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The x component of vector is -20.0 m and the y component is +54.0 m. (a) What is the magnitude of ? (b) What is the angle (in radians) between the direction of and the positive direction of x?

2007-09-11 06:12:52 · 5 answers · asked by Tastytaint 1 in Science & Mathematics Physics

5 answers

Try drawing the x and y components... You'll see you can draw a rectangle or triangle depending on how you do it.

The magnitude of the vector can be figured out by using a pythagorian's therm. m=sqrt(54^2 + (-20)^2). Notice, the sign cancells out when you square -20.

If you want the angle, you'll have to use arctangent function.
r=arctan(y/x) Don't forget the sign.... and what it is really giving you.

2007-09-11 06:26:45 · answer #1 · answered by tkquestion 7 · 0 0

The magnitude of vector V is √(20²+54²) = 57.585 m(approx).
The angle is arctan( -54/20 = arctan(-2.7)
= π -arctan(2.7) = 1.9255 radians.
The last being done because the angle needed
is in the 2nd quadrant.
Mountain Gym's answer for the angle is wrong
because his angle is in the 4th quadrant.
When you use the arctan function on a calculator
it always gives you an angle in the 4th quadrant
if the angle is negative and an angle in the 1st
quadrant if the angle is positive.

2007-09-11 06:28:44 · answer #2 · answered by steiner1745 7 · 0 0

Mag A+B = 57.6 m
Theta = 1.92 radians

2007-09-11 06:31:10 · answer #4 · answered by Anonymous · 0 0

a. sqrt( x^2+y^2) = 57,584720195551875234953509299804
b.sin A = y/57,584720195551875234953509299804 so 0,93774876072370363772240334806683 asin (0,93774876072370363772240334806683) is your answer

2007-09-11 06:28:41 · answer #5 · answered by yoda288 1 · 0 2

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