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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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Calculate the gravitational potential energy of a body of mass 40 kg at a vertical height of 10 m. ( g = 9.8 m/s2)
olganol [36]
Ep= mgh
Ep = 40 x 9.8 x 10
Ep = 3920J
Ep = 3900J (2sf)
8 0
3 years ago
The drag on a pitched baseball can be surprisingly large. Suppose a 145 g baseball with a diameter of 7.4 cm has an initial spee
kupik [55]

Answer:

<h2>Part A)</h2><h2>Acceleration of the ball is 10.1 m/s/s</h2><h2>Part B)</h2><h2>the final speed of the ball is given as</h2><h2>v_f = 35.3 m/s</h2>

Explanation:

Part a)

As we know that drag force is given as

F = \frac{C_d \rho A v^2}{2}

C_d = 0.35

A = \frac{\pi d^2}{4}

A = \frac{\pi(0.074)^2}{4}

A = 4.3 \times 10^{-3} m^2

v = 40.2 m/s

so we have

F = \frac{0.35\times 1.2 (4.3 \times 10^{-3})(40.2)^2}{2}

F = 1.46 N

So acceleration of the ball is

a = \frac{F}{m}

a = \frac{1.46}{0.145}

a = 10.1 m/s^2

Part B)

As per kinematics we know that

v_f^2 - v_i^2 = 2 a d

v_f^2 - 40.2^2 = 2(-10.1)(18.4)

v_f = 35.3 m/s

4 0
3 years ago
Two circular coils are concentric and lie in the same plane. The inner coil contains 170 turns of wire, has a radius of 0.0095 m
PIT_PIT [208]

Answer:

Current in outer circle will be 15.826 A

Explanation:

We have given number of turns in inner coil N_I=170

Radius of inner circle r_i=0.0095m

Current in the inner circle I_i=8.9A

Number of turns in outer circle N_o=150

Radius of outer circle r_o=0.015m

We have to find the current in outer circle so that net magnetic field will zero

For net magnetic field current must be in opposite direction as in inner circle

We know that magnetic field is given due to circular coil is given  by

B=\frac{N\mu _0I}{2r}

For net magnetic field zero

\frac{N_I\mu _0I_I}{2r_I}=\frac{N_O\mu _0I_0}{2r_O}

So \frac{170\times \mu _0\times 8.9}{2\times 0.0095}=\frac{150\times \mu _0I_O}{2\times 0.015}

I_O=15.92A

4 0
3 years ago
wo fixed charges, A and B are located at x axis. A is at x = 0 m, B is at x = 4 m. QA = +4.0 μC and QB = -5.0 μC. Calculate the
lys-0071 [83]

Answer:

10250 N/C leftwards

Explanation:

QA = 4 micro Coulomb

QB = - 5 micro Coulomb

AP = 6 m

BP = 2 m

A is origin, B is at 4 m and P is at 6 m .

The electric field due to charge QA at P is EA rightwards

E_{A}=\frac{KQ_{A}}{AP^{2}}=\frac{9\times10^{9}\times4\times10^{-6}}{6^{2}}=1000 N/C (rightwards)

The electric field due to charge QB at P is EB leftwards

E_{B}=\frac{KQ_{B}}{BP^{2}}=\frac{9\times10^{9}\times5\times10^{-6}}{2^{2}}=11250 N/C (leftwards)

The resultant electric field at P due the charges is given by

E = EB - EA

E = 11250 - 1000 = 10250 N/C leftwards

5 0
3 years ago
PLEASE HELP ME
hammer [34]

Answer:

Element Is The Answer I think

6 0
3 years ago
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