A small plastic ball is connected to a lightweight cord of length L = 1.60 m that istied to a point P, as shown in the figure.y = 0 + 1L.Top viewThe ball, string, and point all lie on a frictionless, horizontal table. The ballhas a mass m= 0.0100 kg, a charge q = +2.50 µc and is released from rest when thestring makes an angle o = 60.0° with a uniform electric field of magnitudeE = 320 V/m. Determine the speed (in m/s) of the ball when the string is parallelto the electric field.m/sNeed Help?Read It

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stion 8 The particle in the picture has a charge q=2 uC and at the location shown (point B) its kimetic energy is K= 100 µJ. 50 V 100V 200V Determine the kinetic energy (in pJ) of the particle at pointA. A Moving to another ouestion wil save this response arch
11. ◆-2 points SerCP11 16.2·P021. My Notes Ask Your Teacher A tiny sphere of mass 7.20 mg and charge-2.80 nC is initially at a distance of 1.64 μm from a fixed charge of +8.45 nC. (a) If the 7.20-mg sphere is released from rest, find its kinetic energy when it is 0.500 μm from the fixed charge (b) If the 7.20-mg sphere is released from rest, find its speed when it is 0.500 μm from the fixed charge. m/s Need Help? ReadIt Submit Answer Save Progress Practice Another Version
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Ch 18 q3 In 1910 Rutherford performed a classic experiment in which he directed a beam of alpha particles at a thin gold foil. He unexpectedly observed a few of the particles scattered almost directly backward. This result was not consistent with then current models of atomic structure and led Rutherford to propose the existence of a very dense concentration of positive charge at the center of an atom—the atomic nucleus. The alpha particle has a charge of +2e and the gold nucleus a charge of +79e. Suppose that an alpha particle is initially a great distance from the gold, has a kinetic energy of 4.20 MeV (4.20 106 eV), and is headed directly at a gold nucleus. How close will the particle come to the center of the nucleus? Treat the nucleus and the alpha particle as point charges.
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