Understanding Our Universe
3rd Edition
ISBN: 9780393614428
Author: PALEN, Stacy, Kay, Laura, Blumenthal, George (george Ray)
Publisher: W.w. Norton & Company,
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Chapter 9, Problem 6QAP
To determine
The reason why Pluto is considered dwarf Planet.
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A proton moves at 5.20 x 105 m/s in the horizontal direction. It enters a uniform vertical electric field with a magnitude of 8.40 × 10³ N/C. Ignore any gravitational effects.
(a) Find the time interval required for the proton to travel 6.00 cm horizontally.
1.15e-7
☑
Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. ns
(b) Find its vertical displacement during the time interval in which it travels 6.00 cm horizontally. (Indicate direction with the sign of your answer.)
5.33e-3
☑
Your response is off by a multiple of ten. mm
(c) Find the horizontal and vertical components of its velocity after it has traveled 6.00 cm horizontally.
| ↑ +
jkm/s
A proton moves at 5.20 105 m/s in the horizontal direction. It enters a uniform vertical electric field with a magnitude of 8.40 103 N/C. Ignore any gravitational effects.
(a) Find the time interval required for the proton to travel 6.00 cm horizontally.
(b) Find its vertical displacement during the time interval in which it travels 6.00 cm horizontally. (Indicate direction with the sign of your answer.)
The figure below shows the electric field lines for two charged particles separated by a small distance.
92
91
(a) Determine the ratio 91/92.
1/3
×
This is the correct magnitude for the ratio.
(b) What are the signs of q₁ and 92?
91 positive
92 negative
×
Chapter 9 Solutions
Understanding Our Universe
Ch. 9.1 - Prob. 9.1CYUCh. 9.2 - Prob. 9.2CYUCh. 9.3 - Prob. 9.3CYUCh. 9.4 - Prob. 9.4CYUCh. 9.5 - Prob. 9.5CYUCh. 9.6 - Prob. 9.6CYUCh. 9 - Prob. 1QAPCh. 9 - Prob. 2QAPCh. 9 - Prob. 3QAPCh. 9 - Prob. 4QAP
Ch. 9 - Prob. 5QAPCh. 9 - Prob. 6QAPCh. 9 - Prob. 7QAPCh. 9 - Prob. 8QAPCh. 9 - Prob. 9QAPCh. 9 - Prob. 10QAPCh. 9 - Prob. 11QAPCh. 9 - Prob. 12QAPCh. 9 - Prob. 13QAPCh. 9 - Prob. 14QAPCh. 9 - Prob. 15QAPCh. 9 - Prob. 16QAPCh. 9 - Prob. 17QAPCh. 9 - Prob. 18QAPCh. 9 - Prob. 19QAPCh. 9 - Prob. 20QAPCh. 9 - Prob. 21QAPCh. 9 - Prob. 22QAPCh. 9 - Prob. 23QAPCh. 9 - Prob. 24QAPCh. 9 - Prob. 25QAPCh. 9 - Prob. 26QAPCh. 9 - Prob. 27QAPCh. 9 - Prob. 28QAPCh. 9 - Prob. 29QAPCh. 9 - Prob. 30QAPCh. 9 - Prob. 31QAPCh. 9 - Prob. 32QAPCh. 9 - Prob. 33QAPCh. 9 - Prob. 34QAPCh. 9 - Prob. 35QAPCh. 9 - Prob. 36QAPCh. 9 - Prob. 37QAPCh. 9 - Prob. 38QAPCh. 9 - Prob. 39QAPCh. 9 - Prob. 40QAPCh. 9 - Prob. 41QAPCh. 9 - Prob. 42QAPCh. 9 - Prob. 43QAPCh. 9 - Prob. 44QAPCh. 9 - Prob. 45QAP
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- Please help me solve this one more detail, thanksarrow_forwardA 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?arrow_forwardIn (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.arrow_forward
- 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
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