In Reverse Chronological Order
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Hour Test #2, 121.4  68.9 +/- 12.7

1.a. 8000 J and 6000 J
1.b. 1450 W and 1200 W
1.c. 3000 s
2.a.  a= 2 m/s.s
2.b. v= 6.5 m/s
2.c. Ff= 50N
2.d.  a= 1 m/s.s
2.e.  Ff= 45 N, a= 0.84 m/s.s
3.a.  Vf= 8.5 m/s
3.b.  K final/K initial =2/7
3.c.  F= 43,000 N
4.a.  500 J
4.b.  V= 10 m/s, h= 5 m
4.c. V2= 8.5 m/s
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Hour Test #2, 121.3, 54.8 +/- 15.8 of 100

1.a.  W = 20N down, Normal = 16 N upwards
1.b.  12 N = ma, a = 6; v = 6.3 m/s from v^2 = 2ax, t = v/a = 1.05 s
1.c.  Ff = mu mg cos = 3.2 N;  Fnet = 12 - 3.2 = ma, a = 4.4 m/s.s

2.a.  10N = 5 a, a = 2 m/s.s; T = 2a = 2x2= 4 N
2.b..  Ff = 0.3 x 2 x g = 6 N, Fnet = 10 - 6 = 5 a, a = 0.8 m/s.s
       T - 6 = 2 a = 2 x 0.8; T = 7.6 N

3.  800 - 700 = (800/g) a, a = 1.2 m/s.s

4.a.  U = mgh = (30,000)(g)(1.5 x 10^5) = 4.5 x 10^10 J
4.b.  v = sqrt (2gh) = 1700 m/s
4.c.  Q = dW = Ff x d;  Ff = U/4 x 10^7 m = 1000N, a = Ff/m = 0.03 m/s.s

5.  x = sqrt (m/k) v = 3.5 m
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Hour Test #1, 121.3,      76.3  +/- 18.0 of 110

1a)  V=0
1b)  a=-10 m/s.s
1c)  v= 50 x cos(45) = 50 x 0.87 = 45 m/s, a = - 10 m/s.s
1d)  V= 50 m/s, at 30 degrees above the Horizontal
1e)  A Force is capable of changing motion (velocity)

2.  F1x = 4 cos 30 = 7.0 N, F2x = 10 cos 45 = -7.1 N, add to - 0.1 N (x), so add F3x = + 0.1 N
x to sum to zero.

3a)  R = Vo.Vo/g  sin(2 x 30) = 30 x 30 x 0.87 / 10 = 80 m
3b)  X = 30 cos 30 x 1 sec = 26 m, y = 30 sin 30 x (1 sec) - (g/2) (1 sec)^2 = 10 m
3c)  Vx = 26 m/s, Vy = 30 sin 30 - g(1 sec) = 5 m

4a)  a = V/t = 110 m/s / 6 s = 20 m/s
4b)  dX = 0.5 (20 m/s.s)(6 s)^2 = 360 m
4c)  V vs. t is linear, origin to 6 s and 110 m/s, 
	x vs t is parabolic origin to (6s, 360 m)

5a)  a= (Vi.Vi)/2x = (200 x 200)/2 x 0.1 = 2 x 10^5 m/s.s
	t = v/a = 200/2x10^5 = 0.001 s

6)  time to get to fence is t = distance/Vx = 70/(30 x cos 45) = 3.3 s
Height at 3.3 sec = y = (30 sin 45)(3.3) - (g/2)(3.3 x  3.3) = 15 m > than 10 m, so a Hrun
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Hour Test #1, 121.4,   66.7 +/- 27.9  of  110  (big spread)

1a)  air drag force cancels out gravity
1b)  Vx = 50 cos40 = 38, Vy = 0
1c)  a = - 10 m/s.s y only
1d)  -80 -(60) = - 140 mph
1e)  10^6 cc/m.m.m

2a)  Vx= 200 = constant, x = 200 T = 200 x 20 = 4000 m 
2b)  y = 2000 = 1/2  g t^2 gives t = 20 s
2c)  Vx = 200 m/s, Vy = gt = 10 x 20 = - 200 m/s

3.  F1x = - 20 x cos 45 = - 14 N, F2x = 50 cos 37 = 40 N
Net Fx = 40 - 14 = 26 N x, so F3x = - 26 N x to keep at rest

4a)  Vi^2 = 2aX = - (2)(-10^5)(5 x 10^-3 m) = 10^3
Vi = 31 m/s
t = Vi/a = 31/10^5 = 3 x 10^-4 s

5a)  Vf^2 = 100^2 = 2aX = 2 x a x 300 m,  gives a = 16 m/s.s
5b)  t = Vf/a = 100/16 = 6 sec.

6a)  at 30 degrees from the perpendicular to the water,  
VDH + Vriver = V at 30 degrees and 1 along the river, 30-60-909 triangle,
The hypotenuse is 2 m/s (DH swims this) and the river is 1 ;m/s.
6b)  t(swim) = 3000 m/v(perpendicular) = 3000/1.7 m/s = 1700 sec.
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