Please answer Questions 001, 002 and 003 thank you This print-out should have 6

Question

Please answer Questions 001, 002 and 003
thank you

This print-out should have 6 questions. Answer in units of m/s. Multiple choice questions may continue on the next column or page find all choices before answering. 005 (part 2 of 3) 10.0 points What is the y component of velocity after 8 s? Answer in units of m/s. 001 10.0 points. A car is parked near a cliff overlooking the 006 (part 3 of 3) 10.0 points ocean on an incline that makes angle what is the magnitude of the displacement with the horizontal. The negligent driver from the origin (z 0 m, y 0 m) after 8 s? leaves the car in neutral, and the emergency Answer in units of m brakes are defective. The car rolls from rest down the incline and has a velocity 5 m/s when it reaches the edge of the cliff. The cliff is 35.7 m above the ocean. How far is the car from the base of the cliff when the car hits the ocean? The acceleration of gravity is 9.8 m/s2 Answer in units of m 002 (part 1 of 2) 10.0 points A particle at rest undergoes an acceleration of 2.5 m/s to the right and 3.9 m/s up What is its speed after 7.8 s? Answer in units of m/s. 003 (part 2 of 2) 10.0 points What is its direction with respect to the hor- izontal at this time? Answer between -180° and Answer in units of 004 (part 1 of 3) 10.0 points A particle moves in the zy plane with constant acceleration, At time zero, the particle is at 7 m, y 4.5 m and has velocity io (7.5 m/s) i (-6,5 m/s)]. The acceleration is given by (7.5 m/s i+ (1.5 after What is the z component of velocity 8 s?

Explanation / Answer

(1)

speed of the car when it reaches the edge of the cliff = 5 m/s

horizontal velocity vx = 5*cos10.3 = 4.91 m/s

vy = -5*sin10.3 = -0.894 m/s

distance of car from the base of cliff is,

x = vx*t ....eq1

from the second equation of motion,

h = h0 + vy*t - (1/2)*g*t^2

0 = 35.7 - 0.894*t - 4.9*t^2

by solving the quadratic equation,

t = 2.60 s

put the value of t eq1,,

x = 4.91*2.60

x = 12.74 m

(2)

given that,

u = 0

a = 2.5 i + 3.9 j m/s^2

t = 7.8 s

vx = ux + ax*t

vx = 0 + 2.5*7.8 = 21.75 m/s

vy = 0 + 3.9*7.8 = 30.42 m/s

(3)

direction = tan^-(vy / vx) = tan^-(30.42 / 21.75)

direction = 54.43 deg

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