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Evgesh-ka [11]
3 years ago
7

A wooden cylinder (0 02 x 0 02 x 0 1m) floats vertically in water with one-third of ts length immersed. a)-Determine the density

of wood b)-Determine if this is a stable condition c)-Determine if the plece of wood would float stably in alcohol with density 700 kg/m3
Engineering
1 answer:
Anuta_ua [19.1K]3 years ago
5 0

Answer:

a)- the density of wood is 333.33 Kg/m³

b)-unstable condition

c)-unstable condition

Explanation:

Given data

wooden cylinder = 0.02m  x 0.02m x 0.1m

floating = 1/3 × Length

to find out

density of wood,  is it stable condition and wood would float stably in alcohol with density 700 kg/m3

Solution

First we find out density of wood

we know density of water is 1000 kg/m³

and we know wood float 1/3 of length so fraction of density will be

density of wood/ density of water = 1/3

density of wood = 1/3 ×  density of water

density of wood  = 1/3 × 1000 = 333.33 Kg/m³

Now in 2nd part we know for stable condition in partially submerged of body the metacentric height is greater than the centroid of body

we check these condition now,

metacentric height (GM)= I/v  

I = ( 0.02 × 0.02³ / 12 )  

v = ( 0.02 × 0.02 × 0.1 )

(GM)= I/v =  ( 0.02 × 0.02³ / 12 ) / ( 0.02 × 0.02 × 0.1 ) = 0.000333

and we know centroid of body (BM) =  0.05 - 0.033 = 0.017

we know height is 0.1m so G act at 0.05 and B act at (0.1 × 0.3 ) = 0.033

we can see that now metacentric height is less than centroid of body so our body is unstable condition

Now in 3rd part we use alcohol so we calculate ratio of density of wood and density of alcohol i.e. = 333.33 / 700 = 0.48

so now our new G will be 0.05 and B will be (0.1 × 0.48 ) = 0.048

metacentric height (GM)= I/v =  ( 0.02 × 0.02³ / 12 ) / ( 0.02 × 0.02 × 0.1 ) = 0.000333

centroid of body (BM) =  0.05 - 0.048 = 0.002

we can see that now metacentric height is less than centroid of body so our body is unstable condition

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Answer:

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Explanation:

Please kindly check attachment for the step by step solution of the given problem

4 0
3 years ago
What is the weight density of a 2.24 in diameter titanium sphere that weights 0.82 lb?
nasty-shy [4]

Answer:

0.14\ lb/in^{3}

Explanation:

Density is defined as mass ler unit volume, expressed as

\rho=\frac {m}{v}

Where m is mass, \rho is density and v is the volume. For a sphere, volume is given as

v=\frac {4\pi r^{3}}{3}

Replacing this into the formula of density then

\rho=\frac {m}{\frac {4\pi r^{3}}{3}}

Given diameter of 2.24 in then the radius is 1.12 in. Substituting 0.82 lb for m then

\rho=\frac {0.82}{\frac {4\pi 1.12^{3}}{3}}=0.13932044952427\approx 0.14 lb/in^{3}

4 0
3 years ago
When subject to an unknown torque, the shear stress in a 2 mm thick rectangular tube of dimension 100 mm x 200 mm was found to b
laila [671]

Answer:

The shear stress will be 80 MPa

Explanation:

Here we have;

τ = (T·r)/J

For rectangular tube, we have;

Average shear stress given as follows;

Where;

\tau_{ave} = \frac{T}{2tA_{m}}

A_m = 100 mm × 200 mm = 20000 mm² = 0.02 m²

t = Thickness of the shaft in question = 2 mm = 0.002 m

T = Applied torque

Therefore, 50 MPa = T/(2×0.002×0.02)

T = 50 MPa × 0.00008 m³ = 4000 N·m

Where the dimension is 50 mm × 250 mm, which is 0.05 m × 0.25 m

Therefore, A_m = 0.05 m × 0.25 m = 0.0125 m².

Therefore, from the following average shear stress formula, we have;

\tau_{ave} = \frac{T}{2tA_{m}}

Plugging in then values, gives;

\tau_{ave} = \frac{4000}{2\times 0.002 \times 0.0125} = 80,000,000 Pa

The shear stress will be 80,000,000 Pa or 80 MPa.

7 0
3 years ago
A wood pole with a diameter of 10 in. has a moisture content of 5%. The fiber saturation point (FSP) for this wood is 30%. The w
Mekhanik [1.2K]

Answer:

a) Δd(change in wood diameter) = 5%

b) The wood would swell since the moisture content is increasing which will also led to increase in the wood's diameter

C) new diameter (D2) = 10.5 in

Explanation:

Wood pole diameter = 10 inches

moisture content = 5%

FSP = 30%

A) The percentage change in the wood's diameter

note : moisture fluctuations from 5% to 30% causes dimensional changes in the wood but above 30% up to 55% causes no change. hence this formula can be used to calculate percentage change in the wood's diameter

Δd/d = 1/5(30 - 5)

Δd/d = 5%  

Δd = 5%

B) would the wood swell or shrink

The wood would swell since the moisture content is increasing which will also led to increase in the wood's diameter

C) The new diameter of the wood

D2 = D + D( \frac{M1}{100} )

D = initial diameter= 10 in , M1 = initial moisture content = 5%

therefore D2 = 10 + 10( 5/100 )

new diameter (D2) = 10.5 in

5 0
3 years ago
Please help me bro come on
melomori [17]
Answer:

12

Explanation:

5 • 3 = 15
3 • 3 = 9
4 • 3 = 12
6 0
3 years ago
Read 2 more answers
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