An electron with an initial velocity v0 = 1.50×105 m/s enters a region 1.0 cm long where it is electrically accelerated (see the figure). It emerges with velocity v = 4.50×106 m/s. What was its acceleration, assumed constant?
This problem is a very easy one. You need just one equation:
(final velocity)^2 = (initial velocity)^2 + 2 * acceleration *
(distance during which acceleration occurred)
also known as Vf^2 = Vi^2 +2ad
Further,
(4.5E6 m/s)^2 = (1.5E5 m/s)^2 + 2*a*(0.01m)
2.03E13 = 2.25E10 + 0.02a
a = 1.011E15 m/s^2
An electron with an initial velocity v0 = 1.50×105 m/s enters a region 1.0 cm long...
An electron with an initial velocity v0 = 2.30×105 m/s enters a region 1.0 cm long where it is electrically accelerated (see the figure). It emerges with velocity v = 5.10×106 m/s. What was its acceleration, assumed constant? (Such a process occurs in the electron gun in a cathode-ray tube, used in television receivers and oscilloscopes.)
An electron with an initial velocity v0 = 2.70×105 m/s enters a region 1.0 cm long where it is electrically accelerated (see the figure). It emerges with v = 6.10×106 m/s. What is its acceleration, assumed constant? (Such a process occurs in the electron gun in a cathode-ray tube, used in television receivers and oscilloscopes.)
An electron with an initial velocity v0 = 2.70×105 m/s enters a region 1.0 cm long where it is electrically accelerated (see the figure). It emerges with v = 6.10×106m/s. What is its acceleration, assumed constant?
An electron with an initial velocity vo = 2.70x105 m/s enters a region 1.0 cm long where it is electrically accelerated (see the figure). It emerges with v = 5.10x106 m/s. What is its acceleration, assumed constant? (Such a process occurs in the electron gun in a cathode-ray tube, used in television receivers and oscilloscopes.) Accelerating region Nonaccelerating region 41.0 cm Path of electron Source of high voltage Submit Answer Tries 0/5
Part A An electron is to be accelerated from a velocity of 1.50×106 m/s to a velocity of 7.50×106 m/s . Through what potential difference must the electron pass to accomplish this? Part B Through what potential difference must the electron pass if it is to be slowed from 7.50×106 m/s to a halt?
An electron travels with velocity < 5 × 105, 0, 0 > m/s. It enters a region in which there is a uniform magnetic field of < 0, 0.6, 0 > T. What is the magnetic force on the electron? F(b)= < , , , > N Despite the magnetic force, the electron continues to travel in a straight line at constant speed. You conclude that there must be another force acting on the electron. Since you know there is...
An elctron enters a region of a uniform magnetic field with velocity v = (5.6 Times 106 m/s) I^ +(-2.8 Times 106 m/s)j^ . If the field, B = (7.8T)^ Calculate the acceleration of the electron Calculate the radius of the electron's trajectory. Calculate the helicalthread of the electron's trajectory.
An electron is fired at a speed v0 = 5.3 ✕ 106 m/s and at an angle θ0 = −45° between two parallel conducting plates that are D = 2.5 mm apart, as in the figure below. The voltage difference between the plates is ΔV = 105 V. (a) Determine how close, d, the electron will get to the bottom plate. mm (b) Determine where the electron will strike the top plate. mm Path of the electron 0
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An electron enters a region of uniform electric field with an initial velocity of 43 km/s in the same direction as the electric field, which has magnitude E = 45 N/C. (a) What is the speed of the electron 1.5 ns after entering this region? _________________m/s (b) How far does the electron travel during the 1.5 ns interval? ___________________m