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3 years ago

Thermoplastics that exhibit a definite TM point have this kind of crystalline structure:

Engineering

1 answer:

geniusboy [140]3 years ago

5 0

Answer:

The correct option is A

Explanation:

Thermoplastics are polymers that can be recycled. They can be melted with heat and moulded into any shape which becomes hard when cooled. The heat applied to it in the process of recycling does not affect its chemical composition.

There are two main classes of thermoplastics; semi-crystalline and amorphous.

An amorphous thermoplastic has a disorderly arranged molecular structure which makes it difficult to have a definite melting temperature (Tm). However, a semi-crystalline thermoplastic has a well arranged molecular structure which makes it melt after a certain/definite amount of heat is absolved. Hence, the correct answer is A

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When was solar power envold

Lyrx [107]

Answer:

1839

Explanation:

the first huge step in solar penal technology came when alxendare becquerel observed the photovolatic effect in 1839,which occurs when a material produces electric current when exposed to light.it was not untill 1888 that the first solar cell was actually built by aleksander stoletov.

5 0

3 years ago

Read 2 more answers

A slight breeze is blowing over the hot tub above and yields a heat transfer coefficient h of 20 W/m2 -K. The air temperature is

patriot [66]

Answer:4050 W

Explanation:

Given

Heat transfer Coefficient(h)= $20 W/m^2-K$

Air temperature =75 F

surface area(A)= $7.5 m^2$

Temperature of hot tube is 102 F

We know heat transfer due to convection is given by

$Q=hA\left ( \Delta T\right )$

$Q=20\times 7.5\left ( 102-75\right )=4050 W$

7 0

3 years ago

a. Determine R for a series RC high-pass filter with a cutoff frequency (fc) of 8 kHz. Use a 100 nF capacitor. b. Draw the schem

Readme [11.4K]

Answer:

a) 199.04 ohms

b) attached in image

c) -0.696dB

Explanation:

We are given:

Fc = 8Khz = 8000hz

$C = 100nF = 100*10^-^9F$

a)Using the formula:

$F_c = \frac{1}{2pie*Rc}$

$8000= \frac{1}{2*3.14*R*100*10^-^9}$

$R =\frac{1}{2*3.14*100*10^-^9*8000}$

R = 199.04 ohms

b) diagram is attached

c) $H(w) = \frac{V_out(w)}{Vin(w)} = \frac{1}{1-j\frac{wc}{w}}$

$H(F) = \frac{1}{1-j\frac{fc}{f}}$

At F = 20KHz and Fc= 8KHz we have:

$H(F)= \frac{1}{1-j\frac{8}{20}} = \frac{1}{1-j(0.4)}$

$|H(F)|= \frac{1}{\sqrt{1^2+(0.4)^2}}$

=0.923

|H(F)| in dB = 20log |H(F)|

=20log0.923

= -0.696dB

5 0

3 years ago

Five bolts are used in the connection between the axial member and the support. The ultimate shear strength of the bolts is 320

lesya [120]

Answer:

The minimum allowable bolt diameter required to support an applied load of P = 450 kN is 45.7 milimeters.

Explanation:

The complete statement of this question is "Five bolts are used in the connection between the axial member and the support. The ultimate shear strength of the bolts is 320 MPa, and a factor of safety of 4.2 is required with respect to fracture. Determine the minimum allowable bolt diameter required to support an applied load of P = 450 kN"

Each bolt is subjected to shear forces. In this case, safety factor is the ratio of the ultimate shear strength to maximum allowable shear stress. That is to say:

$n = \frac{S_{uts}}{\tau_{max}}$