Physics, 05.11.2019 05:31 jonallere04

Amass spectrometer is being used to separate common oxygen-16 from the much rarer oxygen-18, taken from a sample of old glacial ice. (the relative abundance of these oxygen isotopes is related to climatic temperature at the time the ice was deposited.) the ratio of the masses of these two ions is 16 to 18, the mass of oxygen-16 is 2.66 x 10^-26 kg, and they are both singly charged and travel at 5.15 x 10^6 m/s in a 1.2 t magnetic field. what is the separation between their paths when they hit a target after traversing a semicircle? show all steps

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Answer from: billy583887

The path separation is 0.089 m

Solution:

As per the question:

Mass of $O_{16}$ , m = 2.66\times 10^{- 26}\ kg[/tex]

Ratio of the masses of $O_{16}$ to $O_{18}$ , R = 16:18

Velocity of the masses, v = $5.15\times 10^{6}\ m/s$

Magnetic field, B = 1.2 T

Now,

The separation between the path after completing a semicircle can be calculated as:

Mass of $O_{18}$ , m' = mR = $2.66\times 10^{- 26}\times \frac{18}{16} = 2.99\times 10^{- 26}\ kg$

Here, magnetic force provides the necessary centripetal force to traverse the semi-circle:

$F_{B} = F_{c}$

$qvB = \frac{mv^{2}}{r}$

where

$F_{c} = \frac{mv^{2}}{r}$

$F_{B} = qvB$ = Magnetic force

q = e = electronic charge = $1.6\times 10^{- 19}\ C$

v = Velocity

B = Magnetic field

r = Radius

Now, the radius from the above eqn comes out to be:

$r = \frac{mv}{qB}$

Now, for $O_{16}[tex]:[tex]r_{16} = \frac{mv}{eB}$

$r_{16} = \frac{2.66\times 10^{- 26}\times 5.15\times 10^{6}}{1.6\times 10^{- 19}\times 1.2} = 0.713\ m$

Now, for $O_{18}[tex]:[tex]r_{16} = \frac{m'v}{eB}$

$r_{16} = \frac{2.99\times 10^{- 26}\times 5.15\times 10^{6}}{1.6\times 10^{- 19}\times 1.2} = 0.802\ m$

Now, the path separation is given by:

$d = r_{18} - r_{16} = 0.802 - 0.713 = 0.089\ m$

Answer from: Quest

most voltage drop calculators want the distance between the source and destination. then internally, they double it so both the source and return wires are calculated. but you need to know which distance the calculator uses so you're not half of what you should be or double what you should be.

most circuit breakers will not take a #6 directly, so the electrician will have to splice on a smaller pigtail. you also have to upsize the grounding wire when you upsized the ungrounded (hot) wire (see note in 250.122). this will preclude using most cable assemblies, as a cable with #6 copper will most likely only have a #10 grounding conductor. you need a #6 grounding conductor if the circuit is 20a. a #8 cable is no better, as it will have a #10 grounding conductor too. but most 20a breakers will take a #8 directly.

in your related post, you didn't say what type of load this is. if it is a light or heater, voltage drop won't be a big issue. if this is a motor, the current needs are huge at startup and the voltage drop can pose a problem. if this is a motor, knowing if the motor is under much load during startup will answer the voltage drop effect (a saw is easier to start than a compressor).

Answer from: Quest

learn it on the intertant so you can solve

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Amass spectrometer is being used to separate common oxygen-16 from the much rarer oxygen-18, taken f...

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