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Alex73 [517]
3 years ago
8

The angular momentum about the center of the planet and the total mechanical energy will be conserved regardless of whether the

object moves from small R to large R (like a rocket being launched) or from large R to small R (like a comet approaching the earth).
O True O False
Physics
1 answer:
ValentinkaMS [17]3 years ago
5 0

Answer:

True

Explanation:

The angular momentum around the center of the planet and the total mechanical energy will be preserved irrespective of whether the object moves from large R  to small R. But on the other hand the kinetic energy of the planet will not be conserved because it can change from kinetic energy to potential energy.

Therefore the given statement is True.

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Which qualifications are typical for a Manufacturing career? Check all that apply.
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Which describes Michael Faraday’s work with electricity and magnetism?
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8 0
3 years ago
A mass weighting 16 lbs stretches a spring 3 inches. The mass is in a medium that exerts a viscous resistance of 20 lbs when the
const2013 [10]

Answer:

The equation for the object's displacement is u(t)=0.583cos11.35t

Explanation:

Given:

m = 16 lb

δ = 3 in

The stiffness is:

k=\frac{m}{\delta } =\frac{16}{3} =5.33lb/in

The angular speed is:

w=\sqrt{\frac{k}{m} } =\sqrt{\frac{5.33*386.4}{16} } =11.35rad/s

The damping force is:

F_{D} =cu

Where

FD = 20 lb

u = 4 ft/s = 48 in/s

Replacing:

c=\frac{F_{D} }{u} =\frac{20}{48} =0.42lbs/in

The critical damping is equal:

c_{c} =\frac{2k}{w} =\frac{2*5.33}{11.35} =0.94lbs/in

Like cc>c the system is undamped

The equilibrium expression is:

u(t)=u(o)coswt+u'(o)sinwt\\u(o)=7=0.583\\u'(o)=0\\u(t)=0.583coswt\\u(t)=0.583cos11.35t

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3 years ago
(please help i gotta turn this in a few minutes 10 points!)
pogonyaev

Answer:

3a, 2b,4c,1d

Explanation:

what do I need to explain just something you know

7 0
3 years ago
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