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Introduction to SAT II Physics - FreeExamPapers

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miscalculation.<br />

Second, make a rough estimate. At what sorts of distances might a satellite orbit? We can<br />

eliminate A immediately: that answer has our satellite orbiting at 4 cm from the center of the<br />

Earth! That leaves us with a choice between B and C. Those aren’t bad odds for guessing.<br />

Slightly Underhanded Way #2: Work with the Letters<br />

This is a method for those of you who like manipulating equations. From looking at the answer<br />

choices, you know the answer will be in meters. You’ve been given three quantities, one expressed<br />

in m/s, one expressed in kg, and one expressed in N·m 2 /kg 2 . These are the only three quantities<br />

you’ll be asked <strong>to</strong> draw upon in order <strong>to</strong> get your answer. Because F = ma, you know you can<br />

substitute kg·m/s 2 for N. So a quantity expressed in N·m 2 /kg 2 can equally be expressed in m 3 /kg·s 2 .<br />

The trick, then, is <strong>to</strong> combine a quantity expressed in these terms with a quantity expressed in<br />

meters per second and a quantity expressed in kilograms, and wind up with a quantity expressed<br />

solely in meters. To do that, you need <strong>to</strong> get rid of the “kg” and the “s” by canceling them out.<br />

Start by canceling out the “kg”:<br />

Now you need <strong>to</strong> cancel out the “s 2 ” in the denomina<strong>to</strong>r. Let’s divide by the square of our “m/s”<br />

quantity:<br />

There you have it. You didn’t need <strong>to</strong> use a single formula <strong>to</strong> get the answer. You just had <strong>to</strong> be<br />

aware of the terms in which your answer needed <strong>to</strong> be expressed, and manipulate the quantities<br />

you were given in the question.<br />

Word <strong>to</strong> the wise: don’t use this method unless you’re absolutely stumped. It can backfire, and is<br />

of course no substitute for careful reasoning.<br />

Vec<strong>to</strong>rs<br />

VECTORS ARE USUALLY THE FIRST THING you learn in a physics class, and they’re the first<br />

thing you’ll learn here. Vec<strong>to</strong>rs are one of the fundamental mathematical <strong>to</strong>ols the physicist uses,<br />

and one that is frequently misunders<strong>to</strong>od or misapplied by students. Generally, there aren’t more<br />

than one or two questions on <strong>SAT</strong> <strong>II</strong> <strong>Physics</strong> that test your knowledge of vec<strong>to</strong>rs directly, but there<br />

are a host of problems—particularly in mechanics—where arriving at the right solution demands a<br />

solid grasp of how <strong>to</strong> apply and manipulate vec<strong>to</strong>rs. Even if you feel confident with vec<strong>to</strong>rs, we<br />

urge you <strong>to</strong> review this chapter and be absolutely sure you won’t get tripped up on what would<br />

otherwise be some easy questions.<br />

What’s a Vec<strong>to</strong>r?<br />

A vec<strong>to</strong>r is a mathematical object possessing, and fully described by, a magnitude and a<br />

direction. It’s possible <strong>to</strong> talk about vec<strong>to</strong>rs simply in terms of numbers, but it’s often a lot easier<br />

<strong>to</strong> represent them graphically as arrows. The vec<strong>to</strong>r’s magnitude is equal <strong>to</strong> the length of the<br />

arrow, and its direction corresponds <strong>to</strong> where the arrow is pointing. Physicists commonly refer <strong>to</strong><br />

the point of a vec<strong>to</strong>r as its tip and the base as its tail.<br />

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