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stellarik [79]
3 years ago
10

Can someone tell me what car year and model this is please

Engineering
2 answers:
tensa zangetsu [6.8K]3 years ago
8 0

2005 BMW 5 Series , that should be it

Arlecino [84]3 years ago
5 0

Answer:

i think 1844

Explanation:

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Three-dimensional measuring references all of these EXCEPT:
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A rod that was originally 100-cm-long experiences a strain of 82%. What is the new length of the rod?
Ierofanga [76]

Answer: (b)

Explanation:

Given

Original length of the rod is L=100\ cm

Strain experienced is \epsilon=82\%=0.82

Strain is the ratio of the change in length to the original length

\Rightarrow \epsilon =\dfrac{\Delta L}{L}\\\\\Rightarrow 0.82=\dfrac{\Delta L}{100}\\\\\Rightarrow \Delta L=82\ cm

Therefore, new length is given by (Considering the load is tensile in nature)

\Rightarrow L'=\Delta L+L\\\Rightarrow L'=82+100=182\ cm

Thus, option (b) is correct.

8 0
3 years ago
Risks can be changed.<br> True<br> False
Mrrafil [7]

Answer:

true

Explanation:

8 0
3 years ago
Timescale limits knowledge for scientists because it is difficult for them to see much beyond their lifetimes. Question 1 option
vaieri [72.5K]

Answer:

I think true

Explanation:

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8 0
2 years ago
A rigid 14-L vessel initially contains a mixture of liquid water and vapor at 100°C with 12.3 percent quality. The mixture is th
tigry1 [53]

Answer:

Q = 65.388 KJ

Explanation:

To calculate the heat required for the given process Q, we recall the energy balance equation.

Therefore, : Q = Δ U = m (u₂ - u₁) ..................equation (1)

We should note that there are no kinetic or potential energy change so the heat input in the system is converted only to internal energy.

Therefore, we will start the equation with the mass of the water (m) using given the initial percentage quality as x₁ = 0.123 and initial temperature t₁ = 100⁰c , we can them determine the initial specific volume v₁ of the mixture. For the calculation, we will also need the specific volume of liquid vₙ  = 0.001043m³/kg and water vapour (vₐ) = 1.6720m³/kg

Therefore, u₁ = vₙ + x₁ . ( vₐ - vₙ)

                   u₁ = 0.001043m³/kg + 0.123 . ( 1.6720m³/kg - 0.001043m³/kg)

                   u₁ = 0.2066m³/kg

Moving forward, the mass of the vapor can then be calculated using the given volume of tank V = 14 L but before the calculation, we need to convert the volume to from liters to m³.

Therefore, V = 14L . 1m² / 1000L = 0.014 m³

Hence, m = V / u₁

                 0.014m³ / 0.2066 m³/kg

              m = 0. 0677 kg

Also, the initial specific internal energy u₁ can be calculated using the given the initial given quality of x₁ , the specific internal energy of liquid water vₐ = 419.06 kj / kg and the specific internal energy of evaporation vₐₙ = 2087.0 kj/kg.

Therefore, u₁ = vₐ + x₁ . vₐₙ

                   u₁ = 419.06 kj / kg + 0.123  .  2087.0 kj/kg

                    u₁ = 675.76 kj/kg

For the final specific internal energy u₂, we first need to calculate the final quality of the mixture x₂ . The tank is rigid meaning the volume does not change and it is also closed meaning the mass does not change.from this, we can conclude the the specific volume also does not change during the process u₁ = u₂. This allows us to use the given final temperature T₂ = 180⁰c to determine the final quality x₂ of the mixture. for the calculation, we will also need the specific volume of liquid vₙ=0.001091m³/kg and vapor vₐ =  0.39248m³/kg

Hence, x₂ = u₂ - vₙ / uₐ

x₂ = 0.2066 m³/kg - 0.001091m³/kg / 0.39248m³/kg

x₂ = 0.524

Moving forward to calculate the final internal energy u₂, we have :

u₂ = vₙ + x₂ . vₙₐ

u₂ = 631.66 kj/kg + 0.524  . 1927.4 kj/kg

u₂ = 1641.62 kj/kg

We now return to equation (1) to plug in the values generated thus far

Q = m (u₂ - u₁)

0. 0677 kg ( 1641.62 kj/kg - 675.76 kj/kg)

Q = 65.388KJ

7 0
3 years ago
Read 2 more answers
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