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Roman55 [17]
3 years ago
10

An electron of mass 9.11x10^-31 kg has an initial speed of 2.40x10^5 m/s. It travels in a straight line, and its speed increases

to 6.80x10^5 m/s in a distance of 4.80 cm. Assume its acceleration is constant.
(a) Determine the magnitude of the force exerted on the electron.
(b) Compare this force (F) with the weight of the electron (Fg), which we ignored. F/Fg
Physics
1 answer:
egoroff_w [7]3 years ago
7 0

Answer: A) Force = 3.841*10^-18 N.

B) force (f) is 4.30* 10^12 times greater than the weight (Fg).

Explanation: mass of electronic charge = 9.11*10^-31kg

v = final velocity = 6.80*10^5 m/s

u = initial velocity = 2.40 * 10^5 m/s

S= distance covered = 4.8cm = 0.048m

a = acceleration

Since the acceleration of the electron is assumed to be constant, newton's laws of motion are valid.

Thus, recall that

v² = u² + 2aS

(6.80*10^5)² = ( 2.40*10^5)² + 2*a( 0.048)

46.24 * 10^10 = 5.76 * 10^10 + 0.096a

46.24 *10^10 - 5.76* 10^10 = 0.096a

40.48* 10^10 = 0.096a

a = 40.48 * 10^10/0.096

a = 4.2167*10^12m/s².

Force = mass * acceleration

Force = 9.11*10^-31 * 4.2167*10^12

Force = 3.841*10^-18 N.

Weight =Fg= mg where g = acceleration due gravity = 9.8m/s²

Fg= 9.11*10^-31 * 9.8

Fg = 8.9278* 10^-30 N

By comparing the force and the weight, we have that

F/Fg = 3.841 * 10^-18/8.9278 * 10^-30 = 4.30* 10^12.

This implies that the force (f) is 4.30* 10^12 times greater than the weight (Fg).

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The relation between force, mass and acceleration according to the Newton's second law of motion is given as:

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Since, F is constant, the acceleration of the truck is inversely proportional to the mass.

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The tangential speed of the tack is 6.988 meters per second.

Explanation:

The tangential speed experimented by the tack (v), measured in meters per second, is equal to the product of the angular speed of the wheel (\omega), measured in radians per second, and the distance of the tack respect to the rotation axis (R), measured in meters, length that coincides with the radius of the tire. First, we convert the angular speed of the wheel from revolutions per second to radians per second:

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