A proton (mass 1.67 x 10 27 kg, charge +e = +1.60 x 10 19 C) follows a path from point B to point A. If its speed at B is 3.00 x 105 m/s, what is its speed at A? Express your answer with the appropriate units. VA = Value Units Su Request Answer

A proton (mass 1.67 x 10 27 kg, charge +e = +1.60 x 10 19 C) follows a path from point B to point A. If its speed at B is 3.00 x 105 m/s, what is its speed at A? Express your answer with the appropriate units. VA = Value Units Su Request Answer

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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A charged ion with charge of +2.65 x 10-19 C and a mass of 3.15 x 10-27 kg is held halfway between two oppositely charged plates with voltage of 3.06 x 104 V. What is the maximum speed the ion will attain when let go and allowed to accelerate? Provide your answer to three significant digits, without units. Answer:
This transmission electron microscope (TEM) image of coronavirus can be taken using a beam of electrons accelerated from rest through a potential difference of 25 kV. What is the final speed of the electrons? Provide the answer: . x 108 m/s
Suppose 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=
Part A Calculate the speed of a proton after it accelerates from rest through a potential difference of 225 V. Express your answer in meters per second. Up = Submit Previous Answers Request Answer Part B VO ΑΣΦ X Incorrect; Try Again; 5 attempts remaining Ve = Submit Calculate the speed of an electron after it accelerates from rest through a potential difference of 225 V. Express your answer in meters per second. ? ——| ΑΣΦ V Request Answer m/s = ? m/s
Item 14 Rutherford's Planetary Model of the Atom In 1911, Ernest Rutherford developed a planetary model of the atom, in which a small positively charged nucleus is orbited by electrons. The model was motivated by an experiment carried out by Rutherford and his graduate students, Geiger and Marsden. In this experiment, they fired alpha particles with an initial speed of 1.30 x 107m/s at a thin sheet of gold. (Alpha particles are obtained from certain radioactive decays. They have a charge of +2e and a mass of 6.64 x 10-27 kg.) Part A 15. ΑΣΦ r= 4.8 10-14 How close can the alpha particles get to a gold nucleus (charge = +79e), assuming the nucleus remains stationary? (This calculation sets an upper limit on the size of the gold nucleus. See Chapter 31 of the textbook for further details.) ● Submit Previous Answers Request Answer Provide Feedback ? X Incorrect; Try Again; 2 attempts remaining 14 of 15 m Review Next >
need help solving for C A constant electric field accelerates a proton from rest through a distance of 1.55 m to a speed of 1.97 ✕ 105 m/s. (The mass and charge of a proton are mp = 1.67 ✕ 10−27 kg and qp = e = 1.60 ✕ 10−19 C.) HINT (a) Find the change in the proton's kinetic energy (in J). 3.24E-17 J (b) Find the change in the system's electric potential energy (in J). The work done by the electric force and the change in electric potential energy are related by WFe = −ΔPE. This relationship holds for any conservative force: the work done by any conservative force equals the negative of the change in the associated potential energy. -3.24E-17J (c) Calculate the magnitude of the electric field (in N/C). N/C
Positive electric charge QQ is distributed uniformly along a thin rod of length 2a. The rod lies along the x-axis between x=−a and x=+a (Figure 1). Calculate how much work you must do to bring a positive point charge q from infinity to the point x=+L on the x-axis, where L>a. What does your result for the potential energy U(x=+L) become in the limit a→0?
Item 14 Rutherford's Planetary Model of the Atom In 1911, Ernest Rutherford developed a planetary model of the atom, in which a small positively charged nucleus is orbited by electrons. The model was motivated by an experiment carried out by Rutherford and his graduate students, Geiger and Marsden. In this experiment, they fired alpha particles with an initial speed of 1.30 x 107m/s at a thin sheet of gold. (Alpha particles are obtained from certain radioactive decays. They have a charge of +2e and a mass of 6.64 × 10-27 kg.) Part A r = How close can the alpha particles get to a gold nucleus (charge = +79e), assuming the nucleus remains stationary? (This calculation sets an upper limit on the size of the gold nucleus. See Chapter 31 of the textbook for further details.) —| ΑΣΦ Submit Request Answer ? 14 of 15 m Review
Problem A newly discovered light positively charged particle has a mass of m and charge q. Suppose it moves within the vicinity of an extremely heavy (fixed in place) particle with a positive charge Q and mass M. When the light particle is xi distance from the heavy particle, it is moving directly away from the heavy particle with a speed of vi. a) What is the lighter particle's speed when it is xf away from the heavy particle? (Consider the Newtonian Gravitation acting between the two charged particles. Ignore the effects of external forces) Solution: We may solve this using two approaches. One involves the Newton's Laws and the other involving Work-Energy theorem. To avoid the complexity of vector solution, we will instead employ the Work-Energy theorem, more specifically, the Conservation of Energy Principle. Let us first name the lighter particle as object 1 and the heavy particle as object 2. Through work-energy theorem, we will take into account all of the energy of the…
Problem A newly discovered light positively charged particle has a mass of m and charge q. Suppose it moves within the vicinity of an extremely heavy (fixed in place) particle with a positive charge Q and mass M. When the light particle is xi distance from the heavy particle, it is moving directly away from the heavy particle with a speed of vi. a) What is the lighter particle's speed when it is xf away from the heavy particle? (Consider the Newtonian Gravitation acting between the two charged particles. Ignore the effects of external forces) Solution: We may solve this using two approaches. One involves the Newton's Laws and the other involving Work-Energy theorem. To avoid the complexity of vector solution, we will instead employ the Work-Energy theorem, more specifically, the Conservation of Energy Principle. Let us first name the lighter particle as object 1 and the heavy particle as object 2. Through work-energy theorem, we will take into account all of the energy of the…
Problem A newly discovered light positively charged particle has a mass of m and charge q. Suppose it moves within the vicinity of an extremely heavy (fixed in place) particle with a positive charge Q and mass M. When the light particle is xi distance from the heavy particle, it is moving directly away from the heavy particle with a speed of vi. a) What is the lighter particle's speed when it is xf away from the heavy particle? (Consider the Newtonian Gravitation acting between the two charged particles. Ignore the effects of external forces) Solution: We may solve this using two approaches. One involves the Newton's Laws and the other involving Work-Energy theorem. To avoid the complexity of vector solution, we will instead employ the Work-Energy theorem, more specifically, the Conservation of Energy Principle. Let us first name the lighter particle as object 1 and the heavy particle as object 2. Through work-energy theorem, we will take into account all of the energy of the…