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Andrej [43]
3 years ago
10

What forces contribute to density

Physics
1 answer:
mixer [17]3 years ago
5 0
No force contributes to density, the density is a physical quantity that is defined as being \rho = m/V 
 the raport between the mass of the object and its volume. However if you want to measure the density of an object you might want to determine its gravity force (weight) G=mg from which knowing the gravitational acceleratin you can find its mass m=G/g
 
where G is given in Newtons and g=9.81 is given in  (m/s^2)


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An 88 kg person steps into a car of mass 2002 kg, causing it to sink 5.36 cm on itssprings. Assuming no damping, with what fre-q
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Answer:

The required frequency = 0.442 Hz

Explanation:

Frequency f  = ( \dfrac{1}{2 \pi}) \omega

where;

\omega = \sqrt{\dfrac{k}{m} }

Then;

f = \Bigg ( \dfrac{1}{2 \pi}  \Bigg )   \Bigg( \sqrt{\dfrac{k}{m} }  \Bigg )

However;

k  = \dfrac{F}{x} and;

mass m = m_{car } + m_{person}

f = \Bigg ( \dfrac{1}{2 \pi}  \Bigg )   \Bigg( \sqrt{\dfrac{\dfrac{F}{x}}{m_{car}+m_{person}} }  \Bigg )

f = \Bigg ( \dfrac{1}{2 \pi}  \Bigg )   \Bigg( \sqrt{\dfrac{{F}}{x(m_{car}+m_{person})} }  \Bigg )

where;

F = m_{person}g

Then;

f = \Bigg ( \dfrac{1}{2 \pi}  \Bigg )   \Bigg( \sqrt{\dfrac{ {m_{person}g }}{x(m_{car}+m_{person})} }  \Bigg )

replacing the values;

f = \Bigg ( \dfrac{1}{2 \pi}  \Bigg )   \Bigg( \sqrt{\dfrac{ {(88 \ kg)* (9.81 \ m/s^2) }}{(5.36 \times 10^{-2} \ m) (2002 \ kg +88 \ kg)} }  \Bigg )

\mathbf{f = 0.442 \ Hz}

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3 years ago
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Explanation:

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