HW04Sol - directions. We have that ( A + B ) x = A x + B x...

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Jim Guinn’s PHYS2211 Assignment #4 Solutions Due Monday, Feb. 1, 2010 1. A vector A has a magnitude of 20.0m at points at an angle of 15.0 o to the left of the y axis. What are its components? We see from the diagram that A x will be negative and A y will be positive, so A x = -A sin(15.0 o ) = -20.0m sin15.0 o = A x = -5.18m , and A y = A cos(15.0 o ) = 20.0m cos15.0 o = A y = 19.3m . 2. A vector B has components B x = 15.0m, B y = -5.00m . What are its magnitude and direction? We see from the diagram that B = √ [(15.0m) 2 + (5.00m) 2 ] = 15.8m , and θ = tan -1 (5.00m / 15.0m) = 18.4 o . B = 15.8m at an angle of 18.4 o down from the x-axis. 3. Given the vectors A and B from the previous two problems, what are A + B and A B ? The sum and difference can be given in terms of their components, or magnitude and
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Unformatted text preview: directions. We have that ( A + B ) x = A x + B x = -5.18m + 15.0m = 9.8m , and ( A + B ) y = A y + B y = 19.3m + -5.00m = 14.3m , and ( A- B ) x = A x- B x = -5.18m - 15.0m = -20.2m , and ( A- B ) y = A y- B y = 19.3m - -5.00m = 24.3m . So A + B = 9.8m i + 14.3m j and A B = -20.2m i + 24.3m j . Or | A + B | = [(9.8m) 2 + (14.3m) 2 ] = 17.3m and A 15.0 o A x A y B x B x B | A- B | = [(20.2m) 2 + (24.3m) 2 ] = 31.6m , and 1 = tan-1 (14.3m / 9.8m) = 55.6 o , and 2 = tan-1 (24.3m / 20.2m) = 50.3 o . A + B = 17.3m at an angle of 55.6 o above the x-axis, and A B = 31.6m at an angle of 50.3 o above the -x-axis....
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HW04Sol - directions. We have that ( A + B ) x = A x + B x...

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