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Effectus [21]
4 years ago
14

What is the maximum mass that can hang without sinking from a 20-cm diameter Styrofoam sphere in water? Assume the volume of the

mass is negligible compared to that of the sphere.
Physics
1 answer:
alex41 [277]4 years ago
3 0

Answer:

the maximum mass that can hang without sinking is 2.93 kg

Explanation:

Given: details:

sphere diameter  d = 20 cm

so, radius r = 10 cm  = 0.10 m

density of the Styrofoam sphere D = 300 kg/m3

sphere volume  V = \frac{4}{3} \pi r^3

                                                   =\frac{4}{3} \pi 0.10^3

                                                   =4.18*10^{-3} m^3

we know that

Density = \frac{Mass}{Volume}

mass  M = Density * Volume

                                  = (300)(4.18*10^{-3} m3)

                                  =1.25 kg

mass of the water displace = volume *density  of water

                                                 = 4.18*10^{-3} m3 * 1000

                                                 = 4.18 kg

The difference between the mass of water and mass of styrofoam is the amount of mass that the sphere can support

=4.18 kg  -1.25 kg

= 2.93 kg

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A mass weighing 24 pounds, attached to the end of a spring, stretches it 4 inches. Initially, the mass is released from rest fro
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Answer:

the equation of motion is

x(t) =-\frac{1}{6} \cos4\sqrt{6} t

Explanation:

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The weight attached to the spring is 24pounds

Acceleration due to gravity is 32ft/s²

Assume x is the string length, 4inches

convert the length inches to to feet = 4/12 = 1/3feet

From Hookes law , we calculate the spring constant k

k = W / x

k = 24 / (1/3)

k = 24 / 0.33

k = 72lb/ft

If the mass is displace from its equilibrium position by amount x

the differential equation is

m\frac{d^2x}{dt^2} + kx=0\\\\\frac{3}{4 } \frac{d^2x}{dt^2}+72x=0\\

\frac{d^2x}{dt^2} +96x=0

Auxiliary equation is

m^2+96=0\\m=\sqrt{-96} \\m=\_ ^+4\sqrt{6}

Thus the solution is,

x(t) =  c_1cos4\sqrt{6t} +c_2sin\sqrt{6t}

x'(t) =-4\sqrt{6c_1} sin4\sqrt{6t} +c_24\sqrt{6} cos4\sqrt{6t}

The mass is release from rest

x'(0) = 0

-4\sqrt{6c_1 }  \sin4\sqrt{6} (0)+c_24\sqrt{6} \cos4\sqrt{6} (0)=0\\c_24\sqrt{6} =0\\\\c_2=0

Therefore

x(t) = c₁ cos4 √6t

x(0) = -2inches

c₁ cos4 √6(0) = 2/12feet

c₁= 1/6feet

There fore, the equation of motion is

x(t) =-\frac{1}{6} \cos4\sqrt{6} t

7 0
4 years ago
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