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4 years ago

What is the weight density of a 2.24 in diameter titanium sphere that weights 0.82 lb?

Engineering

1 answer:

nasty-shy [4]4 years ago

4 0

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

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

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Determine the slopes and deflections at points B and C for the beam shown below by the moment-area method. E=constant=70Gpa I=50

inn [45]

Answer:

hello your question is incomplete attached below is the complete question

answer :

Slopes : B = 180 mm , C = 373 mm

Deflection: B = 0.0514 rad , C = 0.077 rad

Explanation:

Given data :

I = 500(10^6) mm^4

E = 70 GPa

The M / EI diagram is attached below

Deflection angle at B

∅B = ∅BA = [ 150 (6) + 1/2 (300)*6 ] / EI

= 1800 / ( 500 * 70 ) = 0.0514 rad

slope at B

ΔB = ΔBA = [ 150(6)*3 + 1/2 (300)*6*4 ] / EI

= 6300 / ( 500 * 70 ) = 0.18 m = 180 mm

Deflection angle at C

∅C = ∅CA = [ 1800 + 300*3 ] / EI

= 2700 / ( 500 * 70 )

= 2700 / 35000 = 0.077 rad

Slope at C

ΔC = [ 150 * 6 * 6 + 1/2 (800)*6*7 + 300(3) *1.5 ]

= 13050 / 35000 = 373 mm

3 0

3 years ago

Two concentric helical compression springs made of steel and having the same length when loaded and when unloaded are used to su

Andrews [41]

Answer:

see explaination for all the answers and full working.

Explanation:

deflection=8P*DN/Gd^4

G(for steel)=70Gpa=70*10^9N/m^2=70KN/mm^2

for outer spring,

deflection=8*3*50^3*5/(70*9^4)=32.66mm

for inner spring

deflection=8*3*30^3*10/(70*5^4)=148.11mm

max stress=k*8*P*C/(3.14*d^2)

for outer spring

c=50/9=5.55

k=(4c-1/4c-4)+.615/c=1.2768

max stress=1.2768*8*3*5.55/(3.14*9^2=.66KN.mm^2

for inner spring

c=6

k=1.2525

max stress=2.29KN/mm^2

6 0

4 years ago

Read 2 more answers

A certain part of the cast iron piping of a water distribution system involves a parallel section. Both parallel pipes have a di

Bezzdna [24]

Answer :

Explanation:

Given :

Length of pipe A $L_{A} = 1500$ m

Length of pipe B $L_{B} = 2500$ m

Flow rate through pipe A $Q_{A} = 0.4 \frac{m^{3} }{s}$

Diameter of pipe $D = 30 \times 10^{-2}$ m

Velocity from pipe A,

$V _{A} = \frac{Q_{A} }{A}$

$V _{A} = \frac{0.4 \times 4 }{\pi ( 30 \times 10^{-2} )^{2} }$

$V_{A} = 5.66$ $\frac{m}{s}$

Here, head loss is same because height is same.

$h_{a} = h_{b}$

$L_{A} V_{A} ^{2} = L_{B} V_{B} ^{2}$

$V_{B} = \sqrt{\frac{1500}{2500}} (5.66)$

$V_{B} = 4.38$ $\frac{m}{s}$

Now rate of flow from pipe B is,

$Q_{B} = V_{B} A$

$Q_{B} = \frac{\pi }{4} (0.3)^{2} \times 4.38$

$Q_{B} = 0.3094$ $\frac{m^{3} }{s}$

4 0

3 years ago

For the following gear train, if the blue gear is moving at 50 rpm, what are the speeds of the other gears?

Flauer [41]

Answer:

Explanation:

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5 0

4 years ago

Use the following assumptions for problems 1 and 2:

Salsk061 [2.6K]

Answer:

The text file attached has the detailed solution of all the parts individually.

Download txt

6 0

3 years ago