All categories

3 years ago

Define what a glass transition is and what happens to a polymer because of it

Engineering

1 answer:

lesya692 [45]3 years ago

7 0

Answer:

The glass transition occur when the given temperature gets drop below the temperature of the glass transition for the polymers. During the creases of the long range of the given motion, the polymer start changing from the state of hard into the form of brittle.

The range of the glass transition temperature are between the 0 to -150°C. The basic use of the polymer is basically depend upon the main properties of the polymer.

You might be interested in

. A storm sewer is carrying snow melt containing 1.200 g/L of sodium chloride into a small stream. The stream has a naturally oc

Cerrena [4.2K]

Answer:

Explanation:

In the steady state total volume going out will be equal to total volume coming in , and

total salt coming in = total salt going out

Total volume coming in per minute = 2000 + 2 x 10³ x 60

= 122000 L .

Total salt coming in per minute = 1200 x 2000 + 20 x 2 x 10³ x 60

= 2400000 + 2400000 mg

= 4800000 mg

volume of water going out per minute = 122000 L

Total salt going out per minute = 4800000 mg

concentration of water going out = 4800000 / 122000

= 39.344 mg / L

3 0

3 years ago

If a population has no predadors and plenty of available resources, how might that population change

ExtremeBDS [4]

Answer:

B (exponential growth )

Explanation: wish u the best bby <33

4 0

3 years ago

The water in a large lake is to be used to generate electricity by the installation of a hydraulic turbine-generator at a locati

Lynna [10]

Answer:

a) 0.76

b) 0.80

c) 1964 kW

Explanation:

GIVEN DATA:

$\dot m = 5000 kg/s$

Assume Mechanical energy at exist is negligible

A) Take lake bottom as reference, and then kinetic and potential energy are taken as zero.

change in mechanical energy is givrn as

$e_{in} - e_{out} = \frac{P}{\rho} - 0 = gh = 9.81 \times 50( \frac{1 kJ/kg}{1000 m^2/s^2}$

= 0.491 kJ/kg

$\Delta \dot E_{mec} = \dot m (e_{in} - e_{out}) = 5000 \times 0.491 = 2455 kW$

$\eta_{OVERALL} = \frac{\dot W}{\Delta \dot E_{mec}} = \frac{1862}{2455} = 0.76$

B) $\eta -{gen} = \frac{\eta_{overall}}{\eta_{gen}} = \frac{0.76}{0.95} = 0.80$

c) $\dot W_{shaft} = \eta_{overall} \left | \Delta \dot E_{mec} \right | = 0.80(2455)$

$\dot W_{shaft} = 1964 kW$

7 0

3 years ago

Write a SELECT statement that returns these column names and data from the Products table: product_name The product_name column

I am Lyosha [343]

Answer:

SELECT

product_name, list_price, discount_percent,

ROUND(list_price * (discount_percent / 100), 2) AS discount_amount,

ROUND(list_price - (discount_percent / 100 * list_price), 2) AS discount_price

FROM

Products

ORDER BY (list_price - (discount_percent / 100 * list_price)) DESC

LIMIT 5;

Explanation:

In the above SELECT statement is used to select columns from Products table.

The columns are product_name,list_price and discount_percent

Next a column discount_amount is selected which is the calculated from previous two columns that are list_price and discount_percent. The discount amount is basically calculated by multiplying list_price with discount_percent/100 and the resultant column is named as discount_amount using Alias which is used to give a temporary name to set of columns or a table.

Next another column discount_price is obtained from previous three columns that are list_price , discount_percent and discount_amount as: list_price - (discount_percent / 100 * list_price) This as a whole is given a column name discount_price.

FROM is used to refer to a table from which these columns are to be selected. The table name is Product.

The result set is sorted in descending order by discount_price so ORDER BY is used to order the resultant records and DESC is used to sort these records according to the discount_price in descending order.

LIMIT statement is used to extract the records from Product and limit the number of rows returned as a result based on a limit value which is 5 here.

5 0

3 years ago

Air expands through a turbine operating at steady state. At the inlet p1 = 150 lbf/in^2, T1 = 1400R and at the exit p2 = 14.8 lb

Paraphin [41]

Answer:

The power developed in HP is 2702.7hp

Explanation:

Given details.

P1 = 150 lbf/in^2,

T1 = 1400°R

P2 = 14.8 lbf/in^2,

T2 = 700°R

Mass flow rate m1 = m2 = m = 11 lb/s Q = -65000 Btu/h

Using air table to obtain the values for h1 and h2 at T1 and T2

h1 at T1 = 1400°R = 342.9 Btu/h

h2 at T2 = 700°R = 167.6 Btu/h

Using;

Q - W + m(h1) - m(h2) = 0

W = Q - m (h2 -h1)

W = (-65000 Btu/h ) - 11 lb/s (167.6 - 342.9) Btu/h

W = (-65000 Btu/h ) - (-1928.3) Btu/s

W = (-65000 Btu/h ) * {1hr/(60*60)s} - (-1928.3) Btu/s

W = -18.06Btu/s + 1928.3 Btu/s

W = 1910.24Btu/s

Note; Btu/s = 1.4148532hp

W = 2702.7hp

5 0

3 years ago