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pls let me know which numbers should i put it 7. [0.42/1 Points] DETAILS PREVIOUS ANSWERS SERPSE10 24.AMT.001. MY NOTES | ASK YOUR'I Analysis Model
pls let me know which numbers should i put it
7. [0.42/1 Points] DETAILS PREVIOUS ANSWERS SERPSE10 24.AMT.001. MY NOTES | ASK YOUR'I Analysis Model Tutorial An electron moving parallel to the x axis has an initial speed of 5.75 x 106 m/s at the origin. Its speed is reduced to 1.19 x 105 m/s at the point x = 2.50 cm. Part 5 of 6 - Analyze: (cont.) (4) Based on the correct choice in question (3), substitute numerical values to find the potential difference between the two points. AV = - - Me(v 2 - v 2 ) 2e (9.11 x 10-31 kg) = x 105 m/s) x 106 m/s) 2(1.60 x 10-19 C)8. [0.39/1 Points] PREVIOUS ANSWERS SERPSE10 24.AMT.002. ASK YOUR TEACHER | PRACTICE ANOTHER This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part. Analysis Model Tutorial A particle having charge q = +2.10 MC and mass m = 0.0100 kg is connected to a string that is L = 1.90 m long and tied to the pivot point P in the figure below. The particle, string, and pivot point all lie on a frictionless, horizontal table. The particle is released from rest when the string makes an angle 9 = 60.0 with a uniform electric eld of magnitude E = 290 V/m. Determine the speed of the particle when the string is parallel to the electric field. Part 5 of 5 - Analyze: (cont.) (4) Based on the correct choice in question (3), substitute numerical values to nd the speed with which the particle passes through the point at which the string is parallel to the electric field. vf = u cos 0) m 2( x 10'6 C)( V/m)( ' In") (1 cos 60.0) E m 13. [0.28/1 Points] PREVIOUS ANSWERS SERPSE10 24.AMT.003. ASK YOUR TEACHER PRACTICE ANOTHER This question has several parts that must be completed sequential/y. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part. Analysis Model Tutorial Two insulating spheres have radii 0.300 cm and 0.500 cm, masses 0.200 kg and 0.700 kg, and uniformly distributed charges 2.00 11C and 6.00 uC. They are released from rest when their centers are separated by 1.00 m. Part 6 of 8 - Analyze: (cont.) (5) Based on the correct choice in question (4), substitute numerical values to find the speed of sphere 1 when it collides with sphere 2. 2 ke9192 1 V 1 1 + m1 2(8.988 x 109 N . m2/C2) x 10-12 C2 1 1 = m m kg m/s14. [0.51/1 Points] PREVIOUS ANSWERS SERPSE10 24.AMT.004. ASK YOUR TEACHER PRACTICE ANOTHER This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part. Analysis Model Tutorial In 1911, Ernest Rutherford and his assistants Geiger and Marsden conducted an experiment in which they scattered alpha particles (nuclei of helium atoms) from thin sheets of gold. An alpha particle, having charge +26 and mass 6.64 x 10'27 kg, is a product of certain radioactive decays. The results of the experiment led Rutherford to the idea that most of an atom's mass is in a very small nucleus, with electrons in orbit around it. Assume an alpha particle, initially very far from a stationary gold nucleus, is fired with a velocity of 2.71 x 107 m/s directly toward the nucleus (charge +79e). What is the smallest distance between the alpha particle and the nucleus before the alpha particle reverses direction? Assume the gold nucleus remains stationary. Part 5 of 5 - Analyze: (cont.) (4) Based on the correct choice in question (3), substitute numerical values to find the distance of closest approach of the alpha particle to the gold nucleus. 316ke2 d = m v2 316(8.988 x 109 N . m2/C2) (1.60 x 10-19 C) 2 x 10-27 kg x 10-14 m x 107 m/S =Chase and Kadri consider the following values, which are part of a set of experimental data for a Millikan oil-droplet experiment. b = 1.48 x 1010 kg/s d = 0.016 m p0\" pair = 0.882 g/cm3 r= 3.0x 107 m V: 5000v What is the terminal speed for a droplet of charge 5e? :l m To what value should the voltage be adjusted so that the droplet is suspended between the two plates? :vStep by Step Solution
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