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Tcecarenko [31]
3 years ago
14

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

mass has a velocity of 4 ft/sec. Suppose the object is displaced an additional 7 inches and released. Find an equation for the object's displacement, u(t), in feet after t seconds.
Physics
1 answer:
const2013 [10]3 years ago
3 0

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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A train is accelerating at a rate of 2 km/hr/s.  If its initial velocity is 20 km/hr, what is its velocity after 30 seconds?
Mademuasel [1]
Here it is. *WARNING* VERY LONG ANSWER

________________________________________... 
<span>11) If Galileo had dropped a 5.0 kg cannon ball to the ground from a height of 12 m, the change in PE of the cannon ball would have been product of mass(m),acceleration(g)and height(h) </span>

<span>The change in PE =mgh=5*9.8*12=588 J </span>
<span>______________________________________... </span>
<span>12.) The 2000 Belmont Stakes winner, Commendable, ran the horse race at an average speed = v = 15.98 m/s. </span>

<span>Commendable and jockey Pat Day had a combined mass =M= 550.0 kg, </span>

<span>Their KE as they crossed the line=(1/2)Mv^2 </span>

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<span>______________________________________... </span>
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<span>The tire rolls down the hill with an intial speed = u = 2.00 m/s. </span>

<span>The height of top of the next hill = h = 5.00 m </span>

<span>Initial total mechanical energy =PE+KE=mgH+(1/2)mu^2 </span>

<span>Initial total mechanical energy =mgH+(1/2)mu^2 </span>

<span>Suppose the final speed at the top of second hill is v </span>

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<span>As mechanical energy is conserved, </span>

<span>Final total mechanical energy =Initial total mechanical energy </span>

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<span>EDIT </span>
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<span>Acceleration </span>

<span>Initial velocity = u = 20 km/hr, </span>

<span>Velocity after 30 seconds = v = u + at </span>

<span>Velocity after 30 seconds = v = 20 km/hr + 2 (km/hr/s)*30s = </span>

<span>Velocity after 30 seconds = v = 20 km/hr + 60 km/hr = 80 km/ hr </span>

<span>Velocity after 30 seconds = v = 80 km/hr=22.22 m/s </span>
<span>_______________________________- </span>
<span>2.) A runner achieves a velocity of 11.1 m/s 9 s after he begins. </span>

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7 0
3 years ago
Two billion people jump up in the air at the same time with an average velocity of 7.0 m/sec. If the mass of an average person i
faust18 [17]

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average mass per person = 60 kg.
jump velocity = 7 m/s straight up and away from the block, all in the same direction
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Divide each side by 'm' :     v = (momentum) / (mass) =

The Earth's "recoil" velocity is   (8.4 x 10¹¹) / (5.98 x 10²⁴) = 

                                                               1.405 x 10⁻¹³ m/s =

                                              <em> 0.00000000014 millimeter per second

</em>
I have no intuitive feeling for this kind of thing, so can't judge whether
the answer is reasonable.  But my math and physics felt OK on the
way to the solution, so that's my answer and I'm sticking to it.

4 0
3 years ago
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