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Sever21 [200]
3 years ago
6

Name two factors that can affect the function of an enzyme

Physics
1 answer:
myrzilka [38]3 years ago
3 0

Answer:

1. pH

2. Temperature

Hope this helps

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1. A box contains 10 blue chips. 5 red chips, and 15 yellow chips. Find the odds of choosing the
skelet666 [1.2K]

Answer:

Explanation:

Blue: 10/30

Red: 5/30

Yellow: 15/30

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A professional cyclist rides a bicycle that is 92 percent efficient. For every 100 joules of energy he exerts as input work on t
emmainna [20.7K]
Efficiency =  Work Output / Work Input

92%  =  Work Output / 100

0.92 =   Work Output / 100

Work Output = 0.92 * 100

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8 0
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A 230 kg steel crate is being pushed along a cement floor. The force of friction is 480 N to the left and the applied force is 1
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Read 2 more answers
A heat engine accepts 200,000 Btu of heat from a source at 1500 R and rejects 100,000 Btu of heat to a sink at 600 R. Calculate
diamong [38]

To solve the problem it is necessary to apply the concepts related to the conservation of energy through the heat transferred and the work done, as well as through the calculation of entropy due to heat and temperatra.

By definition we know that the change in entropy is given by

\Delta S = \frac{Q}{T}

Where,

Q = Heat transfer

T = Temperature

On the other hand we know that by conserving energy the work done in a system is equal to the change in heat transferred, that is

W = Q_{source}-Q_{sink}

According to the data given we have to,

Q_{source} = 200000Btu

T_{source} = 1500R

Q_{sink} = 100000Btu

T_{sink} = 600R

PART A) The total change in entropy, would be given by the changes that exist in the source and sink, that is

\Delta S_{sink} = \frac{Q_{sink}}{T_{sink}}

\Delta S_{sink} = \frac{100000}{600}

\Delta S_{sink} = 166.67Btu/R

On the other hand,

\Delta S_{source} = \frac{Q_{source}}{T_{source}}

\Delta S_{source} = \frac{-200000}{1500}

\Delta S_{source} = -133.33Btu/R

The total change of entropy would be,

S = \Delta S_{source}+\Delta S_{sink}

S = -133.33+166.67

S = 33.34Btu/R

Since S\neq   0 the heat engine is not reversible.

PART B)

Work done by heat engine is given by

W=Q_{source}-Q_{sink}

W = 200000-100000

W = 100000 Btu

Therefore the work in the system is 100000Btu

4 0
3 years ago
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