Problem 5.14. The pitch of a screw gauge is 1 mm and there are 100 divisions on the circular scale. While measuring the diameter of a wire, the linear scale reads 1 mm and 47th division on the circular scale coincides with reference line. The length of the wire is 5-6 cm. Find the curved surface area(in cm2) of the wire in appropriate number of significant

Problem 5.14. The pitch of a screw gauge is 1 mm and there are 100 divisions on the circular scale. While measuring the diameter of a wire, the linear scale reads 1 mm and 47th division on the circular scale coincides with reference line. The length of the wire is 5-6 cm. Find the curved surface area(in cm2) of the wire in appropriate number of significant

Principles of Physics: A Calculus-Based Text

5th Edition

ISBN:9781133104261

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter1: Introduction And Vectors

Section: Chapter Questions

Problem 25P: A sidewalk is to be constructed around a swimming pool that measures (10.0 0.1) m by (17.0 0.1) m....

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A sidewalk is to be constructed around a swimming pool that measures (10.0 0.1) m by (17.0 0.1) m. If the sidewalk is to measure (1.00 0.01) m wide by (9.0 0.1) cm thick, what volume of concrete is needed and what is the approximate uncertainty of this volume?
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Check Your Understanding Is the equation v=atdimensionally consistent? One further point thin needs to be mentioned is the effect of the operations of calculus on dimensions. We have seen that dimensions obey the rules of algebra, just like units, but what happens when we take the derivative of one physical quantity with respect to another or integrate a physical quantity over another? The derivative of a function is just the slope of the line tangent to its graph and slopes are ratios, so for physical quantities vand t , we hive that the dimension of the derivative of vwith respect to us just the ratio of the dimension of v over that of t : [dvdt]=[vt] . Similarly, since integrals are just sums of products, the dimension of the integral of vwith respect to t is simply the dimension of vtimes the dimension of t : [vdt]=[v][t] . By the same reasoning, analogous rules hold for the units of physical quantities derived from other quantities by integration or differentiation.
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