straight velocity of the particle line) motion with an 4. A particle initially a

Question

straight velocity of the particle line) motion with an 4. A particle initially at rest undergoes recth and direction. The velocity of acceleration that is constant in magnitu A)etionthat is ret dres rectilinear (i.e, straigh Ais constant in magnitude and direction. B) is constant in direction only. C) is constant in magnitude only D) can change in magnitude and direction. E) is described by none of these. 5. A ball is thrown upward from m an 80-ft tower with an initial vertical speed of 40 f/s. If with an r resistance is ignored, the ball's speed when it reaches the g A) 67 ft/s B) 1.3×10?ft/s C) 1.2 x 102 f/s D) 49 f/s E) 82 ft/s o: The position vector of an object is given by 21+3 and 4 s later, its position vector sition vector is -2i + 7j. The units are in m. The magnitude of the change in the po AF is A) 0 m B) 2/2 m C) 4 2 m D) 10 m 7. A pilot wants the heading of an airplane to be 370 east of north. The wind is from the east at 37 km/h. The airplane has a speed relative to the air of 370 km/h. Approximately what heading should the pilot maintain? (Hint: Sketch the problem graphically and select the best answer from the list given.) A) 318° B) 4.6 C) 122 D) 32 E) 420

Explanation / Answer

4) option E)

5) option E)

      v^2 = u^2 + 2aS

      v^2 = 40^2 + 2*32.74*80 = 6838.4

      v = 82 ft/s

6) option C)

r1 = 2 i^ + 3j^

r2 = -2i^ + 7j^

dr = r2 – r1 = -4i^ + 4j^

Magnitude = sqrt(16 + 16) = 4*sqrt(2)

7) Option E)

plane velocity + wind velocity = desired velocity
where the desired velocity has equal east and north components (37º)
(370km/h*cos - 37) i + (370km/h*sin) j = desired velocity
Then
370cos - 37 = 370sin
Using wolfram, this solves to
= 2*(*0 + arctan((1/11)(199 - 10)) = 41º north of east,
which is 49º east of north.

9) Option C)

    T = 2pi/w = 2pi*r/v

   

10) Option C)

A = w^2*r = (2pi/T)^2*r = 4*pi^2*2/1600 = pi^2/200

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