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Briefly explain how each of the following influences the tensile modulus of a semicrystalline polymer and why:(a) molecular weig

marin [14]

Answer:

(a) Increases

(b) Increases

(c) Increases

(d) Increases

(e) Decreases

Explanation:

The tensile modulus of a semi-crystalline polymer depends on the given factors as:

(a) Molecular Weight:

It increases with the increase in the molecular weight of the polymer.

(b) Degree of crystallinity:

Tensile strength of the semi-crystalline polymer increases with the increase in the degree of crystallinity of the polymer.

(c) Deformation by drawing:

The deformation by drawing in the polymer results in the finely oriented chain structure of the polymer with the greater inter chain secondary bonding structure resulting in the increase in the tensile strength of the polymer.

(d) Annealing of an undeformed material:

This also results in an increase in the tensile strength of the material.

(e) Annealing of a drawn material:

A semi crystalline material which is drawn when annealed results in the decreased tensile strength of the material.

5 0

3 years ago

Hello I need some help with this please. Pick a problem in your school or community that you think could be solved with technolo

bezimeni [28]

Answer:

An AI operated automatic garbage collection system

Explanation:

There is always an issue in my neighbourhood with the garbagemen coming on time so having an automatic system will help in the overall efficiency in the task

7 0

3 years ago

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A 100 ft long steel wire has a cross-sectional area of 0.0144 in.2. When a force of 270 lb is applied to the wire, its length in

blondinia [14]

Answer:

(a) The stress on the steel wire is 19,000 Psi

(b) The strain on the steel wire is 0.00063

(c) The modulus of elasticity of the steel is 30,000,000 Psi

Explanation:

Given;

length of steel wire, L = 100 ft

cross-sectional area, A = 0.0144 in²

applied force, F = 270 lb

extension of the wire, e = 0.75 in

Part (A) The stress on the steel wire;

δ = F/A

= 270 / 0.0144

δ = 18750 lb/in² = 19,000 Psi

Part (B) The strain on the steel wire;

σ = e/ L

L = 100 ft = 1200 in

σ = 0.75 / 1200

σ = 0.00063

Part (C) The modulus of elasticity of the steel

E = δ/σ

= 19,000 / 0.00063

E = 30,000,000 Psi

4 0

3 years ago

Water flows through a pipe at an average temperature of T[infinity] = 70°C. The inner and outer radii of the pipe are r1 = 6 cm

Paul [167]

Answer:

The differential equation and the boundary conditions are;

A) -kdT(r1)/dr = h[T∞ - T(r1)]

B) -kdT(r2)/dr = q'_s = 734.56 W/m²

Explanation:

We are given;

T∞ = 70°C.

Inner radii pipe; r1 = 6cm = 0.06 m

Outer radii of pipe;r2 = 6.5cm=0.065 m

Electrical heat power; Q'_s = 300 W

Since power is 300 W per metre length, then; L = 1 m

Now, to the heat flux at the surface of the wire is given by the formula;

q'_s = Q'_s/A

Where A is area = 2πrL

We'll use r2 = 0.065 m

A = 2π(0.065) × 1 = 0.13π

Thus;

q'_s = 300/0.13π

q'_s = 734.56 W/m²

The differential equation and the boundary conditions are;

A) -kdT(r1)/dr = h[T∞ - T(r1)]

B) -kdT(r2)/dr = q'_s = 734.56 W/m²

6 0

3 years ago

Does a thicker core make an electromagnet stronger?

mel-nik [20]

Answer:

Yes

Explanation:

The core of an electromagnet serves to stabilize the magnetic field created by the wire. The thicker the core, the more metal there is to amplify the current. Therefore, a thicker core does make an electromagnet stronger. Hope this helps!

6 0

3 years ago

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