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Bogdan [553]
3 years ago
10

What do you think of the category of polymer for making a helmet?​

Engineering
2 answers:
Mice21 [21]3 years ago
8 0

Answer: Find the answer in the explanation

Explanation:

Helmets are manufactured by the use of different categories of polymer. The common material used for manufacturing of helmets are:

1. Polycarbonate

2. Styrofoam

3. Foam

4. Acrylic and polycarbonate

5. Nylon

All the polymer used for the manufacturing of helmets fall under synthetic and semi synthetic categories of polymer.

The styrofoam is classified under addition polymerization. While the nylon is classified under condensation polymerization.

The low density polythene used is classified as branch chains polymer. While the polycarbonate is of high tensile strength with strong mechanical properties.

jarptica [38.1K]3 years ago
3 0
I think that is a very good idea

;)
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In the figure show, what's the distance from point H to point C?
padilas [110]

Answer:

B.

Explanation:

I think not sure hehe

4 0
2 years ago
Consider 1.0 kg of austenite containing 1.15 wt% C, cooled to below 727C (1341F). (a) What is the proeutectoid phase? (b) How
pogonyaev

Answer:

a) The proeutectoid phase is known like cementite and its chemical formula is Fe₃C

b) The kilograms of total ferrite and is 0.8311 kg

The kilograms of total cementite is 0.1689 kg

c) The kilograms of total cementite is 0.9343 kg

Explanation:

Given:

1 kg of austenite

1.15 wt% C

Cooled to below 727°C

Questions:

a) What is the proeutectoid phase?

b) How many kilograms each of total ferrite and cementite form, Wf = ?, Wc = ?

c) How many kilograms each of pearlite and the proeutectoid phase form, Wp = ?

d) Schematically sketch and label the resulting microstructure

a) The proeutectoid phase is known like cementite and its chemical formula is Fe₃C

b) To get the mass of the total ferrite form:

W_{f} =\frac{C_{cementite}-C_{2}  }{C_{cementite}-C_{1}  }

Here,

Ccementite = composition of cementite = 6.7 wt%

C₁ = composition of phase 1 = 0.022 wt%

C₂ = composition of alloy = 1.15 wt%

Substituting values:

W_{f} =\frac{6.7-1.15}{6.7-0.022} =0.8311kg

To get the mass of the total cementite:

W_{c} =\frac{C_{2}-C_{1}}{C_{cementite}-C_{1} } =\frac{1.15-0.022}{6.7-0.022} =0.1689kg

c) The mass of pearlite:

W_{p} =\frac{6.7-1.15}{5.94} =0.9343kg

d) In the diagram you can see the different compositions (pearlite, proeutectoid cementite, ferrite, eutectoid cementite)

5 0
3 years ago
Wiring harnesses run
bazaltina [42]

Answer:

C is the answer if not c then b

5 0
3 years ago
Technician a s ays a shorted circuit can generate excessive heat. technician b says a shorted circuit will cause the circuit pro
Lerok [7]

Based on the information provided, the technician who is correct is: C. Both Technician A and Technician B.

<h3>What is an open circuit?</h3>

An open circuit can be defined as a type of electric circuit in which the continuity between the conducting wire (paths) has been broken or cut.

This ultimately implies that, an open circuit is designed and developed to prevent the flow of electric charges (electrons or currents) from one point in an electric circuit to another.

In Electrical engineering, a short usually causes an electric circuit protection device such as a fuse, circuit breaker, etc., to open when higher than normal current flows through the electrical circuit.

Read more on short circuit here: brainly.com/question/25018411

#SPJ1

Complete Question:

Technician A says a shorted circuit can generate excessive heat. Technician B says a shorted circuit will cause the circuit protection device to open. who is correct?

A. Technician A only

B. Technician B only

C. Both Technician A and B

D. Neither Technician A nor B

5 0
1 year ago
Ammonia gas is diffusing at a constant rate through a layer of stagnant air 1 mm thick. Conditions are such that the gas contain
fiasKO [112]

Answer:

The solution to this question is 5.153×10⁻⁴(kmol)/(m²·s)

That is the rate of diffusion of ammonia through the layer is

5.153×10⁻⁴(kmol)/(m²·s)

Explanation:

The diffusion through a stagnant layer is given by

N_{A}  = \frac{D_{AB} }{RT} \frac{P_{T} }{z_{2} - z_{1}  } ln(\frac{P_{T} -P_{A2}  }{P_{T} -P_{A1} })

Where

D_{AB} = Diffusion coefficient or diffusivity

z = Thickness in layer of transfer

R = universal gas constant

P_{A1} = Pressure at first boundary

P_{A2} = Pressure at the destination boundary

T = System temperature

P_{T} = System pressure

Where P_{T} = 101.3 kPa P_{A2} =0, P_{A1} =y_{A}, P_{T} = 0.5×101.3 = 50.65 kPa

Δz = z₂ - z₁ = 1 mm = 1 × 10⁻³ m

R =  \frac{kJ}{(kmol)(K)} ,    T = 298 K   and  D_{AB} = 1.18 \frac{cm^{2} }{s} = 1.8×10⁻⁵\frac{m^{2} }{s}

N_{A} = \frac{1.8*10^{-5} }{8.314*295} *\frac{101.3}{1*10^{-3} }* ln(\frac{101.3-0}{101.3-50.65}) = 5.153×10⁻⁴\frac{kmol}{m^{2}s }

Hence the rate of diffusion of ammonia through the layer is

5.153×10⁻⁴(kmol)/(m²·s)

5 0
3 years ago
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