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3 years ago

8

A seesaw has an irregularly distributed mass of 30 kg, a length of 3.0 m, and a fulcrum beneath its midpoint. It is balanced whe

n a 60-kg person sits on one end and a 78-kg person sits on the other end.
Required:
Find a displacement of the center of mass of the system relatively to the seesaw's midpoint.

1 answer:

gladu [14]3 years ago

4 0

Answer:

x = 0.9 m

Explanation:

For this exercise we must use the rotational equilibrium relation, we will assume that the counterclockwise rotations are positive

∑ τ = 0

60 1.5 - 78 1.5 + 30 x = 0

where x is measured from the left side of the fulcrum

90 - 117 + 30 x = 0

x = 27/30

x = 0.9 m

In summary the center of mass is on the side of the lightest weight x = 0.9 m

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Part c)

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Explanation:

Part a)

As we know that charge density is the ratio of total charge and total volume

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Part b)

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$E_2 = 7992 N/C$

electric field at r = 20 cm

$E_3 = \frac{(9\times 10^9)(71 nC)}{0.20^2}$

$E_3 = 15975 N/C$

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