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4vir4ik [10]
1 year ago
13

The lons entering the mass spectrometer have the same charges. After being accelerated through a potential difference of 8.20 kV

, a
singly charged 12c+ ion moves in a circle of radius 19.4 cm in the magnetic field of a mass spectrometer. What is the magnitude of the
field? Use these atomic mass values: 12C, 12.0 u; 14C, 14.0 u; 160, 15.99 u. The conversion between atomic mass units and kilograms is
1u=1.66 x 10-27 kg.
Physics
1 answer:
Ratling [72]1 year ago
4 0

The calculated magnitude is  6.73 x 10³ V/m.

AMU is described as being one-twelfth the mass of a carbon-12 atom (12C). C makes up more than 98% of the carbon that can be found in nature, making it the most prevalent isotope. The magnitude of the field is the change in potential across a small distance in the indicated direction divided by that distance.

Potential difference = 8.20 kV= 8.20 x 10³ V

radius= 19.4/100=0.194 m

total distance that is circumference of the circle= 2πr =2 x 3.14 x 0.194

                                                                               = 1.218 m

therefore Magnitude= 8.20 x 10³ / 1.218

                                  =6.73 x 10³ V/m

Learn more about Magnitude here-

brainly.com/question/15681399

#SPJ9

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If the maximum height of the object is given in feet, we use the value for g in  ft/s^2  which is : 32\,\,ft/s^2

Now, when the ball reaches its maximum height, the ball's velocity is zero, so that allows us to solve for the time (t) the process of reaching the max height takes:

v=v_0-g \,t\\0=v_0-g \,t\\g\,\,t=v_0\\t=\frac{v_0}{g}

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y-y_0=v_0\,t - \frac{g}{2} \,t^2\\max\,height=v_0\,(\frac{v_0}{g})  - \frac{g}{2} \,(\frac{v_0}{g})^2\\max\,height= \frac{v_0^2}{2\,g}

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1.1= \frac{v_0^2}{2\,32}\\(32)\,(1.1)=v_0^2\\v_0=35.2\\v_0=\sqrt{35.2} \\v_0=5.93\,\,ft/s

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