College Physics
OER 2016 Edition
ISBN: 9781947172173
Author: OpenStax
Publisher: OpenStax College
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Chapter 9, Problem 1TP
To determine
The object which is not an example of an object undergoing a torque.
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Chapter 9 Solutions
College Physics
Ch. 9 - What can you say about the velocity of a moving...Ch. 9 - Under what conditions can a rotating body be in...Ch. 9 - What three factors affect the torque created by a...Ch. 9 - A wrecking ball is being used to knock down a...Ch. 9 - Mechanics put a length of Pipe over the handle of...Ch. 9 - Prob. 6CQCh. 9 - Explain the need for tall towers on a suspension...Ch. 9 - When visiting some countries, you may see a person...Ch. 9 - Prob. 9CQCh. 9 - Prob. 10CQ
Ch. 9 - Why are the forces exerted on the outside world by...Ch. 9 - Explain why the forces in our joints are several...Ch. 9 - Why are the forces exerted on the outside world by...Ch. 9 - Explain why the forces in our joints are several...Ch. 9 - Certain of dinosaurs were bipedal (walked on two...Ch. 9 - Swimmers and athletes during competition need to...Ch. 9 - If the maximum force the biceps muscle can exert...Ch. 9 - Suppose the biceps muscle was attached through...Ch. 9 - Explain one of the reasons why pregnant women...Ch. 9 - (a) When opening a door, you push on it...Ch. 9 - When tightening a bolt, you push perpendicularly...Ch. 9 - Two children push on opposite sides of a door...Ch. 9 - Use the second condition for equilibrium (net =0 )...Ch. 9 - Repeat the seesaw problem in Example 9.1 with the...Ch. 9 - Prob. 6PECh. 9 - Two children of mass 20.0 kg and 30.0 kg sit...Ch. 9 - Prob. 8PECh. 9 - A person carries a plank of wood 2.00 m long with...Ch. 9 - Prob. 10PECh. 9 - Prob. 11PECh. 9 - Prob. 12PECh. 9 - Prob. 13PECh. 9 - Prob. 14PECh. 9 - Prob. 15PECh. 9 - Prob. 16PECh. 9 - To get up on the roof, a person (mass 70.0 kg)...Ch. 9 - Prob. 18PECh. 9 - Prob. 19PECh. 9 - Suppose you needed to raise a 250-kg mower a...Ch. 9 - Prob. 21PECh. 9 - Prob. 22PECh. 9 - Prob. 23PECh. 9 - Prob. 24PECh. 9 - Prob. 25PECh. 9 - Prob. 26PECh. 9 - Prob. 27PECh. 9 - Prob. 28PECh. 9 - Prob. 29PECh. 9 - Prob. 30PECh. 9 - Prob. 31PECh. 9 - Prob. 32PECh. 9 - Prob. 33PECh. 9 - Prob. 34PECh. 9 - Prob. 35PECh. 9 - Integrated Concepts Suppose we replace the 4.0-kg...Ch. 9 - Prob. 37PECh. 9 - You have just planted a sturdy 2-m-tall palm tree...Ch. 9 - Unreasonable Results Suppose two children are...Ch. 9 - Construct Your Own Problem Consider a method for...Ch. 9 - Prob. 1TPCh. 9 - Prob. 2TPCh. 9 - Prob. 3TPCh. 9 - Prob. 4TPCh. 9 - Prob. 5TPCh. 9 - Prob. 6TPCh. 9 - Prob. 7TPCh. 9 - Prob. 8TPCh. 9 - Prob. 9TPCh. 9 - Prob. 10TPCh. 9 - Prob. 11TP
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- Plastic beads can often carry a small charge and therefore can generate electric fields. Three beads are oriented such that system of all three beads is zero. 91 E field lines 93 92 What charge does each bead carry? 91 92 -1.45 = = What is the net charge of the system? What charges have to be equal? μC 2.9 × What is the net charge of the system? What charges have to be equal? μC 93 = 2.9 μС 92 is between and 91 93° The sum of the charge on q₁ and 92 is 91 + 92 = −2.9 μC, and the net charge of thearrow_forwardPlastic beads can often carry a small charge and therefore can generate electric fields. Three beads are oriented such that 92 is between q₁ and 93. The sum of the charge on 9₁ and 92 is 9₁ + 92 = −2.9 µС, and the net charge of the system of all three beads is zero. E field lines 93 92 What charge does each bead carry? 91 92 -1.45 What is the net charge of the system? What charges have to be equal? μC 2.9 ✓ What is the net charge of the system? What charges have to be equal? μC 93 2.9 μεarrow_forwardNo chatgpt pls will upvotearrow_forward
- Point charges of 6.50 μC and -2.50 μC are placed 0.300 m apart. (Assume the negative charge is located to the right of the positive charge. Include the sign of the value in your answers.) (a) Where can a third charge be placed so that the net force on it is zero? 0.49 m to the right of the -2.50 μC charge (b) What if both charges are positive? 0.49 xm to the right of the 2.50 μC chargearrow_forwardFind the electric field at the location of q, in the figure below, given that q₁ =9c9d = +4.60 nC, q = -1.00 nC, and the square is 20.0 cm on a side. (The +x axis is directed to the right.) magnitude direction 2500 x What symmetries can you take advantage of? What charges are the same magnitude and the same distance away? N/C 226 × How does charge sign affect the direction of the electric field? counterclockwise from the +x-axis 9a 9b % 9 9darrow_forwardwould 0.215 be the answer for part b?arrow_forward
- Suppose a toy boat moves in a pool at at a speed given by v=1.0 meter per second at t=0, and that the boat is subject to viscous damping. The damping on the boat causes the rate of speed loss to be given by the expression dv/dt=-2v. How fast will the boat be traveling after 1 second? 3 seconds? 10 seconds? Use separation of variables to solve this.arrow_forwardWhat functional form do you expect to describe the motion of a vibrating membrane without damping and why?arrow_forwardIf speed is tripled, how much larger will air drag become for an object? Show the math.arrow_forward
- What does it tell us about factors on which air drag depends if it is proportional to speed squared?arrow_forwardWhat is the net charge on a sphere that has the following? x (a) 5.75 × 106 electrons and 8.49 × 106 protons 4.39e-13 What is the charge of an electron? What is the charge of a proton? C (b) 200 electrons and 109 protons 1.60e-10 What is the charge of an electron? What is the charge of a proton? Carrow_forwardA spider begins to spin a web by first hanging from a ceiling by his fine, silk fiber. He has a mass of 0.025 kg and a charge of 3.5 μC. A second spider with a charge of 4.2 μC rests in her own web exactly 2.1 m vertically below the first spider. (a) What is the magnitude of the electric field due to the charge on the second spider at the position of the first spider? 8.57e3 N/C (b) What is the tension in the silk fiber above the first spider? 0.125 How does the electric field relate to the force? How do you calculate the net force? Narrow_forward
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