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4 years ago

15

Glass-ceramic is a fine-grained crystalline ceramic formed by the controlled crystallization of a ceramic. ( True , False )

1 answer:

koban [17]4 years ago

5 0

The answer to the question is true

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A pin must be inserted into a collar of the same steel using an expansion fit. The coefficient of thermal expansion of the metal

nirvana33 [79]

Answer:

a) the temperature to which the pin must be cooled for assembly is $T_2 = -101.89^ \ ^0}C$

b) the radial pressure at room temperature after assembly is $P_f = 62.8 \ MPa$

c) the safety factor in the resulting assembly = 6.4

Explanation:

Coefficient of thermal expansion $\alpha = 12.3*10^{-6} \ ^0 C$

Yield strength $\sigma_y$ = 400 MPa

Modulus of elasticity (E) = 209 GPa

Room Temperature $T_1$ = 20°C

outer diameter of the collar $D_o = 95 \ mm$

inner diameter of the collar $D_i = 60 \ mm$

pin diameter $D_p$ = $60.03 \ mm$

Clearance c = 0.06 mm

a)

The temperature to which the pin must be cooled for assembly can be calculated by using the formula:

$(D_i - c )-D_p = \alpha * D_p(T_2-T_1)$

$(60-0.06)-60.03=12.3*10^{-6}*60.03(T_{2}-20^{0}C)$

-0.09 = $7.38369*10^{-4}$ $(T_{2}-20^{0}C)$

-0.09 = $7.38369*10^{-4}T_2$ $\ \ - \ \ 0.01476738$

$-0.09 + 0.01476738$ = $7.38369*10^{-4}T_2$

−0.07523262 = $7.38369*10^{-4}T_2$

$T_2 = \frac{-0.07523262}{7.38369*10^{-4}}$

$T_2 = -101.89^ \ ^0}C$

b)

To determine the radial pressure at room temperature after assembly ;we have:

$P_f = \frac{E * (D_p-D_i)(D_o^2-D_1^2)}{D_i*D_o} \\ \\ \\ P_f = \frac{209*10^9* 0.03(95^2-60^2)}{60*95^2} \\ \\ P_f = 62815789.47 \ Pa \\ \\ P_f = 62.8 \ MPa$

c) the safety factor of the resulting assembly is calculated as:

safety factor = $\frac{Yield \ strength }{walking \ stress}$

safety factor = $\frac{400}{62.8}$

safety factor = 6.4

Thus, the safety factor in the resulting assembly = 6.4

4 0

4 years ago

Consider a 0.15-mm-diameter air bubble in a liquid. Determine the pressure difference between the inside and outside of the air

777dan777 [17]

Answer:

a) $\Delta P= 2666.66 Pa$

b) $\Delta P= 3200 Pa$

Explanation:

Given that

Diameter ,d= 0.15 mm

We know that pressure difference is given as

$\Delta P=\dfrac{4\sigma }{d}$

Now by putting the values

When surface tension 0.1 N/m :

The surface tension , $\sigma=0.1\ N/m$

$\Delta P=\dfrac{4\times 0.1 }{0.15}\times 10^3\ Pa$

$\Delta P= 2666.66 Pa$

When surface tension 0.12 N/m :

The surface tension , $\sigma=0.12\ N/m$

$\Delta P=\dfrac{4\times 0.12 }{0.15}\times 10^3\ Pa$

$\Delta P= 3200 Pa$

a) $\Delta P= 2666.66 Pa$

b) $\Delta P= 3200 Pa$

4 0

4 years ago

The amplitudes of the displacement and acceleration of an unbalanced motor were measured to be 0.15 mm and 0.6 g, respectively.

PilotLPTM [1.2K]

Answer:

N=945.76 RPM

Explanation:

Given that

A= 0.15 m

Acceleration = 0.6 g

a=0.6 x 9.81 m/s²

a= 5.886 m/s²

We know that acceleration a given as

a = ω² A

ω=Angular speed

$\omega=\sqrt{\dfrac{0.6\times 9.81}{0.6\times 10^{-3}}}$

ω=99.04 rad/s

We know that

$\omega=\dfrac{2\pi N}{60}\\\\N=\dfrac{60\times \omega}{2\pi }$

$N=\dfrac{60\times 99.04}{2\pi }$

N=945.76 RPM

Therefore the speed of the motor will be 945.76 RPM.

6 0

4 years ago

What are the dimensions of the base of the pyramid?

Cloud [144]

Answer:

b

Explanation:

bcus it is

7 0

3 years ago

HELP! ASAP SOS WILL MARK BRAINLIEST!!!

liq [111]

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7 0

3 years ago