1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Deffense [45]
3 years ago
14

A biologist wants to increase the rate of his chemical reaction but has a limited amount of enzyme. He continues to increases th

e substrate concentration instead. Eventually, the reaction rate levels off, and he can't get it to go any faster. What prevented the rate from increasing further?
The solution ran out of enzyme
The substrate concentration reached Vmax
The products of the reaction inhibited the enzyme
The solution ran out of reactants
Physics
1 answer:
boyakko [2]3 years ago
3 0

Answer:

The substrate concentration reached Vmax

Explanation:

In this scenerio, further increase in the rate of reaction was prevented because the substrate concentration hit Vmax.

Beyond the Vmax, the substrate can no longer proceed.

  • The substrate is the reactant on which the enzyme works on.
  • Enzymes are natural organic chemical  catalysts.
  • Increasing the concentration of reactants is one of the known and proven ways to speed up the rate of reaction.
  • Since we have limited amount of enzymes, once they all bound to the substrate, no further increase in the substrate will have an effect on the rate.
  • At the optimum reaction point, all the available substrate will bond with the enzyme.
  • Beyond this, the enzymes are saturated will not further any reaction.
You might be interested in
The spring of a spring gun has force constant k = 400 N/m and negligible mass. The spring is compressed 6.00 cm and a ball with
nikdorinn [45]

Answer:

A) v = 6.93 m/s

B) v = 4.9 m/s

C) x_m = 0.015m

D) v_max = 5.2 m/s

Explanation:

We are given;

x = 6 cm = 0.06 m

k = 400 N

m = 0.03 kg

F = 6N

A) from work energy law, work dome by the spring on ball which now became a kinetic energy is;

Ws = K.E = ½kx²

Similarly, kinetic energy of ball is;

K.E = ½mv²

So, equating both equations, we have;

½kx² = ½mv²

Making v the subject gives;

v = √(kx²/m)

Plugging in the relevant values to give;

v = √((400 × 0.06²)/0.03)

v = √48

v = 6.93 m/s

B) If there is friction, the total work is;

Ws = ½kx² - - - (1)

Work of the ball is;

Wb = KE + Wf

So, Wb = ½mv² + fx - - - (2)

Combining both equations, we have;

½mv² + fx = ½kx²

Plugging in the relevant values, we have;

(½ × 0.03 × v²) + (6 × 0.06) = ½ × 400 × 0.06²

0.015v² + 0.36 = 0.72

0.015v² = 0.72 - 0.36

v² = 0.36/0.015

v = √24

v = 4.9 m/s

C) The speed is greatest where the acceleration stops i.e. where the net force on the ball is zero. (ie spring force matches 6.0N friction)

So, from F = Kx;

(x is measured into barrel from end where F = 0)

Thus; 6.0 = 400x

x_m = 6/400

x_m = 0.015m from the end after traveling 0.045m

D) Initial force on ball = (Kx - F) =

[(400 x 0.06) - 6.0] = 18N

Final force on ball = 0N

Mean Net force on ball = ½(18 + 0)

Mean met force, F_m = 9N

Net Work Done on ball = KE = 9N x 0.045m = 0.405 J

Thus;

½m(v_max)² = 0.405J

(v_max)² = 2 x 0.405/0.03

(v_max)² = 27

v(max) = √27

v_max = 5.2 m/s

6 0
2 years ago
We have discovered some exoplanets that are still forming from a nebula. How might those planets change over time?
Rudik [331]
Based on several theories made by scientists, planets are formed because of the accumulation of gases and other particles that are attracted to each other. These accumulated gases form into clumps and eventually the clumps get bigger and turn into a big orbital mass. The exoplanets may experience change over time through the observance of its orbit in a particular axis, and if there are other debris that might affect the planet's continuous growth.
6 0
3 years ago
Suppose that the Mars orbiter was to have established orbit at 155 km and that one group of engineers specified this distance as
MAXImum [283]

Answer:

108 km

Explanation:

The conversion factor between meters and feet is

1 m = 3.28 ft

So the second altitude, written in feet, can be rewritten in meters as

h_2 = 1.55 \cdot 10^5 ft \cdot \frac{1}{3.28 ft/m}=4.7\cdot 10^4 m

or in kilometers,

h_2 = 47 km

the first altitude in kilometers is

h_1 = 155 km

so the difference between the two altitudes is

\Delta h = 155 km - 47 km = 108 km

8 0
3 years ago
A guitarist finds that the pitch of one of her strings is slightly flat—the frequency is a bit too low. Should she increase or d
Yuri [45]

Answer:

The guitarist should increase the tension of the string.

Explanation:

The frequency of a vibrating string is determined by fn=(n/(2L))√T/μ. So if the tension in the string increased, the rate at which it vibrates (the frequency) will also increase.

Therefore it is advisable that she increase the tension of the string.

I hope it helps, please give brainliest if it does

6 0
3 years ago
Fill in the blanks. When the northern hemisphere experiences _________, the southern hemisphere experiences __________.
schepotkina [342]

I choose the letter B

8 0
3 years ago
Other questions:
  • (30 Points) The peak intensity of radiation from a star named Sigma is 2 x 10^6 nm. What is the average surface temperature of S
    14·1 answer
  • A policeman starts giving chase 60 seconds after a stolen car zooms by at 108 km/hr. At what minimum speed should he drive if he
    12·1 answer
  • Why there are not green stars?
    13·1 answer
  • Friction is always undesirable. True False
    5·2 answers
  • In an elastic head-on collision, a 0.60 kg cart moving at 5.0 m/s [W] collides with a 0.80 kg cart moving at 2.0 m/s [E]. The co
    7·1 answer
  • A solid uniform cylinder is rolling without slipping. What fraction of its kinetic energy is rotational?
    10·1 answer
  • An Object with a mass of 40kg is moving at a velocity of 10 m/s. Determine its kinetic energy
    8·1 answer
  • Use technology and the given confidence level and sample data to find the confidence interval for the population mean muμ. Assum
    5·1 answer
  • A 200 kg weather rocket is loaded with 100 kg of fuel and fired straight up. It accelerates upward at 35 m/s2 for 35 s , then ru
    12·1 answer
  • According to Newton’s first law, only ___ has the ability to change motion.
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!