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Sedbober [7]
3 years ago
10

Consider the activity and specificity of the three enzymes. Pepsin is most active in the human ________ and helps to digest ____

______.

Physics
2 answers:
maksim [4K]3 years ago
8 0
I would have to say stomach and proteins. <span />
jeka943 years ago
5 0

Answer:

"Consider the activity and specificity of the three enzymes. Pepsin is most active in the human<u> Stomach or gut </u>and helps to digest<u>  proteins</u>."

<u>Types of Enzymes:</u>

  1. <u>Amylase:</u>For the breakdown of the carbohydrate macro molecules in to small molecules, the amylase is present in the stomach of the human beings and other living beings who ingest the carbohydrates.
  2. <u>Lipase: </u>For, the break down of the lipid molecules(fatty acids and waxes) we have the lipase inside the gut or stomach of the living beings.
  3. <u>Pepsin:</u> The pepsin is secreted inside the gut or stomach of the living beings to digest the protein molecules, changing them into small forms of the basic structural units known as the amino acids.

Explanation:

<u>Enzymes: </u>

The chemical components or molecules that are required to speed up the reactions that happens inside the living body. They are important because with out the presence of the enzyme there will be less cellular activities and we will have less productivity inside our body.

<u>Pepsin:</u>

The pepsin is present and secreted by the mucus layer and the chief cells that are found in the stomach's layers. While, they are present in the gastric juices of the stomach for better digestion of the food molecules.

While it is present or secreted for the digestion of the proteins molecules. As the macro molecules are changed into smaller amino acids for the proper digestion. As the pepsin is found inside the gut or stomach of the different mammals, birds,and other living beings

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A completely inelastic collision occurs between two balls of wet putty that move directly toward each other along a vertical axi
Marrrta [24]

Answer:

h = 2.64 meters      

Explanation:

It is given that,

Mass of one ball, m_1=3\ kg

Speed of the first ball, v_1=20\ m/s (upward)

Mass of the other ball, m_2=2\ kg

Speed of the other ball, v_2=-12\ m/s (downward)

We know that in an inelastic collision, after the collision, both objects move with one common speed. Let it is given by V. Using the conservation of momentum to find it as :

V=\dfrac{m_1v_1+m_2v_2}{m_1+m_2}

V=\dfrac{3\times 20+2\times (-12)}{3+2}

V = 7.2 m/s

Let h is the height reached by the combined balls of putty rise above the collision point. Using the conservation of energy as :

mgh=\dfrac{1}{2}mV^2

h=\dfrac{V^2}{2g}

h=\dfrac{7.2^2}{2\times 9.8}

h = 2.64 meters

So, the height reached by the combined mass is 2.64 meters. Hence, this is the required solution.

5 0
3 years ago
antsy andy tries to push his 6kg desk away from him and finds that it takes him 12 N of force to start the desk from rest and 8
Rama09 [41]

Answer:

static coefficient = 0,203 & kinetic coefficient = 0,14

Explanation:

There are two (2) conditions, when the desk is about to move and when the desk is moving. In the attachements you can see the two free body diagram for each condition.

In the first condition, there is no movement and the force is 12 N, in the image we can see the total forces are equal to 0 and by the definition of the friction force we can get the static friction coefficient.

In the second condition there is movement in the direction of the force which is equal to 8 N, again by the definition of the friction force we can get the kinetic friction coefficient. Since the desk is moving with constant velocity there is not acceleration.

4 0
3 years ago
If the coefficient of static friction between your coffee cup and the horizontal dashboard of your car is µs = 0.800, how fast c
Cloud [144]

Answer:

Before start of slide velocity will be 14.81 m/sec

Explanation:

We have given coefficient of static friction \mu =0.8

Angle of inclination is equal to \Theta =tan^{-1}\mu

\Theta =tan^{-1}0.8=38.65^{\circ}

tan{38.65^{\circ}}=0.8

Radius is given r = 28 m

Acceleration due to gravity g=9.8m/sec^2

We know that tan\Theta =\frac{v^2}{rg}

0.8=\frac{v^2}{28\times 9.8}

v^2=219.52

v=14.816m/sec

So before start of slide velocity will be 14.81 m/sec

3 0
3 years ago
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