A constant electric field accelerates a proton from rest through a distance of 1.20 m to a speed of 1.53 ✕ 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). J (b) Find the change in the system's electric potential energy (in J). J (c) Calculate the magnitude of the electric field (in N/C).
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- An electron acquires 5.65x10-18 J of kinetic energy when it is accelerated by an electric field from plate A to plate B.Two objects (m,- 5.50 kg and m,- 2.70 kg) are connected by a light string passing over a light, frictionless pulley as in the figure below. The 5.50-kg object is released from rest at a point h-4.00 m above the table. (a) Determine the speed of each object when the two pass each other. 3.653 v m/s (b) Determine the speed of each object at the moment the 5.50-kg object hits the table. 6160 V m/s (c) How much higher does the 2.70-kg object travel after the 5.50-kg object hits the table? 1.167 xm ASK YOUR TEA proton and an alpha particle (charge = 2e, mass = 6.64 ✕ 10−27 kg) are initially at rest, separated by 7.09 ✕ 10−15 m. (a) If they are both released simultaneously, explain why you can't find their velocities at infinity using only conservation of energy. This answer has not been graded yet. (b) What other conservation law can be applied in this case? This answer has not been graded yet. (c) Find the speeds of the proton and alpha particle, respectively, at infinity. proton m/s alpha particle m/s
- (a) Calculate the speed of a proton that is accelerated from rest through an electric potential difference of 144 V. 166.11*10^3 (b) Calculate the speed of an electron that is accelerated through the same potential difference. 7.12*10^6Consider an electron, of charge magnitude e = 1.602 × 10-1⁹ C and mass m₂ = 9.11 × 10-31 kg, moving in an electric field with an electric field magnitude E = 4 x 10² N/C, similar to what Thana observed in the simulation. Let the length of the plates be L = 50 cm, and the distance between them be d = 20 cm. Find the maximum speed, v, the electron could be moving if it enters the space halfway between and parallel to the two plates to just barely strike one of the plates. m/s If the field is pointing upward, which plate will Thana conclude the electron strikes at this speed? O The upper plate, because the electron charge magnitude is positive. O The upper plate, because we are only considering the magnitude of the electron charge, and magnitudes are always positive. O The lower plate, because the electron is attracted to the negative plate. O The lower plate, because the electron is negatively charged.A proton (m=1.67 x 10^-27 kg) travels a distance of 4.3 cm parallel to a uniform electric field 3.5 x 10^5 V/m between the plates shown in the figure. If the initial velocity is 3.2 x 10^5 m/s, find the magnitude of its final velocity in m/s. (Ignore gravity)
- Problem 17: An evacuated tube uses a potential difference of AV= 0.56 kV to accelerate electrons, which then hit a copper plate and produce X-rays. . Part (a) Write an expression for the non-relativistic speed of these electrons v in terms of e, AV, and m, assuming the electrons start from rest. v = AV 7 9 HOME a b d. 4 5 e h 1 j k P END m S V VO BACKSPACE CLEAR DEL Submit Hint Feedback I give up! Hints: 0% deduction per hint. Hints remaining: 2 Feedback: 1% deduction per feedback. Part (b) Calculate the non-relativistic speed of these electrons v in m/s.An electron is to be accelerated in a uniform electric field having a strength of 2.106 (a) What energy in keV is given to the electron if it is accelerated through 0.45 m? AKE = keV m (b) Over what distance (in km) would it have to be accelerated to increase its energy by 45 GeV? d = ✔km Hint: How is potential energy, PE, gained by an electron related to the uniform electric field? How is the potential difference, V, related to the uniform electric field?The proton-proton chain is the multi-step process that powers the Sun via fusion of hydrogen into helium. In the first step of the proton-proton chain, a positron is produced when two protons fuse together (with one proton converting to a neutron). The emitted positron quickly collides with its anti-particle, the electron. (Recall that the electron and positron have exactly the same mass, but opposite electric charges.) The electron and positron then completely annihilate, converting all of their rest mass into two gamma-ray photons. Given a single particle of mass m, the amount of energy E produced when all of its mass is converted to energy is given by Einstein's famous formula, E = m c?, where c = 2.9979 x 10° m/s is %3D the speed of light. We also learned that the energy Ephoton of a single photon is related to its frequency f or wavelength A via Ephoton = hf = hc where h = 6.626 x 1034 m² kg/s is a fundamental constant of nature called Planck's constant. In the electron + positron…
- A proton is fired from very far away directly at a fixed particle with charge q = 1.60 ✕ 10−18 C. If the initial speed of the proton is 1.7 ✕ 105 m/s, what is its distance of closest approach to the fixed particle? The mass of a proton is 1.67 ✕ 10−27 kg.If electrons have kinetic energy of 2000 eV, find (a) their speed, (b) the time needed to traverse a distance of 5 cm between plates that our horizontal, and (c) the vertical component of their velocity after passing between the plates if the electric field is 3.33 x 10^3 V/m.To see why, compute the force of Earth’s gravity on an electron and compare it with the force exerted on the electron by an electric field of magnitude 14000 V/m (a relatively small field). Express your answer to two significant figures and include the appropriate units.