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(III) Many sailboats are moored at a marina 4.4 $\mathrm{km}$ away on the opposite side of a lake. You stare at one of the sailboats because, when you are lying flat at the water's cdge, you can just see itsdeck but none of the side of the sailboat. You then go to that $$ \begin{array}{l}{\text { sailboat on the other side of the }} \\ {\text { lake and measure that the deck }} \\{\text { is } 1.5 \mathrm{m} \text { above the level of the }} \\ {\text { water. Using Fig, } 12, \text { where }} \\ {h=15 \mathrm{m}, \text { estimate the radius } R} \\ {\text { of the Farth. }}\end{array} $$

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$6.5 \times 10^{6} m$

Physics 101 Mechanics

Chapter 1

Introduction, Measurement, Estimating

Physics Basics

Cornell University

University of Michigan - Ann Arbor

University of Winnipeg

McMaster University

Lectures

04:16

In mathematics, a proof is a sequence of statements given to explain how a conclusion is derived from premises known or assumed to be true. The proof attempts to demonstrate that the conclusion is a logical consequence of the premises, and is one of the most important goals of mathematics.

09:56

In mathematics, algebra is one of the broad parts of mathematics, together with number theory, geometry and analysis. In its most general form, algebra is the study of mathematical symbols and the rules for manipulating these symbols; it is a unifying thread of almost all of mathematics.

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(III) Many sailboats are d…

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Solve each problem.Boa…

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so we can re draw the diagram in the textbook. Ah, this will be radius are of the earth. We can just, um, label is our We can say that this would be d and then this length or here would be the height of the whole of the sailboat. We can just leave less h and so he rejoined to estimate the radius of the Earth. We can say that H is going to be much less than our which means that H squared could be considered much less than two r h. So when me use path, I rinsed, Erm, in order to estimate the radius of the earth, we can say D squared Plus are spared would be equal to our plus H Square, which would be equal to R squared plus two R h plus h squared. Now, because of these relationships, we can say that d squared my body d squared would be equal to two r h plus eight squared, knowing that are squared r squared. Both cancel out. And so we can say we can also eliminate this term as well due to this relationship and the spirit is going to be approximately equal to two r h. And then we basically saw for our our would be equal to d squared, divided by two times a tch. This is equaling 4400 meters squared, divided by two times a tch of 1.5 meters. And we find that the radius of the earth is approximated is approximately equal to 6.5 times 10 to the six meters. This is, um, somewhat accurate in the the true radius of the earth would be equal to 6.38 times 10 to the six meters. So it is accurate to a kn order of magnitude. However, it is about a point 22 times 10 to the sixth meters off. However, given that we only have approximations, this is somewhat of a good approximation. That is the end of the solution. Thank you for watching

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