Answers: 1
Physics, 21.06.2019 23:20
Initially, a particle is moving at 5.33 m/s at an angle of 37.9° above the horizontal. two seconds later, its velocity is 6.11 m/s at an angle of 54.2° below the horizontal. what was the particle's average acceleration during these 2.00 seconds in the x-direction (enter first) and the y-direction?
Answers: 1
Physics, 22.06.2019 01:10
The x-coordinate of a particle in curvilinear motion is given by x = 3.1t3 - 4.9t where x is in feet and t is in seconds. the y-component of acceleration in feet per second squared is given by ay = 2.3t. if the particle has y-components y = 0 and vy = 5.0 ft/sec when t = 0, find the magnitudes of the velocity v and acceleration a when t = 1.8 sec. sketch the path for the first 1.8 seconds of motion, and show the velocity and acceleration vectors for t = 1.8 sec. answers: v = ft/sec a = ft/sec2
Answers: 2
Physics, 22.06.2019 02:30
Aforce of 9.00 newtons acts at an angle of 19.0 to the horizontal. what is its component in the horizontal direction?
Answers: 2
Which statement best describes the direction of the buoyant force on any object? a. opposite the fo...
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