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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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Force F = (-6.0 N)ihat + (2.0 N) jhatacts on a particle with position vector r =(4.0 m) ihat+ (4.0 m) jhat.
finlep [7]

Answer:

\tau=(-32k)\ N-m

\theta=116.55^{\circ}

Explanation:

Given that.

Force acting on the particle, F=(-6i + 2.0j)\ N

Position of the particle, r=(4i+4j)\ m

To find,

(a) Torque on the particle about the origin.

(b) The angle between the directions of r and F

Solution,

(a) Torque acting on the particle is a scalar quantity. It is given by the cross product of force and position. It is given by :

\tau=F\times r

\tau=(-6i + 2.0j)\times (4i+4j)

\tau=\begin{pmatrix}0&0&-32\end{pmatrix}

\tau=(-32k)\ N-m

So, the torque on the particle about the origin is (32 N-m).

(b) Magnitude of r, |r|=\sqrt{4^2+4^2}=5.65\ m

Magnitude of F, |F|=\sqrt{(-6)^2+2^2}=6.324\ m

Using dot product formula,

F{\circ}\ r=|F|.|r|\ cos\theta

cos\theta=\dfrac{F{\circ} r}{|F|.|r|}

cos\theta=\dfrac{-24+8}{6.324\times 5.65}

\theta=116.55^{\circ}

Therefore, this is the required solution.

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2 years ago
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A glass of root beer with a scoop of ice cream floating on top and a straw sticking out.
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Answer:

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Calculate the radius of the orbit of a proton moving at 2.2x10^6 m/s in a magnetic field 0.7 T where v and B are perpendicular.
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Answer:

3.28 cm

Explanation:

To solve this problem, you need to know that a magnetic field B perpendicular to the movement of a proton that moves at a velocity v will cause a Force F experimented by the particle that is orthogonal to both the velocity and the magnetic Field. When a particle experiments a Force orthogonal to its velocity, the path it will follow will be circular. The radius of said circle can be calculated using the expression:

r = \frac{mv}{qB}

Where m is the mass of the particle, v is its velocity, q is its charge and B is the magnitude of the magnetic field.

The mass and  charge of a proton are:

m = 1.67 * 10^-27 kg

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r =  \frac{1.67 * 10^-27 kg * 2.2*10^6 m/s}{1.6 * 10^-19 C* 0.7 T} = 0.0328 m, or 3.28  cm.

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