Answer:
write the equation of motion go over the centre of mass
Explanation:
the center of mass of a distribution of mass in space (sometimes referred to as the balance point) is the unique point where the weighted relative position of the distributed mass sums to zero. This is the point to which a force may be applied to cause a linear acceleration without an angular acceleration.
Answer:
the distance from charge A to C is r₁₃= 1.216 m
Explanation:
following Coulomb's law , the force exerted by 2 point charges between themselves is:
F= k*q₁*q₂/r₁₂² , where q is charge , r is distance and 1 and 2 represents the charge A and charge B respectively , k=constant
since C ( denoted as 3) is at equilibrium
F₁₃=F₂₃
k*q₁*q₃/r₁₃²=k*q₂*q₃/r₂₃²
q₁/r₁₃²=q₂/r₂₃²
r₁₃²/q₁=r₂₃²/q₂
r₂₃=r₁₃*√(q₂/q₁)
since C is at rest and is co linear with A and B ( otherwise it would receive a net force in either vertical or horizontal direction) , we have
r₁₃+r₂₃=d=r₁₂
r₁₃+r₁₃*√(q₂/q₁)=d
r₁₃*(1+√(q₂/q₁))=d
r₁₃=d/(1+√(q₂/q₁))
replacing values
r₁₃=d/(1+√(q₂/q₁)) = 3.00 m/(1+√(3.10 C/1.44 C)) = 1.216 m
thus the distance from charge A to C is r₁₃= 1.216 m
125 because f=ma so you would use 100=mx0.75
Answer:
the location of the center of gravity for the entire body is 1.08 m
Explanation:
Given the data in the question;
w1 = 458 N, y1 = 1.34 m
w2 = 120 N, y2 = 0.766 m
w3 = 89.8 N, y2 = 0.204 m
The location arrangement of the body part is vertical, locate the overall centre of gravity by simply replacing the horizontal position x by the vertical position y as measured relative to the floor.
so,
= (w1y1 + w2y2 + w3y3 ) / ( w1 + w2 + w3 )
so we substitute in our values
= (458×1.34 + 120×0.766 + 89.8×0.204 ) / ( 458 + 120 + 89.8 )
= 723.9592 / 667.8
= 1.08 m
Therefore, the location of the center of gravity for the entire body is 1.08 m
