Answer:
Radiation
Explanation:
The fire from the burning house is not directly touching the house. Also not convection because there is not water involved
Answer:
so the answer is this because the answer is that
Explanation:
and the reason why the answer is this and that is because the answer is that
Answer:
<em>a. 4.21 moles</em>
<em>b. 478.6 m/s</em>
<em>c. 1.5 times the root mean square velocity of the nitrogen gas outside the tank</em>
Explanation:
Volume of container = 100.0 L
Temperature = 293 K
pressure = 1 atm = 1.01325 bar
number of moles n = ?
using the gas equation PV = nRT
n = PV/RT
R = 0.08206 L-atm-![mol^{-1}](https://tex.z-dn.net/?f=mol%5E%7B-1%7D)
![K^{-1}](https://tex.z-dn.net/?f=K%5E%7B-1%7D)
Therefore,
n = (1.01325 x 100)/(0.08206 x 293)
n = 101.325/24.04 = <em>4.21 moles</em>
The equation for root mean square velocity is
Vrms = ![\sqrt{\frac{3RT}{M} }](https://tex.z-dn.net/?f=%5Csqrt%7B%5Cfrac%7B3RT%7D%7BM%7D%20%7D)
R = 8.314 J/mol-K
where M is the molar mass of oxygen gas = 31.9 g/mol = 0.0319 kg/mol
Vrms =
= <em>478.6 m/s</em>
<em>For Nitrogen in thermal equilibrium with the oxygen, the root mean square velocity of the nitrogen will be proportional to the root mean square velocity of the oxygen by the relationship</em>
= ![\sqrt{\frac{Mnit}{Moxy} }](https://tex.z-dn.net/?f=%5Csqrt%7B%5Cfrac%7BMnit%7D%7BMoxy%7D%20%7D)
where
Voxy = root mean square velocity of oxygen = 478.6 m/s
Vnit = root mean square velocity of nitrogen = ?
Moxy = Molar mass of oxygen = 31.9 g/mol
Mnit = Molar mass of nitrogen = 14.00 g/mol
= ![\sqrt{\frac{14.0}{31.9} }](https://tex.z-dn.net/?f=%5Csqrt%7B%5Cfrac%7B14.0%7D%7B31.9%7D%20%7D)
= 0.66
Vnit = 0.66 x 478.6 = <em>315.876 m/s</em>
<em>the root mean square velocity of the oxygen gas is </em>
<em>478.6/315.876 = 1.5 times the root mean square velocity of the nitrogen gas outside the tank</em>
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Answer:
a) v, v
b) 2mv^2
c) Elastic collion
Explanation:
(a) The velocity of the second particle after the collision is (v2x,v2y)=(v,−v). From momentum conservation in x-direction
Here x, y represent direction.They are not variable. 1 and 2 represent before and after.
2vm=v1xm+v2xm, we find v1x=v.
From momentum conservation in y-direction
0 =v1ym+v2ym, we findv1y=v.
(b) By energy conservation principle
Before: K=1/2m(2v)^2=2mv^2.
After: K=1/2m(v^2(1x)+v^2(1y))+12m(v22x+v22y)=2mv^2
(c) The collision is elastic