PHYS 225 EXAM 1 a closed book examination. The use of notes, books, cell phones,
ID: 1394063 • Letter: P
Question
PHYS 225 EXAM 1 a closed book examination. The use of notes, books, cell phones, or consultation with fellow students or other persons are forbidden. will have 1 hr to complete the exam. Place your answers in the given booklet. Write clearly. Your credit is based on how you reached your swer. Show all necessary work. 1. The coordinates of an object's position as a function of time are given by x -0.310t? 7.20t 28.0 0.220t 9.10t 30.0 a) Atta 15.0 s, what is the object's positionevector in unit vector notation and in magnitude-angle notatio 2. Aboat is travelling upstream in the positive direction of an x axis at 14.0 km/h with respect to the water of a river The water is flowing at 9.00 km/h with respect to the ground. What are the (a) magnitude and (b) direction of boat's velocity with respect to ground? A child on the boat walks from front to rear at 6.00 km/h with respect to the boat. What are the (c) magnitude and direction of the child's velocity with respect to the ground? A rescue plane flies at 198 km/h (55.0 m/s at a constant height h 500 m toward a point directly over a victim at T, where a rescue capsule is to land. (a) what should be the angle a of the pilot's line of sight to the victim when the capsule release is made? 3. A mass m2 is on a frictionless plane inclined a an angle e. It is connected by a string of negligible mass to another mass m1 on a horizontal frictionless surface. The pulley is frictionless and massless. (a)Draw the free-body for mi and m2. (b) Find an expression for the acceleration of the masses. (c) Find an expression for the tension in the string. (b) If m1 3.00 kg, m2 1.00 kg and e 30.0 find the values for acceleration and tension m2 Extra credit: 1. A crate m1 is pushed at a constant speed up the frictionless ramp as shown below. What horizontal force F is required? (m 110 kg, 0 34Explanation / Answer
1)
at t = 15s
x = -0.31*15^2 + 7.2*15 + 28 = 66.25 m
y = 0.22*15^2 - 9.1*15 + 30 = -57 m
in unit vector notation, r = xi + yj
= 66.25 i - 57 j <<<<<<------Answer
magnitude = sqrt(x^2 + y^2)
= sqrt(66.25^2 + 57^2)
= 87.4
angle, theta = tan^-1(y/x)
= tan^-1(-57/66.25)
= 360 - 40.7
= 319.3 degrees
so, in magnitude angle nonation (87.4 m, 319.3 degrees) <<<<<<------Answer
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