Physx141MPch3sol

Physx141MPch3sol - 3.7. Visualize: The figure shows the...

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3.7. Visualize: The figure shows the components v x and v y , and the angle θ . Solve: We have, sin40 , y vv =− ° or 10 m/s sin40 , v ° or 15.56 m/s. v = Thus the x -component is cos40 (15.56 m/s ) cos40 11.9 m/s. x = ° = Assess: Note that we had to insert the minus sign manually with y v since the vector is in the fourth quadrant. 3.8. Visualize: Solve: Vector E r points to the left and up, so the components x E and y E are negative and positive, respectively, according to the Tactics Box 3.1. (a) cos x EE and sin . y = (b) sin x φ and cos . y = Assess: Note that the role of sine and cosine are reversed because we are using a different angle. and are complementary angles. 3.10. Visualize: We will follow the rules given in the Tactics Box 3.1. Solve: (a) (2 km)sin30 1 km x r °=− (2 km)cos30 1.73 km y r = (b) (5 cm/s)sin90 5 cm/s x v (5 cm/s)cos90 0 cm/s y v = (c) 22 (10 m/s )sin40 6.43 m/s x a (10 m/s )cos40 7.66 m/s y a
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(d) (50 N)sin36.9 30.0 N x F = (50 N)cos36.9 40.0 N y F = Assess: The components have the same units as the vectors. Note the minus signs we have manually inserted according to the Tactics Box 3.1. 3.12. Visualize: Solve: The magnitude of the vector is 22 2 2 ( ) ( ) (125 V/m) ( 250 V/m) 280 V/m. xy EE E =+ = + = In the expression for , E v the and + means that ˆ j ˆ i E r is in quadrant IV. The angle θ is below the positive x -axis. We have: || 11 1 250 V/m tan tan tan 2 63.4 125 V/m y x E E −−  == = =   ° Assess: Since | | | | y x E E > , the angle made with the + x -axis is larger than 45 ° . = 45 ° for | | | | y x E E = .
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This note was uploaded on 05/05/2008 for the course PHYS 141 taught by Professor Staff during the Spring '06 term at Cal Poly.

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Physx141MPch3sol - 3.7. Visualize: The figure shows the...

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