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New answer posted
10 months agoContributor-Level 10
1 litre, T = 300K, P = 2 atm, KE = 2*10? J/molecule, no of molecule =?
No. of molecules = (no of moles) * NA = nNA
Also, n = PV/RT = PV/ (NAkT)
KE = (3/2)kT = 2*10? J [Given]
kT = (4/3)*10?
P = 2 atm = 2 * 1.013 * 10? N/m²
vol = 1 lit = 10? ³ m³
No. of molecules = PV/kT = (2*1.013*10? * 10? ³)/ (4/3)*10? ) ≈ 1.5 * 10¹¹
New answer posted
10 months agoContributor-Level 10
Polygon law is applicable in both the situation given but the equation given in the reason is not useful in explaining the assertion.
New answer posted
10 months agoContributor-Level 10
v = 1/√με = 1/√ (µ? µ? ε? ε? ) = c/√ (µ? ε? )
v = 3*10? / √ (1*81)
v = 3*10? / 9 m/s
= 0.33 * 10? m/s
= 3.33 * 10? m/s
New answer posted
10 months agoContributor-Level 10
v = √2gh velocity of efflux.
F = v ( dm/dt ) = v (aρv) = aρv² = 2aρgh
fr = µR = µAhρg
For just sliding, for = F
µAhρg = 2aρgh
or µ = 2a/A
New answer posted
10 months agoContributor-Level 10
1/Ceq = 1/C? + 1/C? + 1/C?
1/Ceq = 1/ (K Aε? /d) + 1/ (3K Aε? /2d) + 1/ (5K Aε? /3d)
1/Ceq = d/ (K Aε? ) + 2d/ (3K Aε? ) + 3d/ (5K Aε? )
1/Ceq = (d/K Aε? ) * (1 + 2/3 + 3/5)
1/Ceq = (d/K Aε? ) * (15+10+9)/15) = 34d / (15K Aε? )
Ceq = 15K Aε? / 34d
New answer posted
10 months agoContributor-Level 10
Let vel of A and B just after collision be VA & VB respectively.
m * 9 + 0 = m * VA + 2mVB
9 = VA + 2VB
again e = (VB - VA)/ (9-0) = 1
9 = VB - VA
From (i) & (ii)
VB = 6 m/s & VA = -3 m/s
Now, for B and C collision (completely inelastic):
2m * 6 + 0 = (2m + 2m)Vc
12m = 4mVc
Vc = 3 m/s
New answer posted
10 months agoContributor-Level 9
t? /? = 3 days = 72 hours
dN/dt = λN = (ln2/t? /? ) N
= (0.693 * 6.02*10²³ * 2*10? ³) / (72 * 3600 * 198)
= 1.618 * 10¹³
= 16.18 * 10¹² disintegration/second
New answer posted
10 months agoContributor-Level 10
V = E (1 - e? /τ)
Where τ = RC = 100 * 10? = 10? sec
50V = 100V (1 - e? /10? )
1/2 = 1 - e? ¹?
1/2 = e? ¹?
-ln2 = -10? t
t = ln2/10?
t = 0.693 * 10? sec
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