Q1)A dog running in an open field has components of velocity v x = 3.5 m/s and v
ID: 776708 • Letter: Q
Question
Q1)A dog running in an open field has components of velocity vx = 3.5 m/s and vy = -1.2 m/s at time t1 = 10.9 s . For the time interval from t1 = 10.9 sto t2 = 21.8 s , the average acceleration of the dog has magnitude 0.49 m/s2 and direction 32.5 measured from the +xaxis toward the +yaxis.
A- At time t2 = 21.8 s , what is the y-component of the dog's velocity?
Express your answer using two significant figures.
B- What is the direction of the dog's velocity (measured from the +xaxis toward the +yaxis)?
Express your answer using two significant figures.
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Q2) A rookie quarterback throws afootball with an initial upward velocity component of 17.0 m/s and a horizontal velocity component of 22.0 m/s . Ignore air resistance.
A- How does this compare with the time calculated in part (a).
Express your answer using three significant figures.
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Q3) A model of a helicopter rotor has four blades, each of length 3.20 m from the central shaft to the blade tip. The model is rotated in a wind tunnel at a rotational speed of 550 rev/min .
A- What is the linear speed of the blade tip?
B-What is the radial acceleration of the blade tip expressed as a multiple of the acceleration of gravity, g?
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Q4) A small toy airplane is flying in the xy-plane parallel to the ground. In the time interval t=0 to t=10.0 s, its velocity as a function of time is given by =(1.30m/s2)ti^+[13.0m/s(2.00m/s2)t]j^.
A- At what value of t is the velocity of the plane perpendicular to its acceleration?
Express your answer to three significant figures and include the appropriate units.
Explanation / Answer
1) v0x = 3.5 m/s
v0y = -1.2 m/s
ax = 0.49 cos32.5 = 0.413 m/s^2
ay = 0.49 sin32.5 = 0.263 m/s^2
t = 21.8 - 10.9 = 10.9 s
(A) vy = v0y + ay t
vy = (-1.2) + (10.9 x 0.263) = 1.7m/s
(B)
vx = (3.5) + (0.413 x 10.9) = 8 m/s
theta = tan^-1(vy /vx) = 12 deg
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