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- 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/sA 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'.In nuclear fission, a nucleus splits roughly in half. A. What is the potential 4.00 x 10-14 m from a fragment that has 50 protons in it? (p=+1.602x10-19 C) →Why 1.8 MV is the answer for this problem?
- Answers: a) 3.3 C c/m² 6. Two electrons are fixed 2.00 cm apart. Another electron is shot from infinity and comes to rest midway between the two. What was its initial speed? Answer: 318 m/sConsider a proton approaching a helium-3 nucleus that has two protons and one neutron. Determine the speed at which the proton must be moving in order to get within 1.0×10^−15 m of the helium-3 nucleus. Assume that the helium nucleus is a point-like object and does not move.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=
- In the science fiction story "The little prince is an android" a second journey of the character from the novel by Antoine de Saint-Exupéry is proposed, in which he visits a planet where respect for the ecosystem is paramount. They have developed a technology capable of storing the energy released in electrical storms where 92 lightning strikes are produced every second. In a simplified way, lightning occurs when the potential difference between the cloud and the planet's surface reaches at least 2.6 cross times 10 to the power of 7 volts. The electrical currents generated in lightning are about 3.4 cross times 10 to the power of 4 amperes on average. Determine the average power of 92 lightning strikes produced with these values of potential difference and an electric current. Select one: O 8.8x 10¹1 W 9.6 × 10⁹ W 8.1 x 10¹3 W 7x 10¹ WCalculate the speed of an electron (in m/s) after it accelerates from rest through a potential difference of 140 V. 7.02e25 m/s The speed of the electron v can be found using the kinetic energy equation, K = 1/2(mv^2). The change in kinetic energy is equal to the work done on the electron, K = W = -U = -qV. Using those two equations, solve for the speed v in terms of the potential V and constants. Remember that the mass of an electron is 9.1e-31 kg and its charge is 1.6e-19 C. Submit Answer Incorrect. Tries 1/2 Previous TriesSuppose 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/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=Suppose an electron (q= -e= -1.6x 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=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 10' m/sIn one model of the hydrogen atom, the electron revolves in a circular orbit of radius 5.3 x 10-11 m. Calculate the speed of the electron.
