FlipIt for College Physics (Algebra Version - Six Months Access)
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Author: Todd Ruskell
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Chapter 6, Problem 33QAP
To determine
The maximum kinetic energy of puma.
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A dielectric-filled parallel-plate capacitor has plate area A = 20.0 ccm2 , plate separaton d = 10.0 mm and dielectric constant k = 4.00. The capacitor is connected to a battery that creates a constant voltage V = 12.5 V . Throughout the problem, use ϵ0 = 8.85×10−12 C2/N⋅m2 . Find the energy U1 of the dielectric-filled capacitor. The dielectric plate is now slowly pulled out of the capacitor, which remains connected to the battery. Find the energy U2 of the capacitor at the moment when the capacitor is half-filled with the dielectric. The capacitor is now disconnected from the battery, and the dielectric plate is slowly removed the rest of the way out of the capacitor. Find the new energy of the capacitor, U3. In the process of removing the remaining portion of the dielectric from the disconnected capacitor, how much work W is done by the external agent acting on the dielectric?
In (Figure 1) C1 = 6.00 μF, C2 = 6.00 μF, C3 = 12.0 μF, and C4 = 3.00 μF. The capacitor network is connected to an applied potential difference Vab. After the charges on the capacitors have reached their final values, the voltage across C3 is 40.0 V. What is the voltage across C4? What is the voltage Vab applied to the network? Please explain everything in steps.
Chapter 6 Solutions
FlipIt for College Physics (Algebra Version - Six Months Access)
Ch. 6 - Prob. 1QAPCh. 6 - Prob. 2QAPCh. 6 - Prob. 3QAPCh. 6 - Prob. 4QAPCh. 6 - Prob. 5QAPCh. 6 - Prob. 6QAPCh. 6 - Prob. 7QAPCh. 6 - Prob. 8QAPCh. 6 - Prob. 9QAPCh. 6 - Prob. 10QAP
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- I need help with these questions again. A step by step working out with diagrams that explains more clearlyarrow_forwardIn a certain region of space the electric potential is given by V=+Ax2y−Bxy2, where A = 5.00 V/m3 and B = 8.00 V/m3. Calculate the direction angle of the electric field at the point in the region that has cordinates x = 2.50 m, y = 0.400 m, and z = 0. Please explain. The answer is not 60, 120, or 30.arrow_forwardAn infinitely long line of charge has linear charge density 4.00×10−12 C/m . A proton (mass 1.67×10−−27 kg, charge +1.60×10−19 C) is 18.0 cm from the line and moving directly toward the line at 4.10×103 m/s . How close does the proton get to the line of charge?arrow_forward
- at a certain location the horizontal component of the earth’s magnetic field is 2.5 x 10^-5 T due north A proton moves eastward with just the right speed so the magnetic force on it balances its weight. Find the speed of the proton.arrow_forwardExample In Canada, the Earth has B = 0.5 mT, pointing north, 70.0° below the horizontal. a) Find the magnetic force on an oxygen ion (O) moving due east at 250 m/s b) Compare the |FB| to |FE| due to Earth's fair- weather electric field (150 V/m downward).arrow_forwardThree charged particles are located at the corners of an equilateral triangle as shown in the figure below (let q = 2.20 µC, and L = 0.810 m). Calculate the total electric force on the 7.00-µC charge. What is the magnitude , what is the direction?arrow_forward
- (a) Calculate the number of electrons in a small, electrically neutral silver pin that has a mass of 9.0 g. Silver has 47 electrons per atom, and its molar mass is 107.87 g/mol. (b) Imagine adding electrons to the pin until the negative charge has the very large value 2.00 mC. How many electrons are added for every 109 electrons already present?arrow_forward(a) Calculate the number of electrons in a small, electrically neutral silver pin that has a mass of 13.0 g. Silver has 47 electrons per atom, and its molar mass is 107.87 g/mol.arrow_forward8 Two moving charged particles exert forces on each other because each creates a magnetic field that acts on the other. These two "Lorentz" forces are proportional to vix (2 xr) and 2 x (vi x-r), where is the vector between the particle positions. Show that these two forces are equal and opposite in accordance with Newton's third law if and only if rx (vi × 2) = 0.arrow_forward
- 6 The force = +3 + 2k acts at the point (1, 1, 1). Find the torque of the force about (a) (b) the point (2, -1, 5). Careful about the direction of ŕ between the two points. the line = 21-+5k+ (i-+2k)t. Note that the line goes through the point (2, -1, 5).arrow_forward5 Find the total work done by forces A and B if the object undergoes the displacement C. Hint: Can you add the two forces first?arrow_forward1 F2 F₁ -F₁ F6 F₂ S A Work done on the particle as it moves through the displacement is positive. True False by the force Farrow_forward
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