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2 years ago

Which type of line is represented by thin, short dashes?

Engineering

1 answer:

Akimi4 [234]2 years ago

6 0

hidden lines

These are dotted lines
These are used to describe hidden spaces .
Like if a cube is drawn then we can't see the diagonal inside from corner to corners so we represent it as dashed lines

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A rigid tank having 25 m3 volume initially contains air having a density of 1.25 kg/m3, then more air is supplied to the tank fr

Hoochie [10]

Answer:

$\Delta m = 102.25\,kg$

Explanation:

The mass inside the rigid tank before the high pressure stream enters is:

$m_{o} = \rho_{air}\cdot V_{tank}$

$m_{o} = (1.25\,\frac{kg}{m^{3}} )\cdot (25\,m^{3})$

$m_{o} = 31.25\,kg$

The final mass inside the rigid tank is:

$m_{f} = \rho \cdot V_{tank}$

$m_{f} = (5.34\,\frac{kg}{m^{3}} )\cdot (25\,m^{3})$

$m_{f}= 133.5\,kg$

The supplied air mass is:

$\Delta m = m_{f}-m_{o}$

$\Delta m = 133.5\,kg-31.25\,kg$

$\Delta m = 102.25\,kg$

4 0

3 years ago

\o you sell them?” "Fivepence farthing for one—Twopence for two,” the Sheep replied. "Then two are cheaper than one?” Alice said

NemiM [27]

Answer:

unusual and strange

Explanation:

3 0

3 years ago

4. An aluminum alloy fin of 12 mm thick, 10 mm width and 50 mm long protrudes from a wall, which is maintained at 120C. The amb

Minchanka [31]

Answer:

a) 84.034°C

b) 92.56°C

c) ≈ 88 watts

Explanation:

Thickness of aluminum alloy fin = 12 mm

width = 10 mm

length = 50 mm

Ambient air temperature = 22°C

Temperature of aluminum alloy is maintained at 120°C

<u>a) Determine temperature at end of fin</u>

m = √ hp/Ka

= √( 140*2 ) / ( 12 * 10^-3 * 55 )

= √ 280 / 0.66 = 20.60

Attached below is the remaining answers

7 0

3 years ago

A cylinder fitted with a frictionless piston contains 2 kg of R-134a at 3.5 bar and 100 C. The cylinder is now cooled so that th

inna [77]

Answer:

The answer to the question is

The heat transferred in the process is -274.645 kJ

Explanation:

To solve the question, we list out the variables thus

R-134a = Tetrafluoroethane

Intitial Temperaturte t₁ = 100 °C

Initial pressure = 3.5 bar = 350 kPa

For closed system we have m₁ = m₂ = m

ΔU = m×(u₂ - u₁) = ₁Q₂ -₁W₂

For constant pressure process we have

Work done = W = $\int\limits^a_b P \, dV$ = P×ΔV = P × (V₂ - V₁) = P×m×(v₂ - v₁)

From the tables we have

State 1 we have h₁ = (490.48 +489.52)/2 = 490 kJ/kg

State 2 gives h₂ = 206.75 + 0.75 × 194.57= 352.6775 kJ/kg

Therefore Q₁₂ = m×(u₂ - u₁) + W₁₂ = m × (u₂ - u₁) + P×m×(v₂ - v₁)

= m×(h₂ - h₁) = 2.0 kg × (352.6775 kJ/kg - 490 kJ/kg) =-274.645 kJ

5 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

3 years ago