When considering free energy change, biochemists usually define a standard state, the biochemical standard state, which is modified from the chemical standard state to fit biochemical applications. Determine which of the phrases describe the biochemical standard state, the chemical standard state, or both.

Answers

Answer 1
Answer:

Answer:

Maximum work under this condition (∆G) = Maximum work under Standard Condition (∆G°) + Activities defining this condition

Explanation:

In this equation, the term DGo provides us with a value for the maximal work we could obtain from the reaction starting with all reactants and products in their standard states, and going to equilibrium. The term DG' provides us with a value for the maximal work we could obtain under the conditions defined by the activities in the logarithmic term. The logarithmic term can be seen as modifying the value under standard conditions to account for the actual conditions. In describing the work available in metabolic processes, we are concerned with the actual conditions in the reaction medium (whether that is a test-tube, or the cell cytoplasm); the important term is therefore DG'. If we measure the actual activities (in practice, we make do with concentrations), and look up a value for DGo in a reference book, we can calculate DG' from the above equation.

Values for DGo provide a useful indication through which we can compare the relative work potential from different processes, because they refer to a standard set of conditions.

Therefore both phrases describe the Biochemical and Chemical Standard State


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Automobiles must be able to sustain a frontal impact. Specifically, the design must allow low speed impacts with little damage, while allowing the vehicle front end structure to deform and absorb impact energy at higher speeds. Consider a frontal impact test of a vehicle with a mass of 1000 kg. a. For a low speed test (v = 2.5 m/s), compute the energy in the vehicle just prior to impact. If the bumper is a pure elastic element, what is the effective design stiffness required to limit the bumper maximum deflection during impact to 4 cm? b. At a higher speed impact of v = 25 m/s, considerable deformation occurs. To absorb the energy, the front end of a vehicle is designed to deform while providing a nearly constant force. For this condition, what is the amount of energy that must be absorbed by the deformation [neglecting the energy stored in the elastic deformation in (a)? If it is desired to limit the deformation to 10 cm, what level of resistance force is required? What is the deacceleration of the vehicle in this condition?

A wall in a house contains a single window. The window consists of a single pane of glass whose area is 0.15 m2 and whose thickness is 5 mm. Treat the wall as a slab of the insulating material Styrofoam whose area and thickness are 10 m2 and 0.15 m, respectively. Heat is lost via conduction through the wall and the window. The temperature difference between the inside and outside is the same for the wall and the window. Of the total heat lost by the wall and the window, what is the percentage lost by the window

Answers

Answer:

The percentage loss of the window is  E  = 97.3%

Explanation:

From the question we are told that

      The area of pane of glass is  A = 0.15 m^2

      The thickness is d = 5mm = (5)/(1000) = 0.005m

       The thickness of the wall is  D = 0.15m

       The area of the wall is  a = 10m^2

Generally the heat lost as a result of conduction of the window is  

              Q_(window) = (j_(glass) * A * (\Delta T) )/(d)

Where j_(glass) is the thermal conductivity of glass which has a constant value of

          j_(glass) = 0.80 J/(s \cdot m \cdot  C^o)

 Substituting values

                 Q_(window) = ( 0.80  * 0.15  * (\Delta T) )/(0.005)

                 Q_(window) = 24 \Delta T

Generally the heat lost as a result of conduction of the wall is  

              Q_(wall) = (j_(styrofoam) * A * (\Delta T) )/(d)

j_(styrofoam) s the thermal conductivity of Styrofoam which has a constant value of  j_(styrofoam) = 0.010J / (s \cdot m \cdot C^o)

       Substituting values

                 Q_(wall) = ( 0.010  * 10  * (\Delta T) )/(0.15)

                 Q_(wall) = 0.667 \  \Delta T

Now the net loss of heat is

         Q_(net) = Q_(window) +  Q_(wall)

  Substituting values

         Q_(net) = 24 + 0.667

         Q_(net) =  24.667

Now the percentage loss by the window is  

            E  = (Q_(window) )/(Q_(net))  * 100

  Substituting value  

           E  = (24)/(24 .667)  * 100

           E  = 97.3%

Write a program to accept a one-line string (maximum of 100 characters) from the keyboard. Edit the string entered in Part 1a (with code that you write) to remove all the white space,digits, punctuation, and other special characters, leaving only the letters. Print out the resulting compressed string to the screen.

Answers

Answer:

// This program is written in C++ programming language

// Comments are used for explanatory purpose

/* The aim of this program is to to remove all the white space,digits, punctuation, and other special characters, leaving only the letters. */

// Program starts here

#include <stdio.h>

#include<iostream>

using namespace std;

int main()

{

// Declare Variable of 100 characters

char word[100];

// Prompt user for input

cout<<"Your input goes here (max, 100 characters)";

cin>>word;

// Iterate through string to check for non alphabetic characters

for (int i = 0; word[i] != '\0'; ++i) {

// Check for uppercase and lowercase letters

while (!((word[i] >= 'a' && word[i] <= 'z') || (word[i] >= 'A' && word[i] <= 'Z') || word[i] == '\0')) {

for (int j = i; word[j] != '\0'; ++j) {

word[j] = word[j + 1];

}

word[j] = '\0';

}

}

cout<<"The resulting compressed string: "<<word;

return 0;

}

Answer:

w = str(input("input your values: "))

values = ' '.join(filter(str.isalpha, w))

while len(w) < 100:

       print(values)

       break

Explanation:

The code is written in python

w = str(input("input your values: "))

This code ask the user to input any string values with characters, numbers, line spaces , letters etc.

values = ' '.join(filter(str.isalpha, w))

This code filters the inputted value to bring only letters. All the letter are then joined together

while len(w) < 100:

The code check if the inputted value is less than 100 characters. While it is less than 100 characters. If it is less than 100 character the next code will function.

print(values)

This code prints the joined letters after checking with  a while loop to confirm the length of character is less than 100

break

The break function breaks the code whether it print the values or not.

Generally, the letters will only be printed if the character inputted is less than 100 and later break the while loop or will not print any letter if the character is greater than 100 and later break.

"Given a nodal delay of 84.1ms when there is no traffic on the network (i.e. usage = 0%), what is the effective delay when network usage = 39.3% ? (Give answer is miliseconds, rounded to one decimal place, without units. So for an answer of 0.10423 seconds you would enter "104.2" without the qu"

Answers

Answer:

Explanation:

effective delay = delay when no traffic x (100)/(100- network\r usage)

effective delay = 84.1 * (100)/(100-39.3)=138.55024711697ms

An aquifer has three different formations. Formation A has a thickness of 8.0 m and hydraulic conductivity of 25.0 m/d. Formation B has a thickness of 2.0 m and a conductivity of 142 m/d. Formation C has a thickness of 34 m and a conductivity of 40 m/d. Assume that each formation is isotropic and homogeneous. Compute both the overall horizontal and vertical conductivities.

Answers

Answer:

The horizontal conductivity is 41.9 m/d.

The vertical conductivity is 37.2 m/d.

Explanation:

Given that,

Thickness of A = 8.0 m

Conductivity = 25.0 m/d

Thickness of B = 2.0 m

Conductivity = 142 m/d

Thickness of C = 34 m

Conductivity = 40 m/d

We need to calculate the horizontal conductivity

Using formula of horizontal conductivity

K_(H)=(H_(A)K_(A)+H_(A)K_(A)+H_(A)K_(A))/(H_(A)+H_(B)+H_(C))

Put the value into the formula

K_(H)=(8.0*25+2,0*142+34*40)/(8.0+2.0+34)

K_(H)=41.9\ m/d

We need to calculate the vertical conductivity

Using formula of vertical conductivity

K_(V)=(H_(A)+H_(B)+H_(C))/((H_(A))/(K_(A))+(H_(B))/(K_(B))+(H_(C))/(K_(C)))

Put the value into the formula

K_(V)=(8.0+2.0+34)/((8.0)/(25)+(2.0)/(142)+(34)/(40))

K_(V)=37.2\ m/d

Hence, The horizontal conductivity is 41.9 m/d.

The vertical conductivity is 37.2 m/d.

A metal crystallizes with a face-centered cubic lattice. The edge of the unit cell is 408 pm. Calculate the number of atoms in the unit cell and diameter of the metal atom.

Answers

Answer:288 pm

Explanation:

Number of atoms(s) for face centered unit cell -

Lattice points: at corners and face centers of unit cell.

For face centered cubic (FCC), z=4.

- whereas

For an FCC lattices √2a =4r =2d

Therefore d = a/√2a = 408pm/√2a= 288pm

I think with this step by step procedure the, the answer was clearly stated.

A 100-mm-diameter cast Iron shaft is acted upon by a 10 kN.m bending moment, a 8 kN.m torque, and a 150 kN axial force simultaneously. The ultimate tensile strength of the shaft material is 210 MPa, and the ultimate compressive strength of the shaft material is 750 MPa, determine the safety factor against failure for the above types of loading using a proper theory of failure.

Answers

Answer:

Are you smart You seem smart

Explanation:

im ur dad