Schaum's Outline of College Physics, Twelfth Edition (Schaum's Outlines)
12th Edition
ISBN: 9781259587399
Author: Eugene Hecht
Publisher: McGraw-Hill Education
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Chapter 25, Problem 38SP
A proton
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Chapter 25 Solutions
Schaum's Outline of College Physics, Twelfth Edition (Schaum's Outlines)
Ch. 25 - 25.29 [I] What happens to the electric potential...Ch. 25 - 25.30 [I] What happens to the electric potential...Ch. 25 - 25.31 [I] What happens to the electric potential...Ch. 25 - 25.32 [I] Determine the electric potential 1.00 cm...Ch. 25 - 25.33 [I] Imagine a +40.0-nC point charge in...Ch. 25 - 25.34 [I] A small metal sphere carrying a charge...Ch. 25 - 25.35 [I] Imagine a charge in an evacuated...Ch. 25 - 25.36 [I] Two metal plates are attached to the two...Ch. 25 - 25.37 [II] The plates described in Problem 25.36...Ch. 25 - 25.38 [II] A proton is accelerated from rest...
Ch. 25 - 25.39 [II] An electron gun shoots electrons at a...Ch. 25 - 25.40 [I] The potential difference between two...Ch. 25 - 25.41 [II] An electron is shot with speed ...Ch. 25 - 25.42 [II] A potential difference of 24 kV...Ch. 25 - 25.43 [II] Compute the magnitude of the electric...Ch. 25 - 25.44 [II] A charge of 0.20 is 30 cm from a point...Ch. 25 - 25.45 [II] A point charge of +2.0 is placed at...Ch. 25 - 25.46 [II] In Problem 25.45, what is the...Ch. 25 - 25.47 [II] An electron is moving in the...Ch. 25 - 25.48 [II] An electron has a speed of as it...Ch. 25 - 25.49 [I] A capacitor with air between its plates...Ch. 25 - 25.50 [I] Determine the charge on each plate of a...Ch. 25 - 25.51 [I] A capacitor is charged with 9.6 nC and...Ch. 25 - 25.52 [I] Compute the energy stored in a 60-pF...Ch. 25 - 25.53 [II] Three capacitors, each of capacitance...Ch. 25 - 25.54 [I] Three capacitors (2.00, 5.00, and 7.00)...Ch. 25 - 25.55 [I] Three capacitors (2.00, 5.00, and 7.00)...Ch. 25 - 25.56 [I] The capacitor combination in Problem...Ch. 25 - 25.57 [II] Two capacitors (0.30 and 0.50 ) are...Ch. 25 - 25.58 [II] A 2.0- capacitor is charged to 50 V and...Ch. 25 - 25.59 [II] Repeat Problem 25.58 if the positive...Ch. 25 - 25.60 [II] (a) Calculate the capacitance of a...Ch. 25 - 25.61 [II] Referring to Fig. 25-2, if the...Ch. 25 - 25.62 [II] Referring to Fig. 25-2, if the...Ch. 25 - 25.63 [II] Referring to Fig. 25-2, if the...Ch. 25 - 25.64 [II] Referring to Fig. 25-10, what is the...Ch. 25 - 25.65 [II] Referring to Fig. 25-12, what is the...Ch. 25 - 25.66 [II] Referring to Fig. 25-13, what is the...
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- If an electron is accelerated from rest through a potential difference of 9.9 kV, what is its resulting speed? (e = 1.60 x 10-19 C, k= 1/4ne0 = 8.99 x 109 N. m2/C2, mel = 9.11 x 10-31 kg) %3D O 5.9 x 107 m/s 2.9 x 107 m/s O 4.9 x 10 m/s 3.9 x 10 m/s None of the given choices.arrow_forwardSuppose an electron (q = - e = - 1.6 x 10¬19 C,m=9.1 × 10¬3' kg) is accelerated from rest through a potential difference of Vab = +5000 V. Solve for the final speed of the electron. Express numerical answer in two significant figures. The potential energy U is related to the electron charge (-e) and potential Vab is related by the equation: U = Assuming all potential energy U is converted to kinetic energy K, K + U = 0 K = -U mv and using the formula for potential energy above, we arrive at an equation for speed: 2 Since K= v = ( 1/2 Plugging in values, the value of the electron's speed is: x 107 m/s V=arrow_forwardSuppose an electron (q = - e= -1.6 × 10¬19 c,m=9.1× 10¬31 kg) is accelerated from rest through a potential difference of Vab = +5000 V. Solve for the final speed of the electron. Express numerical answer in two significant figures. The potential energy U is related to the electron charge (-e) and potential Vab is related by the equation: U = Assuming all potential energy U is converted to kinetic energy K, K+U = 0 K= -U Since K= and using the formula for potential energy above, we arrive at an equation for speed: v = ( 1/2 Plugging in values, the value of the electron's speed is: x 107 m/s V=arrow_forward
- An electron is accelerated from rest through a potential difference of 181 V. What is the electron's final speed? (e = 1.60×10-¹9 C, Melectron 9.11x10-31 kg) =arrow_forwardIf an electron is accelerated from rest through a potential difference of 9.9 kV, what is its resulting speed? (e = 1.60 x 10-19 C. k= 1/4xe0 %3D = 8.99 x 109 N-m2/C2, mel = 9.11 x 10-31 kg) %3D O 5.9 × 10' m/s O 3.9 x 107 m/s O 2.9 x 107 m/s O None of the given choices. O 4.9 x 107 m/sarrow_forwardA proton is accelerated through a potential difference of 6 MV. (megavolts) from rest. Calc. the final velocity in m/s (a) 1.4 X 10' (b) 2.4 X 10' (c) 3.4 X 10' (d) 4.4 X 10'.arrow_forward
- Suppose an electron (q= -e= -1.6 x 10-19 C,m=9.1x 10-31 kg) is accelerated from rest through a potential difference of Vab = +5000 V. Solve for the final speed of the electron. Express numerical answer in two significant figures. The potential energy U is related to the electron charge (-e) and potential Vab is related by the equation: U = Assuming all potential energy U is converted to kinetic energy K, K +U = 0 K = -U Since K- and using the formula for potential energy above, we arrive at an equation for speed: v = ( 51/2 Plugging in values, the value of the electron's speed is: V= x 107 m/sarrow_forwardAn electron is accelerating under a potential difference of 105 volts. (m,=9,108x10 31) a) What is the kinetic energy of the electron?arrow_forwardSuppose an electron (q = -e= - 1.6 × 10¬19 C,m = 9.1 x 10-31 kg) is accelerated from rest through a potential difference of Vab = +5000 V. Solve for the final speed of the electron. Express numerical answer in two significant figures. The potential energy U is related to the electron charge (-e) and potential Vab is related by the equation: U = Assuming all potential energy U is converted to kinetic energy K, K+U = 0 K = -U 1 mv and using the formula for potential energy above, we arrive at an equation for speed: Since K= V = ( 1/2 Plugging in values, the value of the electron's speed is: x 107 m/s V=arrow_forward
- Suppose an electron (q = - e= - 1.6 x 10-19 C.m = 9.1 x 10-31 kg) is accelerated from rest through a potential difference of Vab = +5000 V. Solve for the final speed of the electron. Express numerical answer in two significant figures. The potential energy U is related to the electron charge (-e) and potential Vab is related by the equation: U = Assuming all potential energy U is converted to kinetic energy K, K+U= 0 K= -U 1 Since K=mv and using the formula for potential energy above, we arrive at an equation for speed: 1/2 Plugging in values, the value of the electron's speed is: x 107 m/s V=arrow_forwardA proton is accelerated by a potential difference of 10 kV. How fast is the proton moving if it started from rest? A. 9.41 x 10-6 m/s B. 3.45 x 106 m/s C. 1.38 x 106 m/s D. 2.12 x 106 m/s Select one: О а. А O b. B О с. С d. D Clear my choicearrow_forwardA proton is accelerated from rest through a potential difference of 370 V. What is its final speed?arrow_forward
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