Ask question

Ask question

All categories

3 years ago

7

Not a characteristic property of ceramic material (a) high temperature stability (b) high mechanical strength (c) low elongation

(d) low hardness

2 answers:

love history [14]3 years ago

6 0

Answer:

d. low hardness

Explanation:

The hardness is the resistance to penetration. Low hardness is not a characteristic property of ceramic material.

Nostrana [21]3 years ago

3 0

Low hardness is not a characteristic property of ceramic material.

Answer: Option D

<u>Explanation: </u>

One of the most important properties of ceramic material is the hardness that the material displays. The hardness owes to the joining of brittle fracture and plastic flow that makes the material to defend against penetration.

The hardness can be tested with the Vickers test. The hardness of ceramic makes its dominance in the use of construction purpose and manufacture of products. Other general properties are less conductivity, high melting temperature, etc.

You might be interested in

What are some "vital signs" that we consider to tell us about the economy?

Wittaler [7]

Explanation:

if there is a deficit or a surplus

if the budget is good

7 0

3 years ago

Read 2 more answers

ممكن الحل ............

Roman55 [17]

Answer:

i dont understand

Explanation:

4 0

3 years ago

Find the time-domain sinusoid for the following phasors:_________

sattari [20]

<u>Answer</u>:

a. r(t) = 6.40 cos (ωt + 38.66°) units

b. r(t) = 6.40 cos (ωt - 38.66°) units

c. r(t) = 6.40 cos (ωt - 38.66°) units

d. r(t) = 6.40 cos (ωt + 38.66°) units

<u>Explanation</u>:

To find the time-domain sinusoid for a phasor, given as a + bj, we follow the following steps:

(i) Convert the phasor to polar form. The polar form is written as;

r∠Ф

Where;

r = magnitude of the phasor = $\sqrt{a^2 + b^2}$

Ф = direction = tan⁻¹ ( $\frac{b}{a}$ )

(ii) Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid (r(t)) as follows:

r(t) = r cos (ωt + Φ)

Where;

ω = angular frequency of the sinusoid

Φ = phase angle of the sinusoid

(a) 5 + j4

<em>(i) convert to polar form</em>

r = $\sqrt{5^2 + 4^2}$

r = $\sqrt{25 + 16}$

r = $\sqrt{41}$

r = 6.40

Φ = tan⁻¹ ( $\frac{4}{5}$ )

Φ = tan⁻¹ (0.8)

Φ = 38.66°

5 + j4 = 6.40∠38.66°

(ii) <em>Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid</em>

r(t) = 6.40 cos (ωt + 38.66°)

(b) 5 - j4

<em>(i) convert to polar form</em>

r = $\sqrt{5^2 + (-4)^2}$

r = $\sqrt{25 + 16}$

r = $\sqrt{41}$

r = 6.40

Φ = tan⁻¹ ( $\frac{-4}{5}$ )

Φ = tan⁻¹ (-0.8)

Φ = -38.66°

5 - j4 = 6.40∠-38.66°

(ii) <em>Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid</em>

r(t) = 6.40 cos (ωt - 38.66°)

(c) -5 + j4

<em>(i) convert to polar form</em>

r = $\sqrt{(-5)^2 + 4^2}$

r = $\sqrt{25 + 16}$

r = $\sqrt{41}$

r = 6.40

Φ = tan⁻¹ ( $\frac{4}{-5}$ )

Φ = tan⁻¹ (-0.8)

Φ = -38.66°

-5 + j4 = 6.40∠-38.66°

(ii) <em>Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid</em>

r(t) = 6.40 cos (ωt - 38.66°)

(d) -5 - j4

<em>(i) convert to polar form</em>

r = $\sqrt{(-5)^2 + (-4)^2}$

r = $\sqrt{25 + 16}$

r = $\sqrt{41}$

r = 6.40

Φ = tan⁻¹ ( $\frac{-4}{-5}$ )

Φ = tan⁻¹ (0.8)

Φ = 38.66°

-5 - j4 = 6.40∠38.66°

(ii) <em>Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid</em>

r(t) = 6.40 cos (ωt + 38.66°)

3 0

3 years ago

Xyxyydfufggivivihogcufuf

Genrish500 [490]

Answer:

ummm why is you doing this

Explanation:

It doesnt make sense.

7 0

3 years ago

For which of 'water' flow velocities at 200C can we assume that the flow is incompressible ? a.1000 km per hour b. 500 km per ho

ad-work [718]

Answer:d

Explanation:

Given

Temperature $=200^{\circ}\approc 473 K$

Also $\gamma for air=1.4$

R=287 J/kg

Flow will be In-compressible when Mach no.<0.32

Mach no. $=\frac{V}{\sqrt{\gamma RT}}$

(a) $1000 km/h\approx 277.78 m/s$

Mach no. $=\frac{277.78}{\sqrt{1.4\times 287\times 473}}$

Mach no.=0.63

(b) $500 km/h\approx 138.89 m/s$

Mach no. $=\frac{138.89}{\sqrt{1.4\times 287\times 473}}$

Mach no.=0.31

(c) $2000 km/h\approx 555.55 m/s$

Mach no. $=\frac{555.55}{\sqrt{1.4\times 287\times 473}}$

Mach no.=1.27

(d) $200 km/h\approx 55.55 m/s$

Mach no. $=\frac{55.55}{\sqrt{1.4\times 287\times 473}}$

Mach no.=0.127

From above results it is clear that for Flow at velocity 200 km/h ,it will be incompressible.

5 0

3 years ago