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Gelneren [198K]
2 years ago
13

Compare and contrast mechanical properties of plastics, metals and ceramics.

Engineering
1 answer:
IgorLugansk [536]2 years ago
6 0

Answer:

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Explanation:

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Answer:

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5 0
2 years ago
Everyone why are you reporting my answers i didnt do anything to you
masya89 [10]

Answer:

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Explanation:

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5 0
3 years ago
Read 2 more answers
A circuit has a source voltage of 15V and two resistors in series with a total resistance of 4000Ω .If RI has a potential drop o
anastassius [24]

Answer:

1500Ω

Explanation:

Given data

voltage = 15 V

total Resistance = 4000Ω

potential drop V = 9.375 V

To find out

R2

Solution

we know R1 +R2 = 4000Ω

So we use here Ohm's law to find out current I

current = voltage / total resistance

I = 15 / 4000 = 3.75 × 10^{-3} A

Now we apply Kirchhoffs Voltage Law for find out R2

R2 = ( 15 - V ) / current

R2 = ( 15 - 9.375 ) / 3.75 × 10^{-3}

R2 = 1500Ω

6 0
3 years ago
A particle, originally at rest and located at point (3 ft, 2 ft, 5 ft), is subjected to an acceleration a = {6ti + 12t^2k} ft/s^
Sindrei [870]

Answer:

The particle's position at t = 2 is

r = (11ft, 2ft, 21ft) = (11, 2, 21) ft

Explanation:

r₀ = (3, 2, 5) ft

a = (6t, 0, 12t²) ft/s²

a = dv/dt

dv/dt = (6tî + 0j + 12t²ķ)

dv = (6tî + 0j + 12t²ķ) st

Integrating the left hand side from 0 to v (the particle was originally at rest) and the right hand side from 0 to t,

We obtain,

v = (3t²î + 0j + 4t³ķ) ft/s

v = dr/dt

dr/dt = (3t²î + 0j + 4t³ķ)

dr = (3t²î + 0j + 4t³ķ) st

Integrating the left hand side from r₀ (the original position) to r and the right hand side from 0 to t,

r - r₀ = (t³î + 0j + t⁴ķ) ft

r = (t³î + 0j + t⁴ķ) + r₀

r = (t³î + 0j + t⁴ķ) + (3î + 2j + 5ķ)

At t=2s, t³ = 8 and t⁴ = 16

r = (8î + 0j + 16ķ) + (3î + 2j + 5ķ)

r = (11î + 2j + 21ķ) ft

r = (11ft, 2ft, 21ft)

Hope this helps!

4 0
3 years ago
Refrigerant-134a is to be cooled by water in a condenser. The refrigerant enters
KonstantinChe [14]

Answer:

Refrigerant R-134a is to be cooled by waterin a condenser.The refrigerant enters thecondenser with a mass flow rate of 6 kg/minat 1 MPa and 70 C and exits at 35 C. The cool-ing water enters at 300 kPa and 15 C andleaves at 25 C. Neglecting pressure drops,determine a) the required mass flow rate ofthe cooling water, and b) the heat transferrate from the refrigerant to the water.SolutionFirst consider the condenser as the control volume. The process is steady,adiabatic and no work is done. Thus over any time intervalΔt,ΔEΔt=0and thusXin˙E=Xout˙Ewhere˙E=˙m h+12V2+gz650:351 Thermodynamics·Prof. Doyle Knight37

Background image

I only know this

sorry for errors

4 0
2 years ago
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