Answer:
F = 6.27 x 10 ¹⁹ N
Explanation:
Given
m₁ = 92 kg, m₂ = 46 kg, % = 0.04% N = 6.022 x 10²³ Z = 18, e = 1.6 x 10 ⁻¹⁹ C, M = 0.018 kg/mol
q₁ = % * [m * N * A * e / M ]
q₁ = 0.0004 * [ ( 92 kg * 6.022 x 10²³ * 18 * 1.6 x 10 ⁻¹⁹ ) / (0.018 kg/mol ) ]
q₁ = 3.54 x 10⁶ C
q₂ = 0.0004 * [ ( 46 kg * 6.022 x 10²³ * 18 * 1.6 x 10 ⁻¹⁹ ) / (0.018 kg/mol ) ]
q₂ = 1.773 x 10⁶ C
Now to determine the electrostatic force con use the equation
F = K * q₁ * q₂ / d²
K = 8.99 x 10 ⁹
F = 8.99 x 10 ⁹ * 3.54 x 10⁶ C * 1.773 x 10⁶ C / (30m)²
F = 6.27 x 10 ¹⁹ N
Answer:
Explanation:
The change is as follows
P₁ V₁ to 3P₁, V₁ ( constt volume ) --- first process
3P₁,V₁ to 3P₁ , 5V₁ ( constt pressure ) ---- second process
In the first process Temperature must have been increased 3 times . So if initial temperature is T₁ then final temperature will be 3 T₁
P₁V₁ = n R T₁ , n is no of moles of gas enclosed.
nRT₁ = P₁V₁
Heat added at constant volume = n Cv ( 3T₁ - T₁)
= n x 5/3 R X 2T₁ ( for diatomic gas Cv = 5/3 R)
= 10/3 x nRT₁
= 10/3x P₁V₁
In the second process, Temperature must have been increased 5 times . So if initial temperature is 3T₁ then final temperature will be 15 T₁
Heat added at constant pressure in second case
= n Cp ( 15T₁ - 3T₁)
= n x 7/3 R X 12T₁ ( For diatomic gas Cp = 7/3 R)
= 28 x nRT₁
= 28 P₁V₁
Answer:

Explanation:
Given data:
mass of sphere is 10 g
Angle between string and vertical axis is 
thickness of string 300 mm = 0.3 m

r =0.3 sin 13 = 0.067 m






q = ne


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