Answer:
b= 2.14 m
Explanation:
Given that
Weight of the board ,wt = 40 N
Wight of the first children , wt₁=500 N
Weight of the second children ,wt₂ = 350 N
The distance of the 500 N child from center ,a= 1.5 m
lets take distance of the 350 N child from center = b m
Now by taking the moment about the center of the board
We know that moment = Force x Perpendicular distance from the force
wt₁ x a = wt₂ x b
500 x 1.5 = 350 x b
Therefore the distance of the 350 N weight child from the center is 2.14 m.
Answer: direct linear
part 1
mass = ρ x V
mass = 1739 kg/m³ x 3.8 km³ = 6608.2 kg
PE (potential energy)= mgh
PE = 6608.2 kg x 9.81 x 403
PE = 2.61 x 10⁷ J
part 2
megaton of TNT (Mt) =4.2 x 10¹⁵ J
convert PE to Mt:
2.61 x 10⁷ J : 4.2 x 10¹⁵ J = 6.21 x 10⁻⁹ Mt
Answer:5
Friction is a force that opposes motion.
