Post by Mr Fisher on Jan 6, 2006 7:17:05 GMT -5
Here are the answers to contract 4 - 1st Semester Exam Worksheet.
Velocity
28 velocity
29 slope of the tangent to the curve
30 300 km North
31 50 km/hr
32 force (net) = 0 (or forces are balanced)
33 19,200 m
34 40 m/s (not any joggers I know!)
35 curve would be linear
36 slope of the line
37 velocity is increasing
Acceleration
38 acceleration
39 zero!
40 linear
41 zero
42 9.8 m/s/s
43 300 + 2(10) = 320 m/s
44 d = 1/2gt^2 so d = about 5 (25) = 125 meters (YOU MUST KNOW THE FREE FALL EQUATIONS!)
45 time to top is 46/10 = 4.6 seconds d = 5(4.6)(4.6) = 106 meters
Forces
46 strong or nuclear force
47 W = mg
48 terminal velocity
49 first
50 second
51 W = mg so m = W/g or 2500/10 = 250 kg! About 550 pounds!
52 Applied F = friction force since constant speed! so F = mu(mass)(g) or 480 N
53 84 N
54 F(net) = F(lift) - mg +25M up - 2M down (gravitational force!) = 23M Newtons
55 use a^2+b^2=r^2 m/s
56 F = 500 cos(60) = 250N
57 IV quadrant. Label quadrants counterclockwise from the x-axis
58 skip! tedious, but doable
Velocity
28 velocity
29 slope of the tangent to the curve
30 300 km North
31 50 km/hr
32 force (net) = 0 (or forces are balanced)
33 19,200 m
34 40 m/s (not any joggers I know!)
35 curve would be linear
36 slope of the line
37 velocity is increasing
Acceleration
38 acceleration
39 zero!
40 linear
41 zero
42 9.8 m/s/s
43 300 + 2(10) = 320 m/s
44 d = 1/2gt^2 so d = about 5 (25) = 125 meters (YOU MUST KNOW THE FREE FALL EQUATIONS!)
45 time to top is 46/10 = 4.6 seconds d = 5(4.6)(4.6) = 106 meters
Forces
46 strong or nuclear force
47 W = mg
48 terminal velocity
49 first
50 second
51 W = mg so m = W/g or 2500/10 = 250 kg! About 550 pounds!
52 Applied F = friction force since constant speed! so F = mu(mass)(g) or 480 N
53 84 N
54 F(net) = F(lift) - mg +25M up - 2M down (gravitational force!) = 23M Newtons
55 use a^2+b^2=r^2 m/s
56 F = 500 cos(60) = 250N
57 IV quadrant. Label quadrants counterclockwise from the x-axis
58 skip! tedious, but doable