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An electron traveling at 6 imes 10^(6)(m)/(s) enters a 0.08m region with a uniform electric field Find the magnitude of the acceleration of the electron while in the electric field. The mass of an electron is 9.109 imes 10^(-31)kg and the fundamental charge is 1.602 imes 10^(-19)C. Answer in units of (m)/(s^(2)). Answer in units of (m)/(s^(2)). part 2 of 3 Find the time it takes the electron to travel through the region of the electric field, assuming it doesn't hit the side walls. Answer in units of s. Answer in units of s. 2. 2.21596 imes 10^(13) 3. 1.12557 imes 10^(13) 4. 1.19592 imes 10^(13) 5. 8.96937 imes 10^(12) 6. 3.2536 imes 10^(13) 7. 3.76362 imes 10^(13) 8. 1.26626 imes 10^(13) 9. 3.90431 imes 10^(13) 10. 9.14524 imes 10^(12) 1 imes 10^(-8) 2.66667 imes 10^(-8) 1.2 imes 10^(-8) 2.5 imes 10^(-8) 3 imes 10^(-8) An electron traveling at 6 10m/s enters a 0.08 m region with a uniform electric field of222N/Cas in the figure 0.08m O 2.2.21596 1013 O 3.1.12557 1013 O 4.1.19592 1013 O 5.8.96937 1012 610m/s O 6. 3.2536 1013 +++++++++ Find the magnitude of the acceleration of the electron while in the electric field.The mass of an electron is 9.109 x 10-31 kg and the fundamental charge is 1.602 10-19 C. Answer in units of m/s2. Answer in units of m/s2. O 7.3.76362 1013 O 8.1.26626 1013 O 9.3.90431 1013 O10.9.14524 1012 O1.110-8 part 2of 3 Find the time it takes the electron to travel through the region of the electric field, assum- ing it doesn't hit the side walls. Answer in units of s. Answer in units of s O 2.2.66667 10-8 O3.1.210-8 O 4.2.5 10-8 Q 5. 3 x 10-8

          An electron traveling at 6	imes 10^(6)(m)/(s) enters a 0.08m region with a uniform electric field
Find the magnitude of the acceleration of the electron while in the electric field. The mass of an electron is 9.109	imes 10^(-31)kg and the fundamental charge is 1.602	imes 10^(-19)C.
Answer in units of (m)/(s^(2)). Answer in units of (m)/(s^(2)).
part 2 of 3
Find the time it takes the electron to travel through the region of the electric field, assuming it doesn't hit the side walls.
Answer in units of s. Answer in units of s.
2. 2.21596	imes 10^(13)
3. 1.12557	imes 10^(13)
4. 1.19592	imes 10^(13)
5. 8.96937	imes 10^(12)
6. 3.2536	imes 10^(13)
7. 3.76362	imes 10^(13)
8. 1.26626	imes 10^(13)
9. 3.90431	imes 10^(13)
10. 9.14524	imes 10^(12)
1	imes 10^(-8)
2.66667	imes 10^(-8)
1.2	imes 10^(-8)
2.5	imes 10^(-8)
3	imes 10^(-8)
An electron traveling at 6  10m/s enters a 0.08 m region with a uniform electric field of222N/Cas in the figure 0.08m
O 2.2.21596  1013
O 3.1.12557 1013
O 4.1.19592 1013
O 5.8.96937  1012
610m/s
O 6. 3.2536  1013
+++++++++
Find the magnitude of the acceleration of the electron while in the electric field.The mass of an electron is 9.109 x 10-31 kg and the fundamental charge is 1.602  10-19 C. Answer in units of  m/s2. Answer in units of m/s2.
O 7.3.76362  1013
O 8.1.26626  1013
O 9.3.90431  1013
O10.9.14524 1012
O1.110-8
part 2of 3
Find the time it takes the electron to travel through the region of the electric field, assum- ing it doesn't hit the side walls. Answer in units of s. Answer in units of s
O 2.2.66667  10-8
O3.1.210-8
O 4.2.5  10-8
Q 5. 3 x 10-8
        
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an electron traveling at 6times 106ms enters a 008m region with a uniform electric field find the magnitude of the acceleration of the electron while in the electric field the mass of an ele 94515

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An electron traveling at 6 imes 10^(6)(m)/(s) enters a 0.08m region with a uniform electric field Find the magnitude of the acceleration of the electron while in the electric field. The mass of an electron is 9.109 imes 10^(-31)kg and the fundamental charge is 1.602 imes 10^(-19)C. Answer in units of (m)/(s^(2)). Answer in units of (m)/(s^(2)). part 2 of 3 Find the time it takes the electron to travel through the region of the electric field, assuming it doesn't hit the side walls. Answer in units of s. Answer in units of s. 2. 2.21596 imes 10^(13) 3. 1.12557 imes 10^(13) 4. 1.19592 imes 10^(13) 5. 8.96937 imes 10^(12) 6. 3.2536 imes 10^(13) 7. 3.76362 imes 10^(13) 8. 1.26626 imes 10^(13) 9. 3.90431 imes 10^(13) 10. 9.14524 imes 10^(12) 1 imes 10^(-8) 2.66667 imes 10^(-8) 1.2 imes 10^(-8) 2.5 imes 10^(-8) 3 imes 10^(-8) An electron traveling at 6 10m/s enters a 0.08 m region with a uniform electric field of222N/Cas in the figure 0.08m O 2.2.21596 1013 O 3.1.12557 1013 O 4.1.19592 1013 O 5.8.96937 1012 610m/s O 6. 3.2536 1013 +++++++++ Find the magnitude of the acceleration of the electron while in the electric field.The mass of an electron is 9.109 x 10-31 kg and the fundamental charge is 1.602 10-19 C. Answer in units of m/s2. Answer in units of m/s2. O 7.3.76362 1013 O 8.1.26626 1013 O 9.3.90431 1013 O10.9.14524 1012 O1.110-8 part 2of 3 Find the time it takes the electron to travel through the region of the electric field, assum- ing it doesn't hit the side walls. Answer in units of s. Answer in units of s O 2.2.66667 10-8 O3.1.210-8 O 4.2.5 10-8 Q 5. 3 x 10-8
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An electron of mass 9.11 10-31 kg has an initial speed of 2.10 105 m/s. It travels in a straight line, and its speed increases to 7.30 105 m/s in a distance of 5.90 cm. Assume its acceleration is constant. (a) Determine the magnitude of the force exerted on the electron. (b) Compare this force with the weight of the electron, which we ignored. Part 1 of 4 - Conceptualize Visualize the electron as part of a beam in a vacuum tube, speeding up in response to an electric force. Only a very small force is required to accelerate an electron because of its small mass, and if this force is much greater than the weight of the electron then the gravitational force can be neglected. Part 2 of 4 - Categorize Since this is a linear acceleration problem, we can use Newton's second law to find the force as long as the electron does not approach relativistic speeds, that is, much less than 3 Ɨ 108 m/s. We know the initial and final velocities, and the distance involved. From these given quantities, we can find the acceleration in order to determine the force. Part 3 of 4 - Analyze (a) From F = ma and vf2 = vi2 + 2axf, we can solve for the acceleration and then find the force using the following. a = (vf2 āˆ’ vi2) / 2xf Substituting to eliminate a, we have, F = m (vf2 āˆ’ vi2) / 2xf. Substituting the given values into the equation above, gives us the following. F = (9.11 Ɨ 10āˆ’31 kg) ((7.30 Ɨ 105 m/s)^2 āˆ’ (2.10 Ɨ 105 m/s)^2) / 2(0.0590 m) = Ɨ 10āˆ’18 N

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Transcript

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00:01 Дарам, тут mēs tevam nevajagt y -dizplizmatu.
00:03 LaiŔu to.
00:04 Y -dizplizmatu kā akcelerācija times square time divadu pa 2, un time laiŔu divadu pa time of the flight, pa speed v -initial.
00:22 Nē, laiÅ”u nevajagt.
00:40 Nē, laiÅ”u nevajagt akcelerāciju, tāpēc elektrÄ«ka forca over mass, un Å”op sometimes it is and now you can calculate y...
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