Thanks a million to whoever helps out. I need the answers in 4hours to study for

Question

Thanks a million to whoever helps out. I need the answers in 4hours to study for an exam.

1. An indestructible bullet 2.00 cm long is fired straight througha board that is 10.0 cm thick. The bullet strikes the board with aspeed of 470 m/s and emerges with aspeed of 290 m/s. (To simplify, assumethat the bullet accelerates only while the front tip is in contactwith the wood.) (a) What is the average acceleration of thebullet through the board? (I know this is the right answer)
m/s2

(b) What is the total time that the bullet is in contact with theboard?

(c) What thickness of board (calculated to 0.1 cm) would it take tostop the bullet, assuming that the acceleration through all boardsis the same?
3 cm


2. A mountain climber stands at the top of a 40.0 m cliff that overhangs a calm pool of water.She throws two stones vertically downward 1.00 s apart and observesthat they cause a single splash. The first stone had an initialvelocity of -1.60 m/s.

(a) How long after release of the first stone did the two stoneshit the water?
s

(b) What initial velocity must the second stone have had, giventhat they hit the water simultaneously?
m/s

(c) What was the velocity of each stone at the instant it hit thewater? first stone m/s second stone m/s (a) What is the average acceleration of thebullet through the board? (I know this is the right answer)
m/s2

(b) What is the total time that the bullet is in contact with theboard?

(c) What thickness of board (calculated to 0.1 cm) would it take tostop the bullet, assuming that the acceleration through all boardsis the same?
3 cm


2. A mountain climber stands at the top of a 40.0 m cliff that overhangs a calm pool of water.She throws two stones vertically downward 1.00 s apart and observesthat they cause a single splash. The first stone had an initialvelocity of -1.60 m/s.

(a) How long after release of the first stone did the two stoneshit the water?
s

(b) What initial velocity must the second stone have had, giventhat they hit the water simultaneously?
m/s

(c) What was the velocity of each stone at the instant it hit thewater? first stone m/s second stone m/s first stone m/s second stone m/s

Explanation / Answer

1.a)Let the average acceleration of the bullet through theboard be a therefore we get v2 - u2 = 2ax or a = (v2 - u2/2x) -----------(1) v = 290 m/s,u = 470 m/s and x = 10.0 cm = 10.0 *10-2 m (b)Let the total time that the bullet is in contact with theboard be t therefore we get v = u + at or t = (v - u/a) (c)Let the thickness of board required to stop the bullet bex1 therefore we get v12 - u12 =2ax1 or x1 = (v12 -u12/2a) v1 = 0,u1 = 470 m/s and the value of ais obtained from equation (1) 2.(a)The height of the cliff be H Let the time taken by the first stone to hit the water be ttherefore we get H = ut + (1/2)gt2 H = 40.0 m,u = -1.60 m/s and g = 9.8 m/s2 or 40.0 = -1.60t + (1/2) * 9.8 * t2 or 4.9t2 - 1.60t - 40.0 = 0 solving the above quadratic equation we get t1 = 3.025 s and t2 = -2.698 s We ignore the negative root therefore the time taken is t =3.025 s Therefore,the time after release of the first stone the twostones hit the water is t = 3.025 s (b)Let the initial velocity the second stone had beu2 therefore we get H = u2t + (1/2)gt2 solving the above equation for u2 we get u2 = (H - (1/2)gt2/t) or u2 = (40.0 - (1/2) * 9.8 *(3.025)2/3.025) = -13.2 m/s (c)The velocity of each stone at the instant it hit the waterbe v1 and v2 respectively.Therefore,weget v1 = u1 + gt1 = -1.60 + 9.8 *3.025 = 28.04 m/s and v2 = u2 + gt2 = -13.2 +9.8 * 3.025 = 16.44 m/s solving the above equation for u2 we get u2 = (H - (1/2)gt2/t) or u2 = (40.0 - (1/2) * 9.8 *(3.025)2/3.025) = -13.2 m/s (c)The velocity of each stone at the instant it hit the waterbe v1 and v2 respectively.Therefore,weget v1 = u1 + gt1 = -1.60 + 9.8 *3.025 = 28.04 m/s and v2 = u2 + gt2 = -13.2 +9.8 * 3.025 = 16.44 m/s

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