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College Physics Answers was created by Shaun Dychko, a teacher with more than 12 years experience teaching high school physics and mathematics, mostly at Point Grey Secondary in Vancouver, Canada. I have taught AP Physics, and all levels of high school mathematics. I studied physics at the University of British Columbia, where I obtained both ...
An object is dropped from a height of 75.0 m above ground level. (a) Determine the distance traveled during the first second. (b) Determine the final velocity at which the object hits the ground. (c) Determine the distance traveled during the last second of motion before hitting the ground.
Problem number. 2. 22.4 Magnetic Field Strength: Force on a Moving Charge in a Magnetic Field. 22.5 Force on a Moving Charge in a Magnetic Field: Examples and Applications. 22.7 Magnetic Force on a Current-Carrying Conductor. 22.10 Magnetic Force between Two Parallel Conductors. OpenStax College Physics. Chapter 22: Magnetism.
Problem 19. Do Exercise 3.16 again using analytical techniques and change the second leg of the walk to straight south. (This is equivalent to subtracting B from A —that is, finding R' = A - B) (b) Repeat again, but now you first walk 25.0 m north and then 18.0 m east. (This is equivalent to subtract A from B — that is, to find R'' = B - A.
Consider the 52.0-kg mountain climber in Figure 5.20. (a) Find the tension in the rope and the force that the mountain climber must exert with her feet on the vertical rock face to remain stationary. Assume that the force is exerted parallel to her legs. Also, assume negligible force exerted by her arms.
Exercise 22.48. (a) A 200-turn circular loop of radius 50.0 cm is vertical, with its axis on an east-west line. A current of 100 A circulates clockwise in the loop when viewed from the east. The Earth’s field here is due north, parallel to the ground, with a strength of 3.00 \times 10^ {-5} \textrm { T} 3.00 ×10−5 T .
We solve this for h by multiplying both sides by one over mg and we get h equals kx squared over two mg. So that's 2.5 times ten to the four newtons per meter times 12 times ten to the minus two meters squared, divided by two times 40 kilograms, times 9.8 newtons per kilogram, which is 0.459 meters. That is the height above this reference level ...
19. OpenStax College Physics. Chapter 4: Dynamics: Force and Newton's Laws of Motion. Problem 19. Question. (a) Calculate the tension in a vertical strand of spider web if a spider of mass 8.00 \times 10^ {-5} \textrm { kg} 8.00 ×10−5 kg hangs motionless on it. (b) Calculate the tension in a horizontal strand of spider web if the same spider ...
5. OpenStax College Physics. Chapter 7: Work, Energy, and Energy Resources. Problem 5. Question. Calculate the work done by an 85.0-kg man who pushes a crate 4.00 m up along a ramp that makes an angle of 20^\circ 20∘ with the horizontal. (See Figure 7.34.) He exerts a force of 500 N on the crate parallel to the ramp and moves at a constant speed.
11. OpenStax College Physics. Chapter 3: Two-Dimensional Kinematics. Problem 11. Question. Find the components of v_\textrm {tot} vtot along the x x and y y axes in Figure 3.55. Figure 3.55: The two velocities v_ {A} and v_ {B} add to give a total v_ {tot} Question by OpenStax is licensed under CC BY 4.0. Final Answer.
