Answer:
assume nitrogen is an ideal gas with cv=5R/2
assume argon is an ideal gas with cv=3R/2
n1=4moles
n2=2.5 moles
t1=75°C <em>in kelvin</em> t1=75+273
t1=348K
T2=130°C <em>in kelvin</em> t2=130+273
t2=403K
u=пCVΔT
U(N₂)+U(Argon)=0
<em>putting values:</em>
=>4x(5R/2)x(Tfinal-348)=2.5x(3R/2)x(T final-403)
<em>by simplifying:</em>
Tfinal=363K
True........don't blame me if i'm wrong.<span />
The correct answer is
- +6q is the charge induced on the interior surface
- -6q is the charge induced on the exterior surface
A uniformly distributed negatively charged -q will be induced on the cavity's inner surface and a positively charged +q on its outer surface. Due to inner charges +q and -q, point P will not have a field or potential. Only the outer +q charge will be responsible for P's potential.
An atom is said to be electrically neutral when its total charge is zero. Protons, which are positively charged, electrons, which are negatively charged, and neutrons, which are not positively or negatively charged, make up an atom. An atom will be electrically neutral if it has an equal amount of protons and electrons since the charge from each particle is of equal intensity. Electrical neutrality is the absence of any net electrical charge. An atom will be electrically neutral if it has an equal amount of protons and electrons since the charge from each particle is of equal intensity.
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Complete question:
Suppose a charge -6q is suspended inside a cavity in an electrically neutral cylindrical conductor. What is the charge induced on the exterior surface?