do not obey ohm's law so it's a I believe
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Answer:
Friction force on the bullet is 58.7 N opposite to its velocity
Explanation:
As we know that initial speed of the bullet is 55 m/s
after travelling into the sand bag by distance d = 1.34 m it comes to rest
so final speed

now we can use kinematics top find the acceleration of the bullet

so we have


now by Newton's II law we know that

so we have


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