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anzhelika [568]
4 years ago
14

Physicist S. A. Goudsmit devised a method for measuring the mass of heavy ions by timing their period of revolution in a known m

agnetic field. A singly charged ion makes 5.00 revolutions in a 50.0 mT field in 1.11 ms. Calculate its mass, in atomic mass units.
Physics
1 answer:
Umnica [9.8K]4 years ago
4 0

Answer:

170.36 amu

Explanation:

Ion makes 5 revolutions in 1.11 ms

Frequency = 5 / (1.11 x 10^-3) = 4504.5 rps

B = 50 mT = 0.05 T

q = 1.6 x 10^-19 C

Let m be the mass in kg

Time period is given by

T = (2 π m) / (B q)

Frequency is the reciprocal of time period.

f = 1 / T = B q / 2πm

So,

m = B q / 2 π f

m = (0.05 x 1.6 x 10^-19) / ( 2 x 3.14 x 4504.5) = 2.828 x 10^-25 kg

as we know that

1 amu = 1.66 x 10^-27 kg

So, m = (2.828 x 10^-25 ) / (1.66 x 10^-27) = 170.36 amu

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Impulse is just the product of mass and speed!

I = 1700 * 22.5  = 38,250 kg m/s

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3 years ago
Which type of stress is most likely the cause of irritability?
masha68 [24]

Answer:

The correct answer is - distress.

Explanation:

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Distress is the bad stress caused by anxiety, sadness, grief, and such type of stress that is caused irritability. Irritability is caused not just by stress but many factors.

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3 years ago
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7 0
4 years ago
Indique onde em quantos metros o rapaz chegará com as seguintes condições: S0 = 0 a) v = 3 m/s e t = 2 s b) v = 2 m/s e t = 3,5
Naily [24]

Answer:

I will answer in English.

Here we will use the relation

Velocity*time = distance

So:

a) velocity = 3m/s

time = 2s

Distance = 3m/s*2s = 6m

b) velocity = 2m/s

time = 3.5s

Distance = 2m/s*3.5s = 7m

c) velocity = 10m/s

time = 0.5s

Distance = 10m/s*0.5s = 5m

d) velocity = 4m/s

time = 2.5s

Distance = 4m/s*2.5s = 9m

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time = 5s

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8 0
3 years ago
A 0.10 kg meter stick is held perpendicular to a vertical wall by a 2.5 mm string going from the wall to the far end of the stic
il63 [147K]

Explanation:

a)

Sum of moments = 0 (Equilibrium)

T . cos (Q)*L = m*g*L/2

cos Q = \frac{\sqrt{(2.5^2 - L^2) } }{2.5}

T*\frac{\sqrt{(2.5^2 - L^2) } }{2.5} * L = 0.1 *9.81*L/2

T = \frac{2.4525}{\sqrt{(2.5)^2 - L^2} }

b) If the String is shorter the Q increases; hence, Cos Q decreases which in turn increases Tension in the string due to inverse relationship!

c)

T = \frac{1.962}{\sqrt{(2)^2 - L^2} }

6 0
3 years ago
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