A1.2-kg object moving with a speed of 8.0 m/s collides perpendicularly with a wall and emerges with a speed of 6.0 m/s in the opposite direction. if the object is in contact with the wall for 2.0 ms, what is the magnitude of the average force on the object by the wall? ans: 8400 n
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Ответ:
F=ma, F = Wall force on the object, m = mass of the object, a = Deceleration = Δv/t = (6-8)/2 = -1 m/s
F = 1.2*1 = 1.2 N
Ответ:
Total time elapsed between the start and when he returns with the ball is 7.5s
Explanation:
From the question,
- The dog starts at the origin and runs forward at 6m/s for 1.5s. First, we will determine the distance covered while running forward.
From
Speed = Distance / Time
Distance = Speed × Time
Speed = 6m/s
Time = 1.5s
∴ Distance = 6m/s × 1.5s
Distance = 9m
That is, the dog covered a distance of 9m while running forward.
- The dog turns around and runs backward at 7m/s for 3s. Now, we will also determine the distance the dog covered backwards.
Distance = Speed × Time
Speed = 7m/s
Time = 3s
Distance = 7m/s × 3s
Distance = 21m
The dog's displacement from the origin is 21m - 9m = 12m
Now, to calculate how much time has elapsed between the start if the dog runs back to the origin at 4m/s, we will first determine the time the dog spent back to the origin and then add to the time spent for the first two distances.
To get back to the origin, the dog needs to cover 12m
From
Speed = Distance / Time
Time = Distance / Speed
Distance = 12m
Speed = 4m/s
∴ Time = (12m) / (4m/s)
Time = 3s
Therefore, the dog spent 3s to run back to the origin.
Hence, total time elapsed = 1.5s + 3s + 3s
Total time elapsed = 7.5s