What factors determine whether a species can be present in a place

The process of digestion breaks down proteins into smaller molecules that are used to rebuild other proteins. During digestion,

Len [333]

The process of digestion breaks down proteins into smaller molecules that are used to rebuild other proteins. During digestion, proteins undergo chemical changes.

<u>Option: D</u>

<u>Explanation:</u>

Proteins are broken down into amino acids considered the foundational element. When issued, these tiny molecules can then be consumed into the bloodstream via the gut wall. An enzyme is a protein which can regulate biochemical response rate. An enzyme integrates a water molecule around the bond in enzymatic hydrolysis processes which allows it to split.

The bonds which hold together the amino acids are recognized as peptide bonds. A hydrolysis process comparable to that included in splitting up carbohydrates is required to break the peptide bonds within a protein. Enzymes identified as proteases are required for the protein to disintegrate.

7 0

4 years ago

Two particles each of mass m and charge q are suspended by strings of length / from a common point. Find the angle e that each s

ozzi

Answer:

$\theta =\left (\frac{kq^{2}}{4L^{2}\times mg} \right )^{\frac{1}{3}}$

Explanation:

Let the length of the string is L.

Let T be the tension in the string.

Resolve the components of T.

As the charge q is in equilibrium.

T Sinθ = Fe ..... (1)

T Cosθ = mg .......(2)

Divide equation (1) by equation (2), we get

tan θ = Fe / mg

$tan\theta =\frac{\frac{kq^{2}}{AB^{2}}}{mg}$

$tan\theta =\frac{\frac{kq^{2}}{4L^{2}Sin^{\theta }}}}{mg}$

$tan\theta =\frac{kq^{2}}{4L^{2}Sin^{2}\theta \times mg}$

$tan\theta\times Sin^{2}\theta =\frac{kq^{2}}{4L^{2}\times mg}$

As θ is very small, so tanθ and Sinθ is equal to θ.

$\theta ^{3} =\frac{kq^{2}}{4L^{2}\times mg}$

$\theta =\left (\frac{kq^{2}}{4L^{2}\times mg} \right )^{\frac{1}{3}}$

7 0

3 years ago

A mass is attached to a vertical spring, which then goes into oscillation. At the high point of the oscillation, the spring is i

andrew-mc [135]

Answer:

0.34 sec

Explanation:

Low point of spring ( length of stretched spring ) = 5.8 cm

midpoint of spring = 5.8 / 2 = 2.9 cm

Determine the oscillation period

at equilibrum condition

Kx = Mg

g= 9.8 m/s^2

x = 2.9 * 10^-2 m

k / m = 9.8 / ( 2.9 * 10^-2 ) = 337.93

note : w = $\sqrt{k/m}$ = $\sqrt{337.93}$ = 18.38 rad/sec

Period of oscillation = $2\pi / w$

= 0.34 sec

8 0

3 years ago

Three balls with different masses are shown below.

algol [13]

Answer:

$10 a g

Explanation:

6 0

3 years ago

Read 2 more answers

Ryan is driving home from work and notices a deer leaping onto the road about 25 m in front of his car. He immediately applies t

Anvisha [2.4K]

Answer:

mu = 0.56

Explanation:

The friction force is calculated by taking into account the deceleration of the car in 25m. This can be calculated by using the following formula:

$v^2=v_0^2+2ax\\$

v: final speed = 0m/s (the car stops)

v_o: initial speed in the interval of interest = 60km/h

= 60(1000m)/(3600s) = 16.66m/s

x: distance = 25m

BY doing a the subject of the formula and replace the values of v, v_o and x you obtain:

$a=\frac{v^2-v_o^2}{2x}=\frac{0m^2/s^2-(16.66m/s)^2}{2(25m)}=-5.55\frac{m}{s^2}$

with this value of a you calculate the friction force that makes this deceleration over the car. By using the Newton second's Law you obtain:

$F_f=ma=(1490kg)(5.55m/s^2)=8271.15N$

Furthermore, you use the relation between the friction force and the friction coefficient:

$F_f= \mu N=\mu mg\\\\\mu=\frac{F_f}{mg}=\frac{8271.15N}{(1490kg)(9.8m/s^2)}=0.56$

hence, the friction coefficient is 0.56

6 0

3 years ago