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Yuki888 [10]
3 years ago
8

If the surface energy of a magnesium oxide - nickel oxide (MgO-NiO) solid solution is 1.05 J/m2 and its elastic modulus is 198 G

Pa, what is the maximum length of an internal elliptical flaw, in mm, if the ceramic part must survive an applied tensile stress of 27 MPa normal to this elliptical flaw
Engineering
1 answer:
Nastasia [14]3 years ago
7 0

Answer:

The maximum length is 3.897×10^-5 mm

Explanation:

Extension = surface energy/elastic modulus

surface energy = 1.05 J/m^2

elastic modulus = 198 GPa = 198×10^9 Pa

Extension = 1.05/198×10^9 = 5.3×10^-12 m

Strain = stress/elastic modulus = 27×10^6/198×10^9 = 1.36×10^-4

Length = extension/strain = 5.3×10^-12/1.36×10^-4 = 3.897×10^-8 m = 3.897×10^-8 × 1000 = 3.897×10^-5 mm

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Explain the conductivity results observed for ionic compounds in the solid state and in aqueous solution.i.Solid ionic compounds
Svetach [21]

Answer:

Aqueous solution of ionic compounds conduct electricity while solid ionic compounds don't.

Explanation:

Ionic compound conduct electricity when liquid or in aqueous solution that is resolved in water because the ionic bonds of the compound become weak and the ions are free to move from place to place.

Ionic compounds don't conduct electricity while in solid state because the ionic bonds are to strong and ions cannot move around with lack of space for movement which makes the electric conductivity zero.

8 0
3 years ago
The 150-lb man sits in the center of the boat, which has a uniform width and a weight per linear foot of 3 lb>ft. Determine t
irina1246 [14]

Answer:

M = 281.25 lb*ft

Explanation:

Given

W<em>man</em> = 150 lb

Weight per linear foot of the boat: q = 3 lb/ft

L = 15.00 m

M<em>max</em> = ?

Initially, we have to calculate the Buoyant Force per linear foot (due to the water exerts a uniform distributed load upward on the bottom of the boat):

∑ Fy = 0  (+↑)     ⇒    q'*L - W - q*L = 0

⇒       q' = (W + q*L) / L

⇒       q' = (150 lb + 3 lb/ft*15 ft) / 15 ft

⇒       q' = 13 lb/ft   (+↑)

The free body diagram of the boat is shown in the pic.

Then, we apply the following equation

q(x) = (13 - 3) = 10   (+↑)

V(x) = ∫q(x) dx = ∫10 dx = 10x   (0 ≤ x ≤ 7.5)

M(x) = ∫10x dx = 5x²  (0 ≤ x ≤ 7.5)

The maximum internal bending moment occurs when x = 7.5 ft

then

M(7.5) = 5(7.5)² = 281.25 lb*ft

8 0
3 years ago
Consider a N-channel enhancement MOSFET with VGS = 3V, Vt = 1 V, VDS = 10 V, and lambda =0 (channel length modulation parameter)
AveGali [126]

The current IDS is greater than 0 since the VGS has induced an inversion layer and the transistor is operating in the saturation region.

<u>Explanation:</u>

  • Since V_{ds} > V_{gs} - Vt because V_{gs} > Vt.
  • By the saturation region the MOSFET is operating.
  • A specific source voltage and gate of NMOS, the voltage get drained during the specific level, the drain voltage is rises beyond where there is no effect of current during saturated region.
  • MOSFET is a transistor which is a device of semiconductor vastly used for the electronic amplifying signals and switching in the devices of electronics.
  • The core of this is integrated circuit.
  • It is fabricated and designed in an individual chips due to tiny sizes.
7 0
3 years ago
What is soil texture in civil engineering​
Bond [772]

Answer:

It will indicate the percentage of sand, silt, and clay present in the soil.

Explanation:

4 0
3 years ago
Air with a mass flow rate of 2.3 kg/s enters a horizontal nozzle operating at steady state at 450 K, 350 kPa, and velocity of 3
viktelen [127]

Answer:

Given that

Mass flow rate ,m=2.3 kg/s

T₁=450 K

P₁=350 KPa

C₁=3 m/s

T₂=300 K

C₂=460 m/s

Cp=1.011 KJ/kg.k

For ideal gas

P V = m R T

P = ρ RT

\rho_1=\dfrac{P_1}{RT_1}

\rho_1=\dfrac{350}{0.287\times 450}

ρ₁=2.71 kg/m³

mass flow rate

m= ρ₁A₁C₁

2.3 = 2.71 x A₁ x 3

A₁=0.28 m²

Now from first law for open system

h_1+\dfrac{C_1^2}{200}+Q=h_2+\dfrac{C_2^2}{2000}

For ideal gas

Δh = CpΔT

by putting the values

1.011\times 450+\dfrac{3^2}{200}+Q=1.011\times 300+\dfrac{460^2}{2000}

Q=1.011\times 300+\dfrac{460^2}{2000}-\dfrac{3^2}{200}-1.011\times 450

Q= - 45.49 KJ/kg

Q =- m x 45.49 KW

Q= - 104.67 KW

Negative sign indicates that heat transfer from air to surrounding

4 0
3 years ago
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