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3 years ago

Thermosets burn upon heating. a)-True b)- false?

Engineering

1 answer:

hram777 [196]3 years ago

5 0

Answer:

true

Explanation:

True, there are several types of polymers, thermoplastics, thermosets and elastomers.

Thermosets are characterized by having a reticulated structure, so they have low elasticity and cannot be stretched when heated.

Because of the above, thermosetting polymers burn when heated.

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Sea water with a density of 1025 kg/m3 flows steadily through a pump at 0.21 m3 /s. The pump inlet is 0.25 m in diameter. At the

myrzilka [38]

Answer:

$\dot W_{pump} = 16264.922\,W\,(16.265\,kW)$

Explanation:

The pump is modelled after applying Principle of Energy Conservation, whose form is:

$\frac{P_{1}}{\rho\cdot g}+ \frac{v_{1}^{2}}{2\cdot g} +z_{1} + h_{pump}=\frac{P_{2}}{\rho\cdot g}+ \frac{v_{2}^{2}}{2\cdot g} +z_{2}$

The head associated with the pump is cleared:

$h_{pump} = \frac{P_{2}-P_{1}}{\rho\cdot g}+\frac{v_{2}^{2}-v_{1}^{2}}{2\cdot g}+(z_{2}-z_{1})$

Inlet and outlet velocities are found:

$v_{1} = \frac{0.21\,\frac{m^{3}}{s} }{\frac{\pi}{4}\cdot (0.25\,m)^{2} }$

$v_{1} \approx 4.278\,\frac{m}{s}$

$v_{2} = \frac{0.21\,\frac{m^{3}}{s} }{\frac{\pi}{4}\cdot (0.152\,m)^{2} }$

$v_{2} \approx 11.573\,\frac{m}{s}$

Now, the head associated with the pump is finally computed:

$h_{pump} = \frac{175\,kPa-81.326\,kPa}{(1025\,\frac{kg}{m^{3}} )\cdot (9.807\,\frac{m}{s^{2}} )} +\frac{(11.573\,\frac{m}{s} )^{2}-(4.278\,\frac{m}{s} )^{2}}{2\cdot (9.807\,\frac{m}{s^{2}} )} + 1.8\,m$

$h_{pump} = 7.705\,m$

The power that pump adds to the fluid is:

$\dot W_{pump} = \dot V \cdot \rho \cdot g \cdot h_{pump}$

$\dot W_{pump} = (0.21\,m^{3})\cdot (1025\,\frac{kg}{m^{3}})\cdot (9.807\,\frac{m}{s^{2}})\cdot(7.705\,m)$

$\dot W_{pump} = 16264.922\,W\,(16.265\,kW)$

4 0

3 years ago

A rectangular plate casting has dimensions 200mm x 100mm x 20mm. The riser for this sand casting mold is in the shape of a spher

9966 [12]

Answer:

Diameter of riser =6.02 mm

Explanation:

Given that

Dimensions of rectangular plate is 200mm x 100mm x 20mm.

Volume of rectangle V= 200 x 100 x 20 $mm^3$

Surface area of rectangle A

A=2(200 x 100+100 x 20 +20 x 200) $mm^2$

So V/A=7.69

We know that

Solidification times given as

$t=K\left(\dfrac{V}{A}\right)^2$ -----1

Lets take diameter of riser is d

Given that riser is in spherical shape so V/A=d/6

And

Time for solidification of rectangle is 3.5 min then time for solidificartion of riser is 4.2 min.

Lets take $\dfrac{V}{A}=M$

$\dfrac{M_{rac}}{M_{riser}}=\dfrac{7.69}{\dfrac{d}{6}}$

Now from equation 1

$\dfrac{3.5}{4.2}=\left(\dfrac{7.69}{\dfrac{d}{6}}\right)^2$

So by solving this d=6.02 mm

So the diameter of riser is 6.02 mm.

3 0

3 years ago

Assume you are an observer standing at a point along a three-lane roadway. All vehicles in lane 1 are traveling at 30 mi/h, all

garik1379 [7]

Answer:

Explanation:

Given data;

In line 1, v1 = 30mi/hr

in line 2, v2 = 45mi/hr

in line 3, v3 = 60mi/hr

therefore time mean speed = v1 + v2 + v3 /n

= VT = 45mi/hr

space mean speed ; Vs

harmonic mean = 1/V = 1/v1 + 1/v2 + 1/v3

V = 13.85mi/hr

Hence Vs = V x n = 3 x 13.85 = 41.55mi/hr

5 0

3 years ago

I NEED HELP ASAP WILL AWARD BRAINLIEST

CaHeK987 [17]

Answer:

It was a power supply of a circuit that was horizontal of the ladder where they controlled the circuit.

Explanation:

6 0

3 years ago

Read 2 more answers

How many kg / day of NaOH must be added to neutralize a waste stream generated by an industry producing 90,800 kg / day of sulfu

hram777 [196]

Answer:

74.12kg/day

Explanation:

Equation of reaction for the neutralization reaction of sodium hydroxide and sulfuric acid: 2NaOH + H2SO4 yields Na2SO4 + 2H2O

Mass of sulfuric acid produced per day = 90,800kg

Percentage of sulfuric acid in wastewater = 0.1%

Mass of sulfuric acid that ends up in wastewater per day = 0.1/100 × 90,800 = 90.8kg

From the equation of reaction, 2 moles of NaOH (80kg of NaOH) is required to neutralize 1 mole of H2SO4 (98kg of H2SO4)

80kg of NaOH is required to neutralize 98kg of sulfuric acid

90.8kg of sulfuric acid would be neutralized by (90.8×80)/98kg of NaOH = 74.14kg/day of NaOH

7 0

3 years ago

Read 2 more answers