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Finger [1]
3 years ago
6

Find the linear speed of the bottom of a test tube in a centrifuge if the centripetal acceleration there is 5.4×104 times the ac

celeration of gravity. The distance from the axis of rotation to the bottom of the test tube is 7.2 cm .
Engineering
1 answer:
Luda [366]3 years ago
4 0

Answer: v= 195.2 m/s

Explanation:

The centripetal acceleration acts on the bottom of the test tube trying to take it closer to the axis of rotation, and it is defined by the following expression:

ac = v² / r

We know that ac = 5.4. 10⁴ . g = 5.4.10⁴.9.8 m/s²

Expressing r in m ⇒ r = 0.072 m.

So, the only unknown that remains is v, which is the linear speed of the bottom of the test tube, which we want to find out:

v = √ac. r = √5.4.10⁴.9.8 m/s². 0.072 m = 195.2 m/s

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Compute the theoretical density of ZnS given that the Zn-S distance and bond angle are 0.234 nm and 109.5o, respectively. The at
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Answer: the theoretical density is 4.1109 g/cm³

Explanation:  

first the image of one set of ZnS bonding in the crystal structure, we calculate the value of angle θ

θ + ∅ + 90° = 180°

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θ = 90° - ( 109.5° / 2 )

θ = 35.25°

next we calculate the value of x from the geometry

given that;  distance angle d = 0.234

x = dsinθ

= 0.234 × sin35.25°)

= 0.135 nm = 0.135 × 10⁻⁷ cm

next we calculate the length of the unit cell

a = 4x

a = 4(0.135)

a = 0.54 nm = 0.54 × 10⁻⁷ cm

next we calculate number of formula units

n' = (no of corner atoms in unit ell × contribution of each corner atom in unit cell) + ( no of face center atom in a unit cell × contribution of each face center atom in a unit cell)

n' = 8 × 1/8) + ( 6 × 1/2)

= 1 + 3

= 4

next we calculate the theoretical density using  this equation

P = [n'∑(Ac + AA)] / [Vc.NA]

= [n'∑(Ac + AA)] / [(a)³NA]

where the ∑Ac is sum of atomic weights of all cations in the formula unit( 65.41 g/mol)

∑AA is the sum of weights of all anions in the formula unit( 32.06 g/mol)

Na is the Avogadro’s number( 6.023 × 10²³ units/mole)

so we substitute

P = [4( 65.41 + 32.06)] / [ ( 0.54 × 10⁻⁷ )³ × (6.023 × 10²³)]

= 389.88 / 94.84

= 4.1109 g/cm³

therefore the theoretical density is 4.1109 g/cm³

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2 years ago
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