer models.
  • Serving as a teaching tool to reinforce the concept that physical laws must be expressed in a form independent of the units used.
Dimensional analysis is a powerful concept in physics education because it provides students with a systematic approach to problem-solving and helps them develop an intuition for the natural relationships between physical quantities.
Physics Education
In physics education, students are often introduced to complex concepts through the application of foundational principles, such as unit and dimensional analysis. Exercises like verifying the unit of an expression are not just about knowing formulas, but also about understanding the underlying principles that govern the physical world.

Teaching students the SI units and how to manipulate them provides a tangible way to comprehend abstract physical laws. Solid physics education enables students to:
  • Develop critical thinking and problem-solving skills.
  • Understand the universality and consistency of physical laws across different systems.
  • Recognize the interconnectivity between various physical phenomena through their shared units and dimensions.
It's important for educators to incorporate a wide range of exercises that reinforce these skills, encouraging students to always double-check their work for consistency in units and dimensions to ensure their solutions are physically meaningful.

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Most popular questions from this chapter

Explain why the fluid reaches equal levels on either side of a manometer if both sides are open to the atmosphere, even if the tubes are of different diameters.

The image shows how sandbags placed around a leak outside a river levee can effectively stop the flow of water under the levee. Explain how the small amount of water inside the column of sandbags is able to balance the much larger body of water behind the levee.

Bird bones have air pockets in them to reduce their weight-this also gives them an average density significantly less than that of the bones of other animals. Suppose an ornithologist weighs a bird bone in air and in water and finds its mass is \(45.0 \mathrm{g}\) and its apparent mass when submerged is \(3.60 \mathrm{g}\) (assume the bone is watertight). (a) What mass of water is displaced? (b) What is the volume of the bone? (c) What is its average density?

What is the fluid speed in a fire hose with a \(9.00-\mathrm{cm}\) diameter carrying 80.0 L of water per second? (b) What is the flow rate in cubic meters per second? (c) Would your answers be different if salt water replaced the fresh water in the fire hose?

Water supplied to a house by a water main has a pressure of \(3.00 \times 10^{5} \mathrm{N} / \mathrm{m}^{2}\) early on a summer day when neighborhood use is low. This pressure produces a flow of \(20.0 \mathrm{L} / \mathrm{min}\) through a garden hose. Later in the day, pressure at the exit of the water main and entrance to the house drops. and a flow of only \(8.00 \mathrm{L} / \mathrm{min}\) is obtained through the same hose. (a) What pressure is now being supplied to the house, assuming resistance is constant? (b) By what factor did the flow rate in the water main increase in order to cause this decrease in delivered pressure? The pressure at the entrance of the water main is \(5.00 \times 10^{5} \mathrm{N} / \mathrm{m}^{2},\) and the original flow rate was \(200 \mathrm{L} /\) min. (c) How many more users are there, assuming each would consume \(20.0 \mathrm{L} / \mathrm{min}\) in the morning?

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