Week 2 PH220
1) Net velocity = 3 cos 55°= 1.72, 3 sin 55°-2 = .46 1.72 + .46 = 1.78 m/s
Direction traveled will be 14.97° NW.
2) Total distance from you: √50km2 + 2.1km2 + 1.5km2 = 50.067 km from you.
Relation to x-axis: = 90° + arccos(2.1/50.067) = 177.59°
Relation to y-axis: = arccos(1.5/50.067) = 88.28°
3) Horozontal distance covered is 0+ 1100 * 0.32 = 352m traveled horizontally
Vertical distance traveled is -1/2(-9.8)(.32)2 = .502 m traveled vertically
4) If he is 8m from the pool edge and 20m vertical from the ground, we must first determine how long it would take him to hit the ground without running.
Using formula D=T0+1/2at2 , we can begin to algebraically solve the equation: D=20 T0 = 0 a= 9.8m/s/s t2= unknown 20 = 4.9t2 ……t2 = 4.081 √t2 = √4.081 ….t= 2.02 seconds to hit the ground. To determine the minimum speed required to clear 8m in 2.02 s just use D/T= 8/2.02= 3.96 m/s horizontally to clear 8 m from 20 m vertically.
5) Using acceleration = force * mass, we know the car had a force of 5000N, and a mass of 1500kg.
5000/1500 = -3.33 m/s deceleration. It took 5.6 seconds for the car to decelerate to 0, so multiply the deceleration/acceleration with the amount of time to obtain the initial speed of the vehicle. 3.33 * 5.6 = Initial speed of 18.65 m/s
6) Weight is equated as mass * gravity. On the moon gravity would be g/6, or 9.8/6. The weight of a 100kg man would be 100*(9.8/6) = 163.33 N or 16.66kg. In order to simulate this phenomenon in an elevator here on Earth, the acceleration would have to be downward, or negative direction. We would take the normal acceleration here on earth at 9.8 m/s2 and subtract it from what it would be on the moon, 9.8/6. This would give us the negative acceleration needed in the elevator to reach moon-like conditions. The negative acceleration would be -8.17 m/s2.
7) The box is 7.93 kg. 7.93kg * 9.8 = 77.71 N downward force. 84N at an angle of 47° = 84 cos 47°= 57.3 N horizontal force forward 84 N sin 47°= 61.4 N
Direction traveled will be 14.97° NW.
2) Total distance from you: √50km2 + 2.1km2 + 1.5km2 = 50.067 km from you.
Relation to x-axis: = 90° + arccos(2.1/50.067) = 177.59°
Relation to y-axis: = arccos(1.5/50.067) = 88.28°
3) Horozontal distance covered is 0+ 1100 * 0.32 = 352m traveled horizontally
Vertical distance traveled is -1/2(-9.8)(.32)2 = .502 m traveled vertically
4) If he is 8m from the pool edge and 20m vertical from the ground, we must first determine how long it would take him to hit the ground without running.
Using formula D=T0+1/2at2 , we can begin to algebraically solve the equation: D=20 T0 = 0 a= 9.8m/s/s t2= unknown 20 = 4.9t2 ……t2 = 4.081 √t2 = √4.081 ….t= 2.02 seconds to hit the ground. To determine the minimum speed required to clear 8m in 2.02 s just use D/T= 8/2.02= 3.96 m/s horizontally to clear 8 m from 20 m vertically.
5) Using acceleration = force * mass, we know the car had a force of 5000N, and a mass of 1500kg.
5000/1500 = -3.33 m/s deceleration. It took 5.6 seconds for the car to decelerate to 0, so multiply the deceleration/acceleration with the amount of time to obtain the initial speed of the vehicle. 3.33 * 5.6 = Initial speed of 18.65 m/s
6) Weight is equated as mass * gravity. On the moon gravity would be g/6, or 9.8/6. The weight of a 100kg man would be 100*(9.8/6) = 163.33 N or 16.66kg. In order to simulate this phenomenon in an elevator here on Earth, the acceleration would have to be downward, or negative direction. We would take the normal acceleration here on earth at 9.8 m/s2 and subtract it from what it would be on the moon, 9.8/6. This would give us the negative acceleration needed in the elevator to reach moon-like conditions. The negative acceleration would be -8.17 m/s2.
7) The box is 7.93 kg. 7.93kg * 9.8 = 77.71 N downward force. 84N at an angle of 47° = 84 cos 47°= 57.3 N horizontal force forward 84 N sin 47°= 61.4 N