Class 11th
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New answer posted
2 months agoContributor-Level 9
Using Conservation of Mechanical Energy at point-A and at point-B, we can write
K_B = U_A - U_B [Since K_A = 0]
⇒ (1/2)mv_B² = mg (h_A - h_B)
⇒ v_B = √ (2 * 10 * (10 - 5) = 10m/s
New answer posted
2 months agoContributor-Level 9
For Wire-A
F / (πr_A²) = Y (l_A / 2) . (1)
For Wire-B
F / (πr_B²) = Y (l_B / 4) . (2)
From equations (1) and (2), we can write
(l_A / (2r_A²) = (l_B / (4r_B²) ⇒ l_B/l_A = 2 (r_B/r_A)² ⇒ x = l_B = 32
New answer posted
2 months agoContributor-Level 9
Angle from direction of flow = 90° + θ = 120°
⇒ θ = 30°
sin 30° = u/v ⇒ u/10 = 1/2 ⇒ u = 5 m/s
New answer posted
2 months agoContributor-Level 10
The reaction is: FeCl? + 3H? C? O? + 3KOH → K? [Fe (C? O? )? ] (A) + 3HCl + 3H? O
In the complex [Fe (C? O? )? ]³? , the oxalate ion (C? O? )²? is a bidentate ligand.
There are three bidentate ligands, so the coordination number, or secondary valency, of Fe is 3 * 2 = 6.
New answer posted
2 months agoContributor-Level 10
The cell constant (G) is given by G = κ * R, where κ is conductivity and R is resistance.
Since the cell constant is constant: (κ * R)KCl = (κ * R)HCl
0.14 Sm? ¹ * 4.19 Ω = κ_HCl * 1.03 Ω
κ_HCl = (0.14 * 4.19) / 1.03 = 0.569 Sm? ¹
This is equivalent to 56.9 * 10? ² Sm? ¹.
The answer, rounded off, is 57.
New answer posted
2 months agoContributor-Level 9
Apply conservation of momentum along y-axis, we can write
10v? - 10v? sin 30° = 0
⇒ v? = 20m/s
New answer posted
2 months agoContributor-Level 9
Deceleration, a = u² / 2S = 10² / (2 * 0.5) = 100 m/s².
Retarding force, F = MA = 0.1 * 100 = 10N
New answer posted
2 months agoContributor-Level 10
-SO? H acts as a cation exchanger.
-NH? acts as an anion exchanger.
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