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andrezito [222]

2 years ago

9

Do plastic materials have high or low ductility? Explain why.

1 answer:

Flura [38]2 years ago

3 0

The impact behavior of plastic materials is strongly dependent upon the temperature. At high temperatures, materials are more ductile and have high impact toughness. At low temperatures, some plastics that would be ductile at room temperature become brittle.

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I ran across this symbol in some Electrical wiring documents and I am unaware of what this means. Any help?

Minchanka [31]

Answer:

Opened Push-button Switch (i.e. a PTM Switch)

Explanation:

Tha's just another symbol for a switch, but this one specifies that the switch is a push-button type of switch.

Since it's not touching and completing the line, the state of the switch is initially open.

6 0

3 years ago

Reduce the force F ij = + (2 5 ) kN to point A(2m,3m) that acts on point B( 3m,5m) - .

Alexxx [7]

Given :

Force, $\vec{F}= (2\hat{i} + 5\hat{j})\ kN$ .

Force is acting at point A( 2 m, 3 m ) and B( 3 m, 5 m )

To Find :

The work done by force F .

Solution :

Displacement vector between point A and B is :

$\vec{d} = (3-2)\hat{i} + (5-3)\hat{j}\\\\\vec{d} = \hat{i} + 2\hat{j}$

Now, we know work done is given by :

$W = \vec{F}.\vec{d}\\\\W= (2\hat{i} + 5\hat{j}).(\hat{i}+\hat{2j})\\\\W = (2\times 1) +( 5\times 2) \ kJ\\\\W = 12 \ kJ$

W = 12000 J

Therefore, work done by force is 12000 J .

6 0

3 years ago

Air enters a cmpressor at 20 deg C and 80 kPa and exits at 800 kPa and 200 deg C. The power input is 400 kW. Find the heat trans

aksik [14]

Answer:

The heat is transferred is at the rate of 752.33 kW

Solution:

As per the question:

Temperature at inlet, $T_{i} = 20^{\circ}C$ = 273 + 20 = 293 K

Temperature at the outlet, $T_{o} = 200{\circ}C$ = 273 + 200 = 473 K

Pressure at inlet, $P_{i} = 80 kPa = 80\times 10^{3} Pa$

Pressure at outlet, $P_{o} = 800 kPa = 800\times 10^{3} Pa$

Speed at the outlet, $v_{o} = 20 m/s$

Diameter of the tube, $D = 10 cm = 10\times 10^{- 2} m = 0.1 m$

Input power, $P_{i} = 400 kW = 400\times 10^{3} W$

Now,

To calculate the heat transfer, $Q$ , we make use of the steady flow eqn:

$h_{i} + \frac{v_{i}^{2}}{2} + gH + Q = h_{o} + \frac{v_{o}^{2}}{2} + gH' + p_{s}$

where

$h_{i}$ = specific enthalpy at inlet

$h_{o}$ = specific enthalpy at outlet

$v_{i}$ = air speed at inlet

$p_{s}$ = specific power input

H and H' = Elevation of inlet and outlet

Now, if

$v_{i} = 0$ and H = H'

Then the above eqn reduces to:

$h_{i} + gH + Q = h_{o} + \frac{v_{o}^{2}}{2} + gH + p_{s}$

$Q = h_{o} - h_{i} + \frac{v_{o}^{2}}{2} + p_{s}$ (1)

Also,

$p_{s} = \frac{P_{i}}{ mass, m}$

Area of cross-section, A = $\frac{\pi D^{2}}{4} =\frac{\pi 0.1^{2}}{4} = 7.85\times 10^{- 3} m^{2}$

Specific Volume at outlet, $V_{o} = A\times v_{o} = 7.85\times 10^{- 3}\times 20 = 0.157 m^{3}/s$

From the eqn:

$P_{o}V_{o} = mRT_{o}$

$m = \frac{800\times 10^{3}\times 0.157}{287\times 473} = 0.925 kg/s$

Now,

$p_{s} = \frac{400\times 10^{3}}{0.925} = 432.432 kJ/kg$

Also,

$\Delta h = h_{o} - h_{i} = c_{p}\Delta T =c_{p}(T_{o} - T_{i}) = 1.005(200 - 20) = 180.9 kJ/kg$

Now, using these values in eqn (1):

$Q = 180.9 + \frac{20^{2}}{2} + 432.432 = 813.33 kW$

Now, rate of heat transfer, q:

q = mQ = $0.925\times 813.33 = 752.33 kW$

4 0

3 years ago

A full-adder is a combinational circuit that forms the arithmetic sum of three input bits.

Vinvika [58]

(b) correct it is false

5 0

2 years ago

If the car passes point A with a speed of 20 m>s and begins to increase its speed at a constant rate of at = 0.5 m>s 2 , d

sattari [20]

Answer:

a = 1.68m/ $S^{2}$

Explanation:

Please kindly find the attached file for explanations

3 0

3 years ago