Kaplunovsky - practice 06

# Kaplunovsky - practice 06 - practice 06 – – Due noon 1...

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Unformatted text preview: practice 06 – – Due: Feb 26 2007, noon 1 Question 1, chap 6, sect 1. part 1 of 1 5 points Consider three sheets of paper, all of the same size and weight. One sheet is un-folded, the other is folded in two, and the third is wadded into a ball. When you drop them from the roof of a highrise building, they reach terminal velocities before falling down, but those terminal velocities are different for each sheet. Compare the drag forces acting on each sheet after it reaches its terminal velocity. Which sheet experiences a greater drag force? 1. All three forces are the same. correct 2. The wadded sheet 3. The plain sheet 4. More information needed to answer the question. 5. The folded sheet Explanation: At similar speeds, the three sheets would feel quite different drag forces. But the ter- minal speeds of the three sheets are also dif- ferent, and in each case v T is determined by the condition that the drag force equals the sheet’s weight. Since all three sheets are of the same weight mg , when they reach their respective terminal speeds, the drag force on each sheet is the same F D = mg . Thus, all three drag forces are the same. Question 2, chap -1, sect -1. part 1 of 1 10 points A runaway truck accelerates down a very long hillside. The truck’s breaks do not work, and the gear is stuck in neutral. Un- der circumstances, the friction force between the truck’s tires and the road is negligible. But the air drag on the truck prevents its speed from growing without limis; instead, it asymptotes to a terminal velocity v T . The truck has mass m = 14500 kg, cross- section A = 5 . 39 m 2 , and aerodynamic drag coefficient D = 0 . 251. Also, air density is ρ air = 0 . 969 kg / m 3 , g = 9 . 8 m / s 2 , and the hill’s slope is θ = 2 . 13 ◦ below the horizontal. Calculate the truck’s terminal velocity v T . Correct answer: 89 . 763 m / s (tolerance ± 1 %). Explanation: Note that the truck isn’t falling down but speeds down a 2 . 13 ◦ incline. Consequently, the forward component of the truck’s weight is not mg but only F f = mg × sin θ = 5281 . 42 N . The backward force on the truck is the air drag F D = 1 2 DAρ air × v 2 , and when the truck reaches its terminal speed, this drug force is precisely equal to the for- ward force F f . Thus, 1 2 DAρ air × v 2 T = F f , and therefore v T = radicalBigg 2 F f DAρ air = radicalBigg 2 × (5281 . 42 N) (0 . 251) × (5 . 39 m 2 ) × (0 . 969 kg / m 3 ) = 89 . 763 m / s . Question 3, chap 6, sect 3. part 1 of 1 5 points Consider a conical pendulum, where a string with length ℓ is attached to a mass m . The angle between the string and the ver- tical is θ . The orbit is in the horizontal plane with radius r and tangential velocity vectorv . practice 06 – – Due: Feb 26 2007, noon 2 v r g ℓ m θ The acceleration of gravity is 9 . 8 m / s 2 ....
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Kaplunovsky - practice 06 - practice 06 – – Due noon 1...

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