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Svetlanka [38]
3 years ago
11

At one instant, an electron (charge = –1.6 x 10–19 C) is moving in the xy plane, the components of its velocity being vx = 5.0 x

105 m/s and vy = 3.0 x 105 m/s. A magnetic field of 0.80 T is in the positive y direction. At that instant, what is the magnitude of the magnetic force on the electron?
Physics
1 answer:
Tomtit [17]3 years ago
3 0

Answer:

The direction of force is along negative Z axis and the magnitude of force is

6.4 x 10^-14 N.

Explanation:

q = - 1.6 x 10^-19 c

vx = 5 x 10^5 m/s, vy = 3 x 10^5 m/s, B = 0.8 T along Y axis

The velocity vector is given by

v = 5 x 10^5 i + 3 x 10^5 j

B = 0.8 j

Force on a charged particle place in a magnetic field is given by

F = q (v x B)

F = -1.6 x 10^-19 {(5 x 10^5 i + 3 x 10^5 j) x (0.8 j)}

F = - 1.6 x 10^-19 (5 x 0.8 x 10^5 k)

F = - 6.4 x 10^-14 k

The direction of force is along negative Z axis and the magnitude of force is

6.4 x 10^-14 N.

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A mass weighing pounds is attached to a spring whose constant is lb/ft. The medium offers a damping force that is numerically eq
andreyandreev [35.5K]

Answer:

hello your question has some missing values attached below is the complete question with the missing values

answer :

a) 0.083 secs

b) 0.33 secs

c)  3e^-4/3

Explanation:

Given that

g = 32 ft/s^2 ,  spring constant ( k ) = 2 Ib/ft

initial displacement = 1 ft above equilibrium

mass = weight / g = 4/32 = 1/8

damping force = instanteous velocity  hence  β = 1

a<u>)Calculate the time at which the mass passes through the equilibrium position.</u>

time mass passes through equilibrium = 1/12 seconds = 0.083

<u>b) Calculate the time at which the mass attains its extreme displacement </u>

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<u>c) What is the position of the mass at this instant</u>

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attached below is the detailed solution to the given problem

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