Answer:
5.70 m
Explanation:
3sₐ + sB = L
3∆sₐ = - ∆sB
3vₐ = -vB
3*3 = -vB
vB = -9 ms⁻²
T₁ + V₁ = T₂ + V₂
(0 + 0) + (0 + 0) = 1/2(20)(3)² + 1/2(30)(-9)² + 20*9.81*(sB/3) + 30*9.81* sB
0 = 90 + 1215 - 228.9sB
-1305 = - 228.9sB
sB = 1305/228.9 = 5.70 m
What information in drawing's title block identifies the project?
1 answer:
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Answer:
......................................................
Answer:
q = 0.1086 micro Coulombs
Explanation:
By Coulombs law, we have;
Where;
F = The electric force = 120 mgm
q₁ and q₂ = Charge
r = The separating distance = 30 cm = 0.3 m
k = 8.9876×10⁹ kg·m³/(s²·C²)
Where, q₁ and q₂, we have;
Whereby the force is the force of 120 milligram mass, we have;
0.00012 × 9.81 = 000011772 N
Substituting the values, we have;
q = 0.1086 micro Coulombs.