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3 years ago

Describe what viscoelastic behavior means

1 answer:

5 0

This refers to the mechanical properties of the materials tested by means of a stress stimulus or deformation. The mechanical behavior of a material will depend on its viscosity / elasticity characteristics. Depending on the response to mechanical stimulus, the material may be classified as elastic or viscous. For a perfectly elastic or Hookian solid the deformation is proportional to the applied stress. A viscous material obeys Newton's law, which states that stress and shear rate are related through an intrinsic characteristic that is viscosity.

However, there are materials that correspond to both properties. Polymers have an intermediate mechanical behavior to elastic and viscous, being called viscoelastic. The elastic and viscous contribution to the mechanical behavior of the polymer depends on the temperature and time scale of the experiment.

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You can change lanes during a turn long as there’s no traffic and you driving slowly

Vanyuwa [196]

Your allowed to switch lanes as long as the road is clear and you use signals.

5 0

3 years ago

g Water is being heated in a closed pan on top of a range while being stirred by a paddle wheel. During the process, 30 kJ of he

AVprozaik [17]

Answer:

38kJ

Explanation:

Final Energy= Total Energy at the beginning + Total energy added - energy lost

Final Energy = 12.5 + 500/1000 + 30- 5

= 38kJ

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4 years ago

Perform the following unit conversions. Please do not use an on-line unit converter since this problem is given to you as practi

lutik1710 [3]

Answer:

The answers are below

Explanation:

a. 180 in^3 to L

1 in³ = 0.0164L

180 in³ = $180\ in^3*\frac{0.0164\ L}{1\ in^3}= 2.95\ L$

b. 750 ft-lbf to kJ

1 ft-lbf = 0.00136 kJ

750 ft-lbf = $750\ ft-lbf *\frac{0.00136\ kJ}{1\ ft-lbf} =1.02\ kJ$

c. 75.0 hp to kW

1 hp = 0.746 kW

75 hp = $75\ hp*\frac{0.746\ kW}{1\ hp}=55.95\ kW$

d. 2500.0 lb/h to kg/s

1 lb/h = 0.000126 kg/s

2500.0 lb/h = $2500.0\ lb/h*\frac{0.000126\ kg/s}{1\ lb/h} =0.315\ kg/s$

e. 120 psia to kPa

1 psia = 6.89 kPa

120 psia = $120\ psia*\frac{6.89\ kPa}{1\ psia} =826.8\ kPa$

f. 120 psig to kPa

1 psig = 6.89 kPa

120 psig = $120\ psia*\frac{6.89\ kPa}{1\ psig} =826.8\ kPa$

g. 300 ft/min to m/s

1 ft/min = 0.005 m/s

300 ft/min = $300\ ft/min*\frac{0.005\ m/s}{1\ ft/min} = 1.5\ m/s\\$

h. 125 km/h to miles/h

1 km/h = 0.62 mph

125 km/h = $125\ km/h*\frac{0.62\ mph}{1\ km/h} =77.5\ mph$

i) 6000 N to Ibf

1 N = 0.2248 lbf

6000 N = $6000\ N*\frac{ 0.2248\ lbf}{1\ N}=1348.8\ N$

j. 6000 N to ton

1 N = 0.000102 Ton-force

6000 N = $6000\ N*\frac{ 0.000102\ Ton-force}{1\ N}=0.612\ N$

3 0

3 years ago

A 0.5m diameter sphere containing pollution monitoring equipment is dragged through the Charles River at a relative velocity of

Dmitriy789 [7]

Answer:

$\phi = 155.57$

Explanation:

from figure

taking summation of force in x direction be zero

$\sum x = 0$

$F_D = Tsin \theta$ .....1

$\frac{c_d \rho v^2 A}{2} =Tsin \theta$

taking summation of force in Y direction be zero

$F_B - W- Tcos \theta$

$T = \frac{F_B -W}{cos \theta}$ .........2

putting T value in equation 1

$F_D - \frac{F_B -W}{cos \theta} sin\theta$

$F_D = \rho g V ( 1 -Sg) tan \theta$ .........3

$F_D = \rho g [\frac{\pi d^3}{6}] ( 1 -Sg) tan \theta$

$tan \theta = \frac{6 c_D \rho v&2 A}{ 2 \rho g V \pi D^3 (1- Sg)}$

Water at 10 degree C has kinetic viscosity v = 1.3 \times 10^{-6} m^2/s

Reynold number

$Re = \frac{ VD}{\nu} = \frac{10\times 0.5}{1.3 \times 10^{-6}} = 3.84 \times 10^6$

so for Re = $3.84 \times 10^6$ cd is 0.072

$tan \theta = \frac{3\times 0.072 \times 10^2 \times \pi \times 0.5^2}{ 2\times 9.81 \pi 0.5^3(1- 1.5)}$

$\theta = tan^{-1} [\frac{3\times 0.072 \times 10^2 \times \pi \times 0.5^2}{ 2\times 9.81 \pi 0.5^3(1- 1.5)}]$

$\theta = - 65.57 degree$

$\phi = 90 - (-65.57) = 1557.57$ degree

8 0

3 years ago

Convert a barometric pressure of 28.75 in. Hg at 32 F (0°C) to (a) psia (b) inches of water, (e) meters of liquid whose specific

LenKa [72]

Answer:

a) 14.12psi, b) 390.86inH2O, d) 97.36kPa, d)12.42m

Explanation:

In order to do the conversions we need to use conversion rates:

a) 1psi=2.03602inHg

so:

$28.75inHg*\frac{1psi}{2.03602inHg}=14.12psi$

and the same procedure is used for parts b and d:

b) 1inHg=13.595inH2O

so:

$28.75inHg*\frac{13.595inH2O}{1inHg}=390.86inH2O$

d) 1inHg=3.386kPa

so:

$28.75inHg*\frac{3.386kPa}{1inHg}=97.36kPa$

Now part e is a little tricky since we need to review our specific gravity concept. The specific gravity is defined as the ratio between the specific weight of a substance ofver the specific weight of water.

$Sg=\frac{\gamma_{substance}}{\gamma_{H_{2}O}}$