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tekilochka [14]
3 years ago
5

Quaternary structure is achieved when a protein folds into a compact, three-dimensional shape stabilized by interactions between

side-chain R groups of amino acids.True or False.
Physics
1 answer:
Sever21 [200]3 years ago
6 0

Answer:

False: Quaternary structure is achieved when multiple polypeptide chains in protein come together.

Explanation:

There are four levels of protein structure: primary, secondary, tertiary, and Quaternary structure.

Tertiary structure is a protein structure, which is achieved when a protein folds into a compact, three-dimensional shape stabilized by interactions between side-chain R groups of amino acids.

However, Quaternary structure is achieved when multiple polypeptide chains in protein come together.

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A 438kg car is accelerating east at 2.55m/s^2. What is the total force acting east on the car
lisabon 2012 [21]

Answer:

<h2>1116.9 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

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force = 438 × 2.55

We have the final answer as

<h3>1116.9 N</h3>

Hope this helps you

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3 years ago
The mass of a string is 7.7 × 10-3 kg, and it is stretched so that the tension in it is 190 N. A transverse wave traveling on th
Scilla [17]

Length of the strings = 2.33 m

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2 years ago
Cliff divers at Acapulco jump into the sea macias (fjm793) – Homework 3, 2d motion 19-20 – dowd – (WoffordWPHY11920 2) 4 from a
Stells [14]

Answer:

v = 7.67 m/s

Explanation:

Given data:

horizontal distance 11.98 m

Acceleration due to gravity 9.8 m/s^2

Assuming initial velocity is zero

we know that

h = \frac{gt^2}{2}

solving for t

we have

t = \sqrt{\frac{2h}{g}}

substituing all value for time t

t = \sqrt{\frac{2\times 11.98}{9.8}}

t = 1.56 s

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v = \frac{d}{t}

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7 0
3 years ago
A-10A twin-jet close-support airplane is approximately rectangular with a wingspan (the length perpendicular to the flow directi
Sidana [21]

Solution :

Given :

Rectangular wingspan

Length,L = 17.5 m

Chord, c = 3 m

Free stream velocity of flow, $V_{\infty}$ = 200 m/s

Given that the flow is laminar.

$Re_L=\frac{\rho V L}{\mu _{\infty}}$

      $=\frac{1.225 \times 200 \times 3}{1.789 \times 10^{-5}}$

    $= 4.10 \times 10^7$

So boundary layer thickness,

$\delta_{L} = \frac{5.2 L}{\sqrt{Re_L}}$

$\delta_{L} = \frac{5.2 \times 3}{\sqrt{4.1 \times 10^7}}$

    = 0.0024 m

The dynamic pressure, $q_{\infty} =\frac{1}{2} \rho V^2_{\infty}$

                                           $ =\frac{1}{2} \times 1.225  \times 200^2$

                                          $=2.45 \times 10^4 \ N/m^2$

The skin friction drag co-efficient is given by

$C_f = \frac{1.328}{\sqrt{Re_L}}$

     $=\frac{1.328}{\sqrt{4.1 \times 10^7}}$

     = 0.00021

$D_{skinfriction} = \frac{1}{2} \rho V^2_{\infty}S C_f$

                  $=\frac{1}{2} \times 1.225 \times 200^2 \times 17.5 \times 3 \times 0.00021$

                  = 270 N

Therefore the net drag = 270 x 2

                                      = 540 N

7 0
2 years ago
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