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4 years ago

What's the monomer? Show the structure.

Engineering

1 answer:

ivolga24 [154]4 years ago

7 0

In order to understand a monomer let´s first see the structure of a polymer. As an example, in the first figure polyethylene (or polyethene) is shown. This polymer, like every other one, is composed of many repeated subunits, these subunits are called monomer. In the second figure, polyethylene's monomer is shown.

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The first thing you should do is develop a ____________________ to determine what vehicle you can afford.

swat32

The first thing you should do is develop a <u>budget</u> to determine what vehicle you can afford.

<h3>What is an automobile?</h3>

An automobile is also referred to as a vehicle, car or motorcar and it can be defined as a four-wheeled vehicle that is designed and developed to be propelled by an internal-combustion (gasoline) engine, especially for the purpose of transportation from one location to another.

<h3>What is a budget?</h3>

A budget can be defined as a financial plan that is typically used for the estimation of revenue and expenditures of an individual, business organization or government for a specified period of time, often one year.

In this context, we can reasonably infer and logically deduce that the first thing anyone should do is to develop a <u>budget</u> in order to determine what vehicle they can afford.

Read more on budget here: brainly.com/question/13964173

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7 0

1 year ago

For a fluid with a Prandtl Number of 1000.0, the hydrodynamic layer is thinner than the thermal boundary layers. a) True b) Fals

kvv77 [185]

Answer:

(b)False

Explanation:

Given:

Prandtl number(Pr) =1000.

We know that $Pr=\dfrac{\nu }{\alpha }$

Where $\nu$ is the molecular diffusivity of momentum

$\alpha$ is the molecular diffusivity of heat.

Prandtl number(Pr) can also be defined as

$Pr=\left (\dfrac{\delta }{\delta _t}\right )^3$

Where $\delta$ is the hydrodynamic boundary layer thickness and $\delta_t$ is the thermal boundary layer thickness.

So if Pr>1 then hydrodynamic boundary layer thickness will be greater than thermal boundary layer thickness.

In given question Pr>1 so hydrodynamic boundary layer thickness will be greater than thermal boundary layer thickness.

So hydrodynamic layer will be thicker than the thermal boundary layer.

8 0

3 years ago

Ammonia contained in a piston-cylinder assembly, initially saturated vapor at 0o F, undergoes an isothermal process during which

Rudik [331]

ANSWERS:

$-P_{2(a)} =15.6lbf/in^2\\-P_{2(b)} =30.146lbf/in^2\\ T_{2(a)} =0^oF\\T_{2(b)} =0^oF\\x_{2(b)} =49.87percent$

Explanation:

Given:

Piston cylinder assembly which mean that the process is constant pressure process P=C.

<u>AMMONIA </u>

state(1)

saturated vapor $x_{1} =1$

The temperature $T_{1} =0^0 F$

Isothermal process $T=C$

$-V_{2} =2V_{1}$ ( double)

$-V_{2} =.5V_{2}$ (reduced by half)

To find the final state by giving the quality in lbf/in we assume the friction is neglected and the system is in equilibrium.

state(1)

using PVT data for saturated ammonia

$-P_{1} =30.416 lbf/in^2\\-v_{1} =v_{g} =9.11ft^3/lb$

then the state exists in the supper heated region.

a) from standard data

$-v_{1(a)} =2v_{1} =18.22ft^3/lb\\-T_{1} =0^oF$

$at\\P_{x} =14lbf/in^2\\-v_{x} =20.289 ft^3/kg$

$at\\P_{y} =16 lbf/in^2\\-v_{y} =17.701ft^3/kg$

assume linear interpolation

$\frac{P_{x}-P_{2(b)} }{P_{x}- P_{y} } =\frac{v_{x}-v_{1(a)} }{v_{x}-v_{y} }$

$P_{1(b)}=P_{x} -(P_{x} -P_{y} )*\frac{v_{x}- v_{1(b)} }{v_{x}-v_{y} }\\ \\P_{1(b)} =14-(14-16)*\frac{20.289-18.22}{20.289-17.701} =15.6lbf/in^2$

$-v_{2(a)} =2v_{1} =4.555ft^3/lb\\v_{g}$

from standard data

$-v_{f} =0.02419ft^3/kg\\-v_{g} =9.11ft^3/kg\\v_{f}$

then the state exist in the wet zone

$-P_{s} =30.146lbf/in^2\\v_{2(a)} =v_{f} +x(v_{g} -v_{f} )$

$x=\frac{v_{2(a)-v_{f} } }{v_{g} -v_{f} } \\x=\frac{4.555-0.02419}{9.11-0.02419} =49.87%$

3 0

3 years ago

A person walks into a refrigerated warehouse with head uncovered. Model the head as a 25- cm diameter sphere at 35°C with a surf

galina1969 [7]

Answer:

Hello some parts of your question is missing below is the missing part

Convection coefficient = 11 w/m^2. °c

answer : 44.83 watts

Explanation:

Given data :

surface emissivity ( ε )= 0.95

head ( sphere) diameter( D ) = 0.25 m

Temperature of sphere( T ) = 35° C

Temperature of surrounding ( T∞ ) = 25°C

Temperature of surrounding surface ( Ts ) = 15°C

б = ( 5.67 * 10^-8 )

Determine the total rate of heat loss

First we calculate the surface area of the sphere

As = $\pi D^{2}$

= $\pi * 0.25^2$ = 0.2 m^2

next we calculate heat loss due to radiation

Qrad = ε * б * As( $T^{4} - T^{4} _{s}$ ) ---- ( 1 )

where ;

ε = 0.95

б = ( 5.67 * 10^-8 )

As = 0.2 m^2

T = 35 + 273 = 308 k

Ts = 15 + 273 = 288 k

input values into equation 1

Qrad = 0.95 * ( 5.67 * 10^-8 ) * 0.2 ( (308)^4 - ( 288)^4 )

= 22.83 watts

Qrad ( heat loss due to radiation ) = 22.83 watts

calculate the heat loss due to convection

Qconv = h* As ( ΔT )

= 11*0.2 ( 35 -25 ) = 22 watts

Hence total rate of heat loss

= 22 + 22.83

= 44.83 watts

5 0

3 years ago

John wants to construct a device using quartz crystal, Which device can he construct?

tatiyna

Answer: Option D, piezoelectric pressure guage

Explanation: Quartz crystal possess a very useful quality in science as they can generate small charges when pressure is applied to them or when they are hit. This property can be harnessed to construct a piezoelectric pressure gauge which would be used to measure and indicate changes in pressure, the quartz crystal releases little voltage each time there is an applied pressure . This device would be able to sense changes in pressure as there would voltage proportional to the applied pressure.

4 0

3 years ago

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