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azamat
4 years ago
15

A copper sphere 10 mm in diameter is dropped into a 1-m-deep drum of asphalt. The asphalt has a density of 1150 kg/m3 and a visc

osity of 105 N.s/m2 . Estimate the time it takes for the sphere to reach the bottom of the drum
Physics
1 answer:
kvasek [131]4 years ago
6 0

Answer:

t = 1964636.542 sec

Explanation:

Given data:

sphere diameter is 10 mm

Density is 1150 kg/m^3

viscosity 105 N s/m^2

We knwo that time taken by sphere can be calculated by following procedure

\tau = \mu \frac{du}{dy}

\frac{F}{A} =  \mu \frac{du}{r}

\frac{\rho_C -\rho_{asphalt} gv}{2 \pi rL} = 10^5 \frac{du}{r}

Solving for du

du = \frac{ (8933 - 1150) 9.81 \frac{4}{3} \pi (10\times 10^{-3})^3}{2\pi \times 1\times 10^5}

du = u = 5.09\times 10^{-7}

u = \frac{1}{t}

t = \frac{1}{5.09\times 10^{-7}} = 1964636.542 sec

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coldgirl [10]
Below is the solution:

Let us say that the disk goes through a vertical elevation change of one meter. 

<span>The change in potential energy will equal the change in kinetic energy </span>

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<span>mgh = ½mv² + ½Iω² </span>

<span>for a uniform disk, the moment of inertia is </span>
<span>I = ½mr² </span>
<span>and </span>
<span>ω = v/r </span>

<span>mgh = ½mv² + ½(½mr²)(v/r)² </span>
<span>mgh = ½mv² + ¼mv² </span>
<span>gh = ¾v² </span>

<span>v² = 4gh/3 </span>

<span>v² = u² + 2as </span>

<span>if we assume initial velocity is zero </span>

<span>v² = 2as </span>
<span>a = v² / 2s </span>

<span>s(sinθ) = h </span>
<span>s = h/sinθ </span>

<span>a = 4gh/3 / 2(h/sinθ) </span>
<span>a = ⅔gsinθ </span>

<span>a = ⅔(9.8)sin25 </span>
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3 0
4 years ago
PLEASE HELP!! Salmon often jump waterfalls to reach their
AlexFokin [52]

Answer:

7.13781 m/s

Explanation:

X-direction             | Y-direction

x=v_{xo}t+\frac{1}{2}a_{x}t^2      | y=v_{yo}t+\frac{1}{2}a_{y}t^2

3.41=v_{o}cos(28.4)t  | 0.397=v_{yo}sin(28.4)t+\frac{1}{2} (-9.81)t^2

3.41=v_{o}(0.87964)t  | 0.397=v_{yo}sin(28.4)(\frac{3.87658}{v_{o} })+\frac{1}{2} (-9.81)(\frac{3.87658}{v_{o} })^2

\frac{3.41}{0.87964}=v_{o}t             | 0.397=1.84379-\frac{73.71171}{v^2}

\frac{3.87658}{v_{o} } =t                | 7.13781=v

8 0
3 years ago
Q3. At an altitude of 6679 km above the center of the Earth, the
EastWind [94]

Answer:

5524.8 m/s

Explanation:

Given that,

Altitude above the surface of earth is 6679 km, the Space Station can complete one orbit in 1.5 hours. We need to find the average speed of the ISS. The average speed of its orbit is given by :

v=\sqrt{\dfrac{GM}{R}}

R is distance from Earth

R = r + d, r is the radius of Earth

R = 6371  + 6679

R = 13050 km

So,

v=\sqrt{\dfrac{6.67\times 10^{-11}\times 5.972 \times 10^{24}}{13050\times 10^3}}\\\\v=5524.8\ m/s

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Hence, the average speed of ISS is 5524.8 m/s.

4 0
3 years ago
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Answer:

1,681,680 J.

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5 0
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AveGali [126]

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6 0
4 years ago
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