Here as we can see that weight on the right side of ruler given as D is applying torque about the fulcrum this is counter balanced by weight on left side given as A
now by torque balance we can say
now we will have
now here as per given setup we can see that
so here we will have
so the weight required to hold D will be more and correct answer will be
<em>The setup increases the amount of force required to lift the block and, therefore, decreases your mechanical advantage.</em>
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<h3>During a chemical reaction,one ore more new substances forwed</h3>
Answer:
v = 1.08 m/s
Explanation:
What is the linear speed of the 0.0500-kg sphere as its passes through its lowest point?
The decrease in PE is
d = 80.0cm * 1 / 1000m = 0.80m
h = 0.80 m /2 = 0.40 m
ΔPE = m*g*h
ΔPE = (0.0500 - 0.0200)kg * 9.8m/s² * 0.400 m
ΔPE = 0.1176 J
The moment of inertia of the assembly is
I = 1/12*m*L² + (m1 + m2)*(L/2)²
I = 1/12*0.390kg*(0.800m)² + 0.0700kg*(0.400m)²
I = 0.032 kg·m²
KE = ½Iω²
0.1176 J = ½ * 0.032kg·m² * ω²
ω = 2.71 rad/s
v = ωr = 2.71 rad/s * 0.400m
The linear velocity
v = 1.08 m/s
F)
size , texture , and color hope this helps
