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madreJ [45]
3 years ago
5

Exercise 4 (Module 6.9) Glycolysis is the first of three steps in cellular respiration. Review glycolysis by matching each phras

e on the right with a term on the left. Some terms are used twice.
_____ 1. Compound formed between glucose and pyruvic acid
_____2. Not involved in glycolysis _____3. Fuel molecule broken down in glycolysis
_____ 4. Produced by substrate-level phosphorylation
_____5. Invested to energize glucose molecule at start of process
_____ 6. Reduced as glucose is'oxddized _____7. Glucose converted to two molecules of this
_____8. Assembled to make ATP
_____9. "Splitting of sugar"
_____10. Carries hydrogen and electrons from oxidation of glucose
A NADH
B. Pyruvic acid
C. ATP
D. NAD+
E Glucose
F. Glycolysis
G. ADP and P
H. Oxygen
I. Intermediate
​
Biology
1 answer:
astraxan [27]3 years ago
6 0

Answer:

Explanation:

?????????? i don't undrssanted

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A record turntable is rotating at 33 rev/min. A watermelon seed is on the turntable 7.8 cm from the axis of rotation. (a) Calcul
Novosadov [1.4K]

Given data:

● Angular velocity of the turntable ω = 33 rev/min

Therefore,

ω = 33 rev/min

ω = 33rev/min × 2π rad/rev × 1 min/60 sec = 3.45 rad/s

● The distance of the watermelon seed from the axis of rotation r = 7.8 cm = 0.078 m

● μs = the coefficient of static friction

Section a:

The seed is undergoes circular motion and it is been effected by centripetal acceleration.

ac = rω^2

ac = 0.078 × 3.45^2

ac = 0.9284 m/s^2

Therefore,

the centripetal acceleration of the seed is 9.274 m/s^2

Section b:

If the seed is observed not to slip at the course of the circular motion, then the supplied frictional force given by the seed and surface of turntable would at least be equivalent to the centripetal force working on the seed.

Centripetal Force = Frictional Force

mrω^2 = μsmg

μs = rω^2 /g

μs = 0.078 × 3.45^2

------------

9.81

μs = 0.09464

Thus,

the coefficient of static friction is 0.09464.

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