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Physics Webassign Help?
I need help with my physics webassign.
a) A car makes a 64.0 km trip with an average velocity of 40.0 km/h in a direction due north. The trip consists of three parts. The car moves with a constant velocity of 23 km/h due north for the first 15 km and 61 km/h due north for the next 32 km. With what constant velocity (magnitude and direction) does the car travel for the last 17 km segment of the trip?
b) You are on a train that is traveling at 3.0 m/s along a level straight track. Very near and parallel to the track is a wall that slopes upward at a 12° angle with the horizontal. As you face the window (0.98 m high, 1.9 m wide) in your compartment, the train is moving to the left, as the drawing indicates. The top edge of the wall first appears at window corner A and eventually disappears at window corner B. How much time passes between appearance and disappearance of the upper edge of the wall?
c) Two runners start one hundred meters apart and run toward each other. Each runs ten meters during the first second. During each second thereafter, each runner runs ninety percent of the distance he ran in the previous second. Thus, the velocity of each person changes from second to second. However, during any one second, the velocity remains constant. Determine how much time passes before the runners collide, and determine the speed with which each is running at the moment of collision
2 Answers
- bonoboLv 71 decade agoFavorite Answer
Total Time = (64 Km) ⁄ (40 Km/Hr) = 1.6 Hr
Ta = (15 Km) ⁄ (23 Km/Hr) = (15 ⁄ 23) Hr ... Part A of trip
Tь = (32 Km) ⁄ (61 Km/Hr) = (32 ⁄ 61) Hr ... Part B of trip
Total Time = Ta + Tь + Tc
1.6 = (15 ⁄ 23) + (32 ⁄ 61) + Tc
Tc = 0.4232 sec
Tc = (17 Km) ⁄ (Vc) ... Part C of trip
0.4232 = (17 Km) ⁄ (Vc)
Vc = 40.17 Km/Hr due north
- MicheleLv 45 years ago
The horizontal component of velocity is constant (7.55m/s). Since time = distance/speed for constant velocity, considering the horizontal motion only gives: t = 7.25/7.55 = 0.9603s That means the time of flight = 0.9603s. Considering vertical motion, the initial velocity = 0, the distance fallen is 1.40m and the time taken is 0.9603s. Taking downwards as positive (so all vectors are positive): x = vi.t + ½gt² 1.40 = (0 x 0.9603) + ½ x g x 0.9603² g = 2.80/0.9221 = 3.04m/s² If that isn't clear, try the video-lesson in the link.(though it uses different symbols, e.g. 'u' for initial velocity).