- #1
moonlit
- 57
- 0
I'm having problems understanding 5 of my homework problems and was wondering if anyone could help me with them.
1) Two forces are applied to an old tree stump, to pull it out of the ground. One rope pulls due east with a force of 2100 N. A second force is 45 degrees South of East with a magnitude of 2800 N. Calculate the magnitude and direction of the resultant force.
My solution: c^2 = a^2 + b^2
sqrt 2100^2 + 2800^2 = 3500 N
theta = tan^-1 (2800/2100)= 53.13 degrees
2) A roofer working 12.0 meters above the ground tosses a piece of scrap wood off the roof. Just for fun he tosses it nearly straight up with a velocity of 4.3 m/s.
a) Calculate the velocity when it hits the ground.
b) Calculate its height above the ground 1.1 seconds after it's tossed.
My solution: a) 12.0/4.3 = 2.79 m/s
b) Vx=Vox + axt = 12+4.3(1.1)=16.73 m
3) A person jogs around a circular road with a radius of 2.30 km. The jogger goes half way around in 17 minutes then stops because of a muscle cramp.
a) What is the total distance run?
b) Calculate the magnitude of the jogger's displacement.
c) Calculate the magnitude of the average velocity in m/s (C=2pir)
My solution: a) 2.30^2 + 2.30^2 = 10.58 km
b) Vx^2=Vox^2+2axX = 2.30^2+2(17)=39.29 km/min
c) C=2pir = 14.45 m/s
4) A car leaves Westown heading for Easton, 120 miles away at an average velocity of 55 mi/h east. At the same time a car pulls out of Easton at 40 mi/h heading west.
a) How long will it take for them to meet (in hrs)?
b) Find the distance in miles from Westown where the two cars cross.
My solution: a) 120/55 = 2.66
120/50 = 3
3+2.66/2=2.8 hrs
b) 55 + 40 = 95
120 - 95 = 25 miles
5) A ball is thrown straight up and reaches a maximum height of 15.3 m. At what height from the launch point is the velocity equal to one half the initial velocity?
My solution: 15.3/9.8 = 1.56
1.56/.5 = 3.12 m
1) Two forces are applied to an old tree stump, to pull it out of the ground. One rope pulls due east with a force of 2100 N. A second force is 45 degrees South of East with a magnitude of 2800 N. Calculate the magnitude and direction of the resultant force.
My solution: c^2 = a^2 + b^2
sqrt 2100^2 + 2800^2 = 3500 N
theta = tan^-1 (2800/2100)= 53.13 degrees
2) A roofer working 12.0 meters above the ground tosses a piece of scrap wood off the roof. Just for fun he tosses it nearly straight up with a velocity of 4.3 m/s.
a) Calculate the velocity when it hits the ground.
b) Calculate its height above the ground 1.1 seconds after it's tossed.
My solution: a) 12.0/4.3 = 2.79 m/s
b) Vx=Vox + axt = 12+4.3(1.1)=16.73 m
3) A person jogs around a circular road with a radius of 2.30 km. The jogger goes half way around in 17 minutes then stops because of a muscle cramp.
a) What is the total distance run?
b) Calculate the magnitude of the jogger's displacement.
c) Calculate the magnitude of the average velocity in m/s (C=2pir)
My solution: a) 2.30^2 + 2.30^2 = 10.58 km
b) Vx^2=Vox^2+2axX = 2.30^2+2(17)=39.29 km/min
c) C=2pir = 14.45 m/s
4) A car leaves Westown heading for Easton, 120 miles away at an average velocity of 55 mi/h east. At the same time a car pulls out of Easton at 40 mi/h heading west.
a) How long will it take for them to meet (in hrs)?
b) Find the distance in miles from Westown where the two cars cross.
My solution: a) 120/55 = 2.66
120/50 = 3
3+2.66/2=2.8 hrs
b) 55 + 40 = 95
120 - 95 = 25 miles
5) A ball is thrown straight up and reaches a maximum height of 15.3 m. At what height from the launch point is the velocity equal to one half the initial velocity?
My solution: 15.3/9.8 = 1.56
1.56/.5 = 3.12 m