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Nutka1998 [239]
3 years ago
15

A 2-kg plastic tank that has a volume of 0.2m^3 is filled with water. Assuming the density of water is 999 kg/m 3 ; determine th

e weight of the combined system.
Engineering
1 answer:
siniylev [52]3 years ago
7 0

Answer:

Mass of the combined system = 201.8 Kg

Weight of the combined system = 1977.64 N

Explanation:

Given:

Mass of the plastic tank = 2 kg

Volume of the plastic tank = 0.2 m³

Density of water = 999 Kg/m³

Now, the volume of water in the tank =  volume of tank = 0.2 m³

Also,

Mass = Density × Volume

therefore,

Mass of water = 999 × 0.2 = 199.8 Kg

Thus, the combined mass = Mass of tank + Mass of water in the tank

or

the combined mass = 2 + 199.8 = 201.8 Kg

Also,

Weight = Mass × Acceleration due to gravity

or

Weight = 201.8 × 9.8 = 1977.64 N

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A series of experiments is conducted in which a thin plate is subjected to biaxial tension/compression, σ1, σ2 , the plane surfa
Andrew [12]

Answer:

                                               

Explanation:

3 0
3 years ago
Q4. (20 points) For a bronze alloy, the stress at which plastic deformation begins is 271 MPa and the modulus of elasticity is 1
babunello [35]

Answer:

a) P = 86720 N

b) L = 131.2983 mm

Explanation:

σ = 271 MPa = 271*10⁶ Pa

E = 119 GPa = 119*10⁹ Pa

A = 320 mm² = (320 mm²)(1 m² / 10⁶ mm²) = 3.2*10⁻⁴ m²

a) P = ?

We can apply the equation

σ = P / A     ⇒    P = σ*A = (271*10⁶ Pa)(3.2*10⁻⁴ m²) = 86720 N

b) L₀ = 131 mm = 0.131 m

We can get ΔL applying the following formula (Hooke's Law):

ΔL = (P*L₀) / (A*E)    ⇒  ΔL = (86720 N*0.131 m) / (3.2*10⁻⁴ m²*119*10⁹ Pa)

⇒  ΔL = 2.9832*10⁻⁴ m = 0.2983 mm

Finally we obtain

L = L₀ + ΔL = 131 mm + 0.2983 mm = 131.2983 mm

3 0
4 years ago
Estimate the endurance strength, Se, of a 37.5-mm- diameter rod of AISI 1040 steel having a machined finish and heat-treated to
7nadin3 [17]

Answer:

endurance length is 236.64 MPa

Explanation:

data given:

d = 37.5 mm

Sut = 760MPa

endurance limit is

Se = 0.5 Sut

   = 0.5*760 = 380 MPa

surface factor is

Ka = a*Sut^b

where

Sut is ultimate strength

for AISI 1040 STEEL

a = 4.51, b = -0.265

Ka = 4.51*380^{-0.265}

Ka = 0.93

size factor is given as

Kb =1.29 d^{-0.17}

Kb = 0.669

Se = Sut *Ka*Kb

    = 380*0.669*0.93

Se = 236.64 MPa

therefore endurance length is 236.64 MPa

4 0
3 years ago
Decide whether the function is an exponential growth or exponential decay function, and find the constant percentage rate of gro
sasho [114]

Answer:

Just answered this to confirm my profile.

Explanation:

I dont have a clue, this is just to confirm my profile.

8 0
3 years ago
A non-licensed person may be the SOLE owner of a civil, electrical, or mechanical engineering business under which of the follow
kotegsom [21]

Answer:

(d) None. No provisions exist.

Explanation:

B&P Code § 6738 prohibits a non-licensed person from being the sole proprietor of an engineering business. The non-licensed can be a partner in an engineering business that offers civil, electrical, or mechanical services. It is mandatory that at least one licensed engineer must be a co-owner of the business.

5 0
3 years ago
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