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netineya [11]
2 years ago
10

If an electron moves in a circle of radius 21 cm perpendicular to a B field of 0.4 T, what are the speed of the electron and the

frequency of motion?
Physics
1 answer:
kodGreya [7K]2 years ago
8 0

Answer:

a)

v = 4.048 *10^6 m/s

b)  

Angular frequency =  1.92 * 10^7

Explanation:

As we know

v =  \frac{qBr}{m}

q is the charge on the electron = 3.2 * 10^{-19} C

B is the magnetic field in Tesla = 0.4 T

r is the radius of the circle = 0.21 m

mass of the electrons = 6.64 * 10^{-27} Kg

a)

Substituting the given values in above equation, we get -

v = \frac{3.2 * 10^{-19}*0.4*0.21}{6.64 * 10^{-27}} \\v = 4.048 *10^6m/s

b)  

Angular frequency =

\frac{4.048 * 10^6 }{0.21} \\1.92 * 10^7

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The design speed of a multilane highway is 60 mi/hr. What is the minimum stopping sight distance that should be provided on the
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Part a: When the road is level, the minimum stopping sight distance is 563.36 ft.

Part b: When the road has a maximum grade of 4%, the minimum stopping sight distance is 528.19 ft.

Explanation:

Part a

When Road is Level

The stopping sight distance is given as

SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}

Here

  • SSD is the stopping sight distance which is to be calculated.
  • u is the speed which is given as 60 mi/hr
  • t is the perception-reaction time given as 2.5 sec.
  • a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
  • G is the grade of the road, which is this case is 0 as the road is level

Substituting values

                              SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0)}\\SSD=220.5 +342.86 ft\\SSD=563.36 ft

So the minimum stopping sight distance is 563.36 ft.

Part b

When Road has a maximum grade of 4%

The stopping sight distance is given as

SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}

Here

  • SSD is the stopping sight distance which is to be calculated.
  • u is the speed which is given as 60 mi/hr
  • t is the perception-reaction time given as 2.5 sec.
  • a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
  • G is the grade of the road, which is given as 4% now this can be either downgrade or upgrade

For upgrade of 4%, Substituting values

                              SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35+0.04)}\\SSD=220.5 +307.69 ft\\SSD=528.19 ft

<em>So the minimum stopping sight distance for a road with 4% upgrade is 528.19 ft.</em>

For downgrade of 4%, Substituting values

                              SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0.04)}\\SSD=220.5 +387.09 ft\\SSD=607.59ft

<em>So the minimum stopping sight distance for a road with 4% downgrade is 607.59 ft.</em>

As the minimum distance is required for the 4% grade road, so the solution is 528.19 ft.

3 0
3 years ago
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