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exam 1 - midterm 01 PATEL VISHAL Due 10:00 pm Question 1...

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midterm 01 – PATEL, VISHAL – Due: Feb 14 2007, 10:00 pm 1 Question 1 part 1 of 1 10 points A 5 . 3 g bullet leaves the muzzle of a rifle with a speed of 334 m / s. What total constant force is exerted on the bullet while it is traveling down the 0 . 85 m long barrel of the rifle? 1. 271 . 562 N 2. 280 . 193 N 3. 289 . 132 N 4. 298 . 116 N 5. 307 . 429 N 6. 316 . 96 N 7. 327 . 044 N 8. 337 . 171 N 9. 347 . 792 N correct 10. 358 . 56 N Explanation: Given : m = 5 . 3 g = 0 . 0053 kg , v = 334 m / s , and Δ x = 0 . 85 m . The acceleration of the bullet is given by v 2 = v 2 i + 2 a x ) = 2 a x ) since v i = 0. a = v 2 2 (Δ x ) and the force exerted on the bullet is summationdisplay F = m a = m v 2 2 (Δ x ) = (0 . 0053 kg)(334 m / s) 2 2 (0 . 85 m) = 347 . 792 N . Question 2 part 1 of 1 10 points A rock thrown straight up with a velocity of 29 m / s from the edge of a building just misses the building as it comes down. The rock is moving at 55 m / s when it strikes the ground. The acceleration of gravity is 9 . 8 m / s 2 . How tall was the building? 1. 108 . 01 m 2. 111 . 429 m correct 3. 115 . 459 m 4. 119 . 388 m 5. 124 . 082 m 6. 128 . 112 m 7. 133 . 01 m 8. 137 . 143 m 9. 142 . 245 m 10. 146 . 939 m Explanation: v 2 f = v 2 i 2 g ( y f y 0 ) y f y 0 = v 2 f v 2 i 2 g so the height is H = y 0 y f = v 2 f v 2 i 2 g = (55 m / s) 2 (29 m / s) 2 2 (9 . 8 m / s 2 ) = 111 . 429 m . Question 3 part 1 of 2 8 points Consider a river with a uniform current velocity vectorv c . A boat crosses the river. When viewed from the shore, the boat’s motion is perpendicular to the river’s banks. However, the heading of the boat — the direction of its motion relative to the water — is at angle θ = 37 upstream from the perpendicular. Use these data to determine the boat’s speed v b relative to the water, or rather its ratio v b /v c to the current speed: 1. v b v c = cos 2 θ 2. v b v c = 1 sin θ correct
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midterm 01 – PATEL, VISHAL – Due: Feb 14 2007, 10:00 pm 2 3. v b v c = 1 cos θ 4. v b v c = 1 cos 2 θ 5. v b v c = cos θ 6. v b v c = sin 2 θ 7. v b v c = 1 tan θ 8. v b v c = sin θ 9. v b v c = tan θ 10. v b v c = 1 sin 2 θ Explanation: Let vectorv ( W ) b be the boat’s velocity vector rel- ative to the water while vectorv ( L ) b is its velocity vector relative to the dry land. The two vec- tors are related according to vectorv ( L ) b = vectorv ( W ) b + vectorv c (1) where vectorv c is the river current’s velocity vector. For the boat in question, the direction of vectorv ( L ) b is perpendicular to the river, while vectorv ( W ) b is directed θ = 37 upstream of the perpen- dicular. This gives us the following vector diagram: vectorv ( L ) b vectorv ( W ) b vectorv c θ The 3 vectors here form a right triangle, hence bardbl vectorv ( L ) b bardbl = bardbl vectorv ( W ) b bardbl × cos θ, (2) bardbl vectorv c bardbl = bardbl vectorv ( W ) b bardbl × sin θ. (3) In our notations, v b = bardbl vectorv ( W ) b bardbl and v c = bardbl vectorv c bardbl , therefore according to eq. (2), v b v c bardbl vectorv ( W ) b bardbl bardbl vectorv c bardbl = 1 sin θ .
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