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nadezda [96]
3 years ago
9

You are asked to construct a mobile with four equal m = 141 kg masses, and three light rods of negligible mass and equal lengths

. The rods are of length 55 cm. (a) At what location on the level 1 rod should the free end of rod 2 be attached?
Physics
1 answer:
GarryVolchara [31]3 years ago
8 0

Answer:

The free end must be attached at a distance of 27.5 cm

Solution:

Mass, m = 141 kg

Length of the rods, L= 55 cm

Now,

As clear from fig. 1:

The free end of the rod 2 must be attached at:

F = 2 W

WL = W(55 - L)

2L = 55

L = 27.5 cm

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Identical isolated conducting spheres 1 and 2 have equal charges and are separated by a distance that is large compared with the
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Answer:

F = 0.1575 N

Explanation:

When the third sphere touches the first sphere, the charge is distributed between both spheres, then now the first sphere has only half of his original charge.

In this moment then

Sphere one has a charge = Q/2

Sphere three has a charge = Q/2

Now when the third sphere touches the second sphere again the charge is distributed in a manner that both sphere has the same charge.

How the total charge is Q = Q/2 + Q = 3/2Q, when the spheres are separated each one has 3/4Q

Sphere two has a charge = 3/4Q

Sphere three has a charge = 3/4Q

The electrostatic force that acts on sphere 2 due to sphere 1 is:

F = \frac{kQ_{1}Q_{2} }{r^{2} }

F= \frac{K(Q/2)(3Q/4)}{r^{2} }

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during a crash, the acceleration was less than 30g. Calculate the force on a 70 kg person accelerating at this rate.
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Answer:

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Explanation:

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