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nadya68 [22]
2 years ago
14

At 1 atm, nitrogen (N2) molecules become a liquid below 77 K. Thus the vapor pressure of liquid nitrogen is 1 atm at 77 K. Use t

his information to estimate the binding energy of a nitrogen molecule within the liquid. Note: You may assume that the equilibrium condition for this system is NG/V = nre-A/KT', with ny =1.93 x 1031 m-3, and that the nitrogen gas is ideal.
a. 0.081 eV
b. 81 eV
c. 0.0081 eV
d. 0.81 eV
e. 8.1 eV
Engineering
1 answer:
Bogdan [553]2 years ago
6 0
What did he sayyyyyyy
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A 1-w, 350-ω resistor is connected to 24 v. Is this resistor operating within its power rating?
seraphim [82]

Answer:

No.

Explanation:

P_r = Power rating = 1 W

R = Resistance = 350\ \Omega

V = Voltage = 24\ \text{V}

Power is given by

P=\dfrac{V^2}{R}\\\Rightarrow P=\dfrac{24^2}{350}\\\Rightarrow P=1.65\ \text{W}

1.65\ \text{W}>1\ \text{W}

So

P>P_r

Hence, the resistor is not operating within its power rating.

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2 years ago
Copper spheres of 20-mm diameter are quenched by being dropped into a tank of water that is maintained at 280 K . The spheres ma
Ivenika [448]

Answer:

The height of the water is 1.25 m

Explanation:

copper properties are:

Kc=385 W/mK

D=20x10^-3 m

gc=8960 kg/m^3

Cp=385 J/kg*K

R=10x10^-3 m

Water properties at 280 K

pw=1000 kg/m^3

Kw=0.582

v=0.1247x10^-6 m^2/s

The drag force is:

F_{D} =\frac{1}{2} Co*p_{w} A*V^{2}

The bouyancy force is:

F_{B} =V*p_{w} *g

The weight is:

W=V*p_{c} *g

Laminar flow:

v_{T} =\frac{p_{c}-p_{w}*g*D^{2}   }{18*u} =\frac{(8960-1000)*9.8*(20x10^{-3})^{2}  }{18*0.00143} =1213.48 m/s

Reynold number:

Re=\frac{1000*1213.48*20x10^{-3} }{0.00143} \\Re>>1

Not flow region

For Newton flow region:

v_{T} =1.75\sqrt{(\frac{p_{c}-p_{w}  }{p_{w} })gD }=1.75\sqrt{(\frac{8960-1000}{1000} )*9.8*20x10^{-3} }  =2.186m/s

Re=\frac{1000*2.186*20x10^{-3} }{0.00143} =30573.4

Pr=\frac{\frac{u}{p} }{\frac{K}{pC_{p} } } =\frac{u*C_{p} }{k} =\frac{0.0014394198}{0.582} =10.31

Nu=2+(0.4Re^{1/2} +0.06Re^{2/3} )Pr^{2/5} (u/us)^{1/4} \\Nu=2+(0.4*30573.4^{1/2}+0.06*30573.4^{2/3}  )*10.31^{2/5} *(0.00143/0.00032)^{1/4} \\Nu=476.99

Nu=\frac{h*d}{K_{w} } \\h=\frac{476.99*0.582}{20x10^{-3} } =13880.44W/m^{2} K

\frac{T-T_{c} }{T_{w}-T_{c}  } =e^{-t/T} \\T=\frac{m_{c}C_{p}  }{hA_{c} } =\frac{8960*10x10^{-3}*385 }{13880.44*3} =0.828 s

e^{-t/0.828} =\frac{320-280}{360-280} \\t=0.573\\heightofthewater=2.186*0.573=1.25m

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Which of the following is NOT true about drawings and specifications?
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