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Ad libitum [116K]
3 years ago
14

I'm stuck on number 4... help please?

Physics
2 answers:
lora16 [44]3 years ago
8 0
V= \frac{\Delta S}{\Delta t} = \frac{5}{15} = \frac{1}{3} m/s
Paha777 [63]3 years ago
3 0

Average <u>speed</u> = (distance covered) / (time to cover the distance) =
 
                                     (5m)  /  (15 sec) =

                                     (5/15) (m/s)  =  <em>1/3 m/s</em> .

Average <u>velocity</u> = 

         (displacement) / (time spent traveling)  in the direction of the displacement

Average velocity =  (5m) / (15 sec)  left =

                           (5/15) / (m/sec)  left  =

                               <em>1/3  m/s  left</em>.

A number without a direction is a speed, not a velocity.


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The suspension system of a 1700 kg automobile "sags" 7.7 cm when the chassis is placed on it. Also, the oscillation amplitude de
spin [16.1K]

Answer:

the spring constant k = 5.409*10^4 \ N/m

the value for the damping constant \\ \\b = 1.518 *10^3 \ kg/s

Explanation:

From Hooke's Law

F = kx\\\\k =\frac{F}{x}\\\\where \ F = mg\\\\k = \frac{mg}{x}\\\\given \ that:\\\\mass \ of \ each \ wheel = 425 \ kg\\\\x = 7.7cm = 0.077 m\\\\g = 9.8 \ m/s^2\\\\Then;\\\\k = \frac{425 \ kg * 9.8 \ m/s^2}{0.077 \ m}\\\\k = 5.409*10^4 \ N/m

Thus; the spring constant k = 5.409*10^4 \ N/m

The amplitude is decreasing 37% during one period of the motion

e^{\frac{-bT}{2m}}= \frac{37}{100}\\\\e^{\frac{-bT}{2m}}= 0.37\\\\\frac{-bT}{2m} = In(0.37)\\\\\frac{-bT}{2m} = -0.9943\\\\b = \frac{2m(0.9943)}{T}\\\\b = \frac{2m(0.9943)}{\frac{2 \pi}{\omega}}\\\\b = \frac{m(0.9943) \ ( \omega) )}{ \pi}

b = \frac{m(0.9943)(\sqrt{\frac{k}{m})}}{\pi}\\\\b = \frac{425*(0.9943)(\sqrt{\frac{5.409*10^4}{425}) }    }{3.14}\\\\b = 1518.24 \ kg/s\\\\b = 1.518 *10^3 \ kg/s

Therefore; the value for the damping constant \\ \\b = 1.518 *10^3 \ kg/s

5 0
3 years ago
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