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Tpy6a [65]
3 years ago
15

uppose that a car weighing 2000 pounds is supported by four shock absorbers Each shock absorber has a spring constant of 6250 lb

s/foot, so the effective spring constant for the system of 4 shock absorbers is 25000 lbs/foot. Assume no damping and determine the period of oscillation of the vertical motion of the car. Hint: g= 32 ft/sec2 .
Physics
1 answer:
marusya05 [52]3 years ago
3 0

Explanation:

According to the given data, formula for the period of oscillation is as follows.

            T = 2 \pi \sqrt{\frac{m}{k}}

or,         T = 2 \pi \sqrt{\frac{mg}{kg}}

                = 2 \times 3.14 \times \sqrt{\frac{2000 pounds}{25000 lbs/foot \times 32 ft/s^{2}}}

                 = 0.314 s

Thus, we can conclude that period of oscillation of the vertical motion of the car is 0.314 s.

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Now imagine these waves are moving through a rope. If blue waves will try to move the rope in positive direction, the red wave will pull it down, and thus the two waves will cancel the effect of each other. Thus resulting in a destructive interference. 

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It takes me 12s to lift a box upward with 20 N of force to a height of 45m. What was my power output ?
Finger [1]

Answer:

75Watts

Explanation:

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Time  = 12s

Force applied  = 20N

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Unknown:

Power output  = ?

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6 0
3 years ago
The driver of a car traveling at 31.3 m/s applies the brakes and undergoes a constant deceleration of 1.6 m/s2.How many revoluti
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Answer:

R=156.99\operatorname{Re}vs

Explanation: The equations used are as follows:

\begin{gathered} x(t)=x_o+v_ot+\frac{1}{2}at^2\Rightarrow(1) \\ v(t)=v_o+at\Rightarrow(2) \end{gathered}

By using equation (2), the time needed for the car to come to rest is calculated as follows:

\begin{gathered} v(t)=(31.3ms^{-1})_{}+(-1.6ms^{-2})t=0 \\ t=\frac{31.3ms^{-1}}{1.6ms^{-2}}=19.56s \\ t=19.563s \end{gathered}

By using equation (1), The total distance traveled in that time would be as:

\begin{gathered} x(19.563s)=_{}(31.3ms^{-1})\cdot(19.563s)+\frac{1}{2}(-1.6ms^{-2})\cdot(19.563s)^2\Rightarrow(1) \\ x(19.563s)=612.31-306.17=306.14m \\ \therefore\Rightarrow \\ x(19.563s)=306.14m \end{gathered}

The revolutions taken by the tire before the car comes to rest would be:

\begin{gathered} C=2\pi\cdot(0.31m)=1.95m \\ R=\frac{x(19.563s)}{C}=\frac{306.14m}{1.95m}=156.99\operatorname{Re}v \\ R=156.99\operatorname{Re}vs \end{gathered}

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2 years ago
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