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MrRissso [65]
3 years ago
9

A mass spectrometer is being used to monitor air pollutants. it is difficult, however, to separate molecules with nearly equal m

ass such as Co (28.0106 u) and N2 (28.0134 u).
How large a radius of curvature must a spectrometer have if these two molecules are to be separated on the film by 0.25 mm ?
Physics
1 answer:
Fynjy0 [20]3 years ago
4 0

Answer:

The answer is 1250 mm

Explanation:

The path that particles move in the spectrometer is semicircular. Each of the particles has a displacement of twice the radius (2r) from the entrance to where the film is hit. According to the exercise, if the separation between the two molecules is 0.25 mm, then the difference in the radius of the molecules is equal to 0.125 mm. The ratio of mass/radius is equal for molecules, and therefore is equal to:

m = q * B * r / v

m/r = constant

(m/r)CO = (m/r)N = (28.0106 u/r) = (28.0134 u/(r + 0.000125 m))

Clearing and solving r:

r = 1.25 m = 1250 mm

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Answer:

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R2= resistance of 8ohms

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7 0
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irga5000 [103]

Answer:

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Explanation:

The given mass of the ball, m = 86 g = 0.089 kg

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Mathematically, we have;

Δp = F·Δt

Where;

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F = The applied force

Δt = The time of contact with the force

The velocity of the ball just before it touches the ground, v₁ = -√(2·g·H)

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The change in momentum, Δp = m·(v₂ - v₁)

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The impulse, Δp, received by the ball from the floor during the bounce is given as follows;

Δp = 0.089 kg × (√(2 × 9.8 m/s² × 1.28 m) + √(2 × 9.8 m/s² × 2.87 m)) ≈ 1.11329438 m·kg/s

The impulse received by the ball from the floor during the bounce, Δp ≈ 1.11329438 m·kg/s

6 0
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We are asked to calculate force required.

\longrightarrow F = ma

\longrightarrow F = (50 × 3) N

\longrightarrow <u>F</u><u> </u><u>=</u><u> </u><u>1</u><u>5</u><u>0</u><u> </u><u>N</u>

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