3 years ago

What measurements will be made to determine the magnitude of the test-mass centripetal acceleration?

1 answer:

3 0

Answer:period, spring constant, radius of circular part, velocity of the test mass, mass of the test-mass, mass of the hanging mass

Explanation:

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Explanation:

r = Radius of ball = 0.11 m

$\rho$ = Density of fluid = $1.1\times 10^3\ kg/m^3$ (Assumed)

g = Acceleration due to gravity = 9.81 m/s²

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The weight of the bowling ball will balance the buouyant force

$W=F_b\\\Rightarrow mg=V\rho g\\\Rightarrow m=\frac{V\rho g}{g}\\\Rightarrow m=V\rho\\\Rightarrow m=\frac{4}{3}\pi 0.11^3\times 1.1\times 10^3\\\Rightarrow m=6.1328\ kg$

The mass of the bowling ball will be 6.1328 kg

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A 70 kg driver gets into an empty taptap to start the day's work. The springs compress 2.3×10−2 m . What is the effective spring

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Answer:

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Explanation:

$m=Mass\ of\ diver=70\ kg\\x=Length\ compressed\ by\ spring=2.3\times 10^{-2}\ m\\a=Acceleration\ due\ to\ gravity=9.81\ m/s^2\\F=Force\ exerted\ by\ diver=m\times a\\\Rightarrow F=70\times 9.81\\\Rightarrow F=686.7\ N\\k=spring\ constant\\F=k\times x\\\Rightarrow k=\frac {F}{x}\\\Rightarrow k=\frac {686.7}{2.3\times 10^{-2}}\\\therefore k=29856.521\ N/m$

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