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2 years ago

Chemical materials that are transported are called..

Engineering

1 answer:

svetlana [45]2 years ago

3 0

Answer:

Transport Molecules

Explanation:

I believe the answer is transport molecules because development of a cell membrane that could allow some materials to pass.

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An insulated piston-cylinder device contains 0.15 of saturated refrigerant-134a vapor at 0.8 MPa pressure. The refrigerant is no

Stells [14]

Answer:

Assumption:

1. The kinetic and potential energy changes are negligible

2. The cylinder is well insulated and thus heat transfer is negligible.

3. The thermal energy stored in the cylinder itself is negligible.

4. The process is stated to be reversible

Analysis:

a. This is reversible adiabatic(i.e isentropic) process and thus $s_{1} =s_{2}$

From the refrigerant table A11-A13

$P_{1} =0.8MPa$ $\left \{ {{ {{v_{1}=v_{g} @0.8MPa =0.025645 m^{3/}/kg } } \atop { {{u_{1}=u_{g} @0.8MPa =246.82 kJ/kg } - also {{s_{1}=s_{g} @0.8MPa =0.91853 kJ/kgK } } \right.$

sat vapor

m= $\frac{V}{v_{1} } =\frac{0.15}{0.025645} =5.8491 kg\\and \\\\P_{2} =0.2MPa \left \{ {{x_{2} =\frac{s_{2} -s_{f} }{s_{fg }}=\frac{0.91853-0.15449}{0.78339} = 0.9753 \atop {u_{2} =u_{f} +x_{2} }(u_{fg}) = 38.26+0.9753(186.25)= 38.26+181.65 =219.9kJ/kg \right. \\s_{1} = s_{2}$

$T_{2} =T_{sat @ 0.2MPa} = -10.09^{o} C$

b.) We take the content of the cylinder as the sysytem.

This is a closed system since no mass leaves or enters.

Hence, the energy balance for adiabatic closed system can be expressed as:

$E_{in} - E_{out} =$ ΔE

$w_{b, out} =$ ΔU

$w_{b, out} =m([tex]u_{1} -u_{2$ )

$w_{b, out} =$ workdone during the isentropic process

=5.8491(246.82-219.9)

=5.8491(26.91)

=157.3993

=157.4kJ

4 0

2 years ago

Design process 8 steps with definition

Troyanec [42]

Answer:

Step 1: Define the Problem.

Step 2: Do Background Research. .

Step 3: Specify Requirements. .

Step 4: Brainstorm, Evaluate and Choose Solution.

Step 5: Develop and Prototype Solution.

Step 6: Test Solution.

Step 7: Does Your Solution Meet the Requirements?

Step 8: Communicate Results.

can u tell me the definition tho?

palled correctly as “though” which is an alternate form of “although”) at the end is informal usage. It's better placed before “she seems better today

8 0

2 years ago

A packet weighs 40kg in air but when it is totally submerged into a 1mx1m square tank the weight of the packet is only 18kg. How

Irina18 [472]

Answer:

water rise = 22 mm

Explanation:

weight of packet IN AIR = 40 *9.81 =392.4 N

weight of packet IN WATER= 18 *9.81 =176.58 N

by Archimedi's principle

difference in weight = weight of displaced water

w_a - w_w = \rho_w v_d g

392.4 - 176.58 = 1000* v_d* 9.81

v_d = 0.022 m^3

v_d = A*H_rise

0.022 =1*H_rise

H_rise = 0.022 m = 22 mm

water rise = 22 mm

5 0

3 years ago

The bulk modulus of a fluid if it undergoes a 1% change in volume when subjected to a pressure change of 10,000 psi is (a) 0.01

Veseljchak [2.6K]

Answer:

The required bulk modulus is $10^{6}$ Psi. So, the answer is non of these.

Explanation:

Change in pressure of the fluid is directly proportional to the volumetric strain. The constant of proportionality is the bulk modulus of the fluid.

Step1

Given:

Percentage change in volume is 1%.

Change in pressure is 10000 Psi.

Calculation:

Step2

Volumetric strain is calculated as follows:

$\frac{\bigtriangleup V}{V}=\frac{1}{100}$

$\frac{\bigtriangleup V}{V}=0.01$

Step3

Bulk modulus is calculated as follows:

\frac{\bigtriangleup V}{V}=0.01

$\frac{\bigtriangleup V}{V}=0.01$

$10000=K\times0.01$

K = 1000000 Psi.

Thus, the required bulk modulus is $10^{6}$ Psi.

3 0

2 years ago

Write a program that adds the numbers 1 through 5 into the numbers ArrayList and then prints out the first element in the list.

nekit [7.7K]

Answer:

Answer code is given below along with comments to explain each step

Explanation:

Method 1:

// size of the array numbers to store 1 to 5 integers

int Size = 5;

// here we have initialized our array numbers, the type of array is integers and size is 5, right now it is empty.

ArrayList<Integer> numbers = new ArrayList<Integer>(Size);

// lets store values into the array numbers one by one

numbers.add(1)

numbers.add(2)

numbers.add(3)

numbers.add(4)

numbers.add(5)

// here we are finding the first element in the array numbers by get() function, the first element is at the 0th position

int firstElement = numbers.get(0);

// to print the first element in the array numbers

System.out.println(firstElement);

Method 2:

int Size = 5;

ArrayList<Integer> numbers = new ArrayList<Integer>(Size);

// here we are using a for loop instead of adding manually one by one to store the values from 1 to 5

for (int i = 0; i < Size; i++)

{

// adding the values 1 to 5 into array numbers

numbers.add(i);

}

//here we are finding the first element in the array numbers

int firstElement = numbers.get(0);

// to print the first element in the array

System.out.println(firstElement);

6 0

2 years ago