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Tcecarenko [31]
3 years ago
12

The graph illustrates the activity level of three common digestive enzymes, across a range of pH values. Which enzyme is likely

to be the most active in the acidic environment of the stomach?
A) pepsin
B) trypsin
C) amylase
D) pepsin and trypsin

Physics
1 answer:
vaieri [72.5K]3 years ago
7 0

Answer:

(A) Pepsin

Explanation:

From the graph it is clear that pepsin is the only enzyme which works in highly acidic condintion in the digestive system.

  • less than 7 the liquid is acidic
  • above 7 the liquid is basic
  • at 7 the liquid is neutral

It has an optimum pH of about 1.5 at which its activity level is 8.5 as shown in graph.

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What is the term for a group of organisms that are able to interbreed and produce fertile offspring?
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<span>A biological species is the term that is used to describe organisms that are able to interbreed and consistently produce fertile (capable of reproducing themselves) offspring. </span>
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2 years ago
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Ball A Mass: 0.45 kg Velocity: 50 m/s Ball B Mass: 0.45 kg Velocity: 80 m/s Ball C Mass: 0.45 kg Velocity: 25 m/s Which ball has
charle [14.2K]
Definition formula for momentum: P = mv

So P(A) = 0.45 * 50 = 22.5 kgm/s

P(B) = 0.45 * 80 = 36 kgm/s

P(C) = 0.45 * 25 = 11.25 kgm/s

B has the greatest momentum
4 0
3 years ago
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The work done by an external force to move a -8.50 μC charge from point a to point b is 6.10×10−4 J . If the charge was started
bekas [8.4K]

Answer:

-54.12 V

Explanation:

The work done by this force is equal to the difference between the final value and the initial value of the energy. Since the charge starts from the rest its initial kinetic energy is zero.

W=\Delta E\\W=\Delta K+\Delta U\\W=K_f+\Delta U\\\Delta U=W-K_f\\\Delta U=6.10*10^{-4}J-1.50*10^{-4}J\\\Delta U=4.60*10^{-4}J

The change in electrostatic potential energy \Delta U, of one point charge q is defined as the product of the charge and the potential difference.

\Delta U=qV\\V=\frac{\Delta U}{q}\\V=\frac{4.60*10^{-4}J}{-8.50*10^{-6}C}\\V=-54.12 V

5 0
3 years ago
A runner is moving at a constant speed of 8.00 m/s around a circular track. If the distance from the runner to the center of the
Genrish500 [490]

Answer: Last option

2.27 m/s2

Explanation:

As the runner is running at a constant speed then the only acceleration present in the movement is the centripetal acceleration.

If we call a_c to the centripetal acceleration then, by definition

a_c =w^2r = \frac{v^2}{r}

in this case we know the speed of the runner

v =8.00\ m/s

The radius "r" will be the distance from the runner to the center of the track

r = 28.2\ m

a_c = \frac{8^2}{28.2}\ m/s^2

a_c = 2.27\ m/s^2

The answer is the last option

3 0
3 years ago
A 40-cmcm-long tube has a 40-cmcm-long insert that can be pulled in and out. A vibrating tuning fork is held next to the tube. A
Licemer1 [7]

Answer:

1070 Hz

Explanation:

First, I should point out there might be a typo in the question or the question has inconsistent values. If the tube is 40 cm long, standing waves cannot be produced at 42.5 cm and 58.5 cm lengths. I assume the length is more than the value in the question then. Under this assumption, we proceed as below:

The insert in the tube creates a closed pipe with one end open and the other closed. For a closed pipe, the difference between successive resonances is a half wavelength \frac{\lambda}{2}.

Hence, we have

\dfrac{\lambda}{2}=58.5-42.5=16 \text{ cm}

\lambda=32\text{ cm}=0.32 \text{ m}.

The speed of a wave is the product of its wavelength and its frequency.

v=f\lambda

f=\dfrac{v}{\lambda}

f=\dfrac{343}{0.32}=1070 \text{ Hz}

7 0
3 years ago
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