Class 11th
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New answer posted
4 months agoContributor-Level 10
P? - P? = 4T/a
P? - P? = 4T/b
P? - P? = 4T (1/a - 1/b)
Also, P? - P? = 4T/r
4T/r = 4T (1/a - 1/b)
1/r = (b-a)/ab
r = ab / (b-a)
New answer posted
4 months agoContributor-Level 10
η = 1 - T_C / T_H
η = 1 - 400/800 = 1 - ½ = ½
η = W/Q_H ⇒ ½ = W/Q_H ⇒ Q_H = 2W = 2 * 1200 = 2400 J
New answer posted
4 months ago
Contributor-Level 10
B is least basic as lone pair of electron is present in resonance so as to make the system aromatic in nature
D is most basic as it results in formation of equivalent resonating structures upon attack of H?
Among A and C, the former is less basic as sp² hybridistion of nitrogen decrease its basic strength.
Hence option 3 follows
New answer posted
4 months agoContributor-Level 10
f = f_Translational + f_Rotational + f_Vibrational
f = 3 + 3 + 48 = 54 (Since each vibrational mode has two degrees of freedom)
γ = 1 + 2/f = 1 + 2/54 = 1 + 1/27 = 28/27 ≈ 1.03
New answer posted
4 months agoContributor-Level 10
According to Aufbau's principal, the increasing energy of atomic orbitals for sixth period element follows the order
6s < 4f < 5d < 6p
New question posted
4 months agoNew answer posted
4 months ago
Contributor-Level 10
N = mg - F_L
f_s = mv²/R ≤ μsN = μs (mg - F_L)
F_L = m (v²/μsR - g)
New answer posted
4 months agoContributor-Level 10
K? = ½ mv² + ½ Iω²
Since I =? mr² and v=rω,
K? = ½ mv² + ½ (? mr²) (v/r)² = ½ mv² +? mv² = (7/10)mv² = 140 J
K_f = 0.05 K?
(7/10)mv_f² = 0.05 * (140)
(7/10)mv_f² = 7
v_f² = 20
v_f = √20 ≈ 4.47 m/s
New answer posted
4 months agoContributor-Level 10
t? + t? = t
v? /t? = α and v? /t? = β
t? = v? /α and t? = v? /β
v? /α + v? /β = t
v? (1/α + 1/β) = t
v? = (αβt) / (α + β)
Distance traveled = Area under speed-time graph
Distance = ½ * base * height = ½ * t * v?
Distance = ½ * t * (αβt) / (α + β) = (αβt²) / (2 (α + β)
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