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Alex73 [517]
4 years ago
6

Polymer ropes and lines for use on water are often designed to float, to aid in their retrieval and to avoid applying a downward

s load to an object or person attached to them in the water. Excessive stretch is undesirable, so a lower limit of 0.5 GPa is also imposed on Young's modulus. Identify suitable polymers, using Young's modulus-density chart.

Engineering
1 answer:
andrew-mc [135]4 years ago
8 0

Answer:

We choose PTFE

Explanation:

Attached are the modulus density and modulus strength chart.

Due to its young modulus, the density is near 0.5 GPa, as seen in the chart and support water gliding. The PTFE density is between 1 and 10 Mg / cubic meter (see module and chart of density), and the resistance is between 10 and 100 Mpa (see module and chart of strength). Therefore, the finest ploymer will be PTFE that meets the requirements.

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alexira [117]

Answer:

The flow of a charge is called​ electric current.

Explanation:

8 0
3 years ago
Water circulates throughout a house in a hot water heating system. If the water is pumped at a speed of 0.50m/s through a 4.0-cm
igor_vitrenko [27]

Answer:

velocity and pressure in a 2.6-cm:

P2 = 2.53x10^5Pa, v2 = 1.18m/s

Explanation:

Pressure = P, Velocity = v, Height = h, Diameter = d, Radius= r, Area = A

Area = πr^2

From the question:

v1 = 0.5m/s

d1 = 4cm = 0.04m

r1 = d1/2 = 0.04/2 = 0.02m

Since water was pumped from basement, h1 = 0m

P1 = 3.03x10^5 Pa

A1 = π×0.02×0.02

A1 = 0.0004πm^2

v2 = unknown

d2 = 2.6cm = 0.026m

r2 = d2/2 = 0.026/2 = 0.013m

h2 = 5m

P2 = unknown

A2 = π×0.013×0.013

A2 = 0.000169πm^2

Using continuity equation:

A1v1 = A2v2

0.0004π * 0.5 = 0.000169π * v2

v2 = (0.0004π * 0.5)/(0.000169π)

v2 = 1.18m/s

Applying a Bernoulli principle

P + 1/2*density*v^2 + density*g*h =C

C = constant

P1 + 1/2*density*v1^2 + density*g*h1

= P2 + 1/2*density*v2^2 + density*g*h2

Let g = 9.81m/s

density of water = 1000kg/m^3

(P1-P2) = 1/2* density(v2^2 - v1^2) +(density*g*h2) - (density*g*h1)

(P1-P2) = 1/2* density(v2^2 - v1^2) +

density* g(h2-h1)

(3.03x10^5 - P2)= 1/2*1000 (1.18^2-0.5^2) + 1000(9.81(5-0))

(3.03x10^5 - P2) = 500(1.3924-0.25) + 49050

3.03x10^5 - P2 = 571.2 + 49050

3.03x10^5 - P2 = 49621.2

3.03x10^5 - 49621.2 = P2

P2 = 253378.8

P2 = 2.53x10^5Pa

P2 = 2.53x10^5Pa, v2 = 1.18m/s

7 0
3 years ago
To make an even better electrical junction, what should you do?
docker41 [41]

Answer:

A:Solder it.

Explanation:

Hopefully this helps!

4 0
3 years ago
Creating vacancies in ceramics. Consider doping ZrO2 with small concentrations of Nb205. The valence of Nb is 5. Assume that Nb
Murljashka [212]

Answer:

Creating vacancies in ceramics. Consider doping ZrO₂ with small concentrations of Nb205. The valence of Nb is 5. Assume that Nb ions sit in Zr ion sites

a. A substitutional defect will be produced.

b. With this dopping, the Nb increases electron band conduction and decreases oxygen anion conduction in ZrO₂.

Explanation:

(a) The defect produced by dopping a little concentration of Nb₂O5 with Nb in the +5 charge state is known as a substitutional defect.

(b) With this dopping, the Nb increases electron band conduction and decreases oxygen anion conduction in ZrO₂.

Moreover, if oxygen vacancies are rate-limiting defect, the corrosion of ZrO₂ decreases and if electrons are rate-limiting then the corrosion of ZrO₂ is accelerated.

7 0
4 years ago
What should the resistance value be on a size 5 motor starter coil
Katen [24]

Answer:

1000 Mega Ohms.

Explanation:

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