Recent content by IAmPat
-
I
A two-side 'A' shaped ladder and then tension in the wire that holds it together
So should I consider just one ladder and double the tension in order to get the tension in the 'A' ladder- IAmPat
- Post #3
- Forum: Introductory Physics Homework Help
-
I
A two-side 'A' shaped ladder and then tension in the wire that holds it together
Homework Statement A ladder is made in the shape of the letter A. Treat the two sides of the ladder as identical uniform rods, each weighing 455 N, with a length of 3.60 m. A frictionless hinge connects the two ends at the top, and a horizontal wire, 1.20 m long, connects them at a distance...- IAmPat
- Thread
- Tension Wire
- Replies: 3
- Forum: Introductory Physics Homework Help
-
I
Weight of a block in equilibrium in a system.
Homework Statement The picture below shows a mobile in equilibrium. Each of the rods is 0.16m long and each hangs from a supporting string that is attached one fourth of the way across it. The mass of each rod is 0.10kg. The mass of the strings connecting the blocks to the rods is...- IAmPat
- Thread
- Block Equilibrium System Weight
- Replies: 3
- Forum: Introductory Physics Homework Help
-
I
Work needed to stop a rolling hoop
Homework Statement A thin hoop of mass 7.8kg rolls on a horizontal floor with a speed (at its center of mass) of 0.23 m/s. How much work must be done on the hoop to bring it to a stop. Homework Equations m = 7.8kg Vcm = 0.23m/s KE = 1/2*m*Vcm + 1/2Iw2 I = m*r2 The Attempt...- IAmPat
- Thread
- Rolling Work
- Replies: 2
- Forum: Introductory Physics Homework Help
-
I
Work done by the torque on a pulley
Found that the moment of inertia of a uniform disc is 1/2*mr2 Plugged that in and got the correct answer. Thanks.- IAmPat
- Post #3
- Forum: Introductory Physics Homework Help
-
I
Work done by the torque on a pulley
Homework Statement The pulley in the illustration is a uniform disk of mass 2.40 kg and radius 0.220 m. The block applies a constant torque to the pulley, which is free to rotate without friction, resulting in an angular acceleration of magnitude 0.180 rad/s2 for the pulley. As the block falls...- IAmPat
- Thread
- Pulley Torque Work Work done
- Replies: 2
- Forum: Introductory Physics Homework Help
-
I
Windmill rotates with constant ang. accel. What time does Tangetial = Centripetal
Homework Statement A windmill starts from rest and rotates with a constant angular acceleration of 0.25 rad/s2. How many seconds after starting will the magnitude of the tangential acceleration of the tip of a blade equal the magnitude of the centripetal acceleration at the same point...- IAmPat
- Thread
- Centripetal Constant Time Windmill
- Replies: 1
- Forum: Introductory Physics Homework Help
-
I
Coin rolling with constant deceleration, how much time to go 1.6m
Homework Statement A coin is rolled without slipping on a table top in a straight line. It starts rolling at 3.4 rad/s and slows down at a constant angular acceleration. It is rolling at 1.2 rad/s when it falls off the table edge. If the radius of the coin is 0.011m, and the edge of the...- IAmPat
- Thread
- Constant Deceleration Rolling Time
- Replies: 1
- Forum: Introductory Physics Homework Help
-
I
Magnitude of the average force of air resistance on a rocket
Homework Statement An engine supplies an upward force of 9.00N to an initially stationary toy rocket, of mass 54.0g, for a distance of 25.0m. The rocket rises to a height of 339 meters before falling back to the ground. What was the magnitude of the average force of air resistance on the...- IAmPat
- Thread
- Air Air resistance Average Average force Force Magnitude Resistance Rocket
- Replies: 1
- Forum: Introductory Physics Homework Help
-
I
Average power needed to lift an object
Homework Statement A weightlifter raises a 115kg weight from the ground to a height of 1.95m in 1.25 seconds. What is the average power of this maneuver. Homework Equations m = 115kg Change in Y = 1.95m time = 1.25 seconds The Attempt at a Solution V = 1.95/1.25 = 1.56 m/s a...- IAmPat
- Thread
- Average Lift Power
- Replies: 1
- Forum: Introductory Physics Homework Help
-
I
Power needed to push an object vertical at constant velocity
Then I end up with the same answer as my second try though. Unless I'm calculating Force wrong? Velocity = 3.96m/s Acceleration = .396m/s2 Fy = m*a Fy = (1.20×105) * .396m/s2 Fy = 47,520 P = F * v P = (47520)(3.96) P = 188179.2 watts- IAmPat
- Post #3
- Forum: Introductory Physics Homework Help
-
I
Power needed to push an object vertical at constant velocity
Homework Statement Stuntman's Freefall, a ride at Six Flags Great Adventure in NewJersey, stands 39.6 meters high. Ignoring the force of friction, what is the minimum power rating of the motor that raises the 1.20×105kg ride from the ground to the top in 10.0 seconds at a constant velocity...- IAmPat
- Thread
- Constant Constant velocity Power Push Velocity Vertical
- Replies: 6
- Forum: Introductory Physics Homework Help
-
I
Distance traveled during speeding-up phase of a propelled sled
The only reason I'm confused as to if it's correct or not is because when I enter in "111.111" as my answer, the grading system is marking it wrong.- IAmPat
- Post #5
- Forum: Introductory Physics Homework Help
-
I
Distance traveled during speeding-up phase of a propelled sled
This is what I get for working after being completely mentally drained. Ok so, it should have been 400,000 DIVIDED by 3600 which gives me 111.1111~ m/s -> a = 111.111~ / 2.5 -> a = 44.4444~ So now I have my (correct?) acceleration. But once again I'm stuck. If I multiply the...- IAmPat
- Post #3
- Forum: Introductory Physics Homework Help
-
I
Distance traveled during speeding-up phase of a propelled sled
Homework Statement At the Aircraft Landing Dynamics Facility located at NASA's Langley Research Center in Virginia, a water-jet nozzle propels a 46,000 kg sled from zero to 400 km/hour in 2.5 seconds. Assuming a constant acceleration for the sled, what distance does it travel during the...- IAmPat
- Thread
- Distance traveled Phase
- Replies: 9
- Forum: Introductory Physics Homework Help