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-Dominant- [34]
4 years ago
14

You are driving on a public road and need to tum your vehicle around. You can use a three-point turn

Engineering
1 answer:
harina [27]4 years ago
5 0
Only where U-turns are permitted, and only if the road is too narrow for your vehicle to make a U-turn and you cannot go.
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Kam
jolli1 [7]
I think D i’m not sure
7 0
3 years ago
The production process of rods from machine "A" yields specimen with the following specs. Mean: µ(LA)=20.00mm, STD: s(LA)=0.50mm
Oxana [17]

Answer: the standard deviation STD of machine B is s (Lb) = 0.4557

Explanation:

from the given data, machine A and machine B produce half of the rods

Lt = 0.5La + 0.5Lb

so

s² (Lt) = 0.5²s²(La) + 0.5²s²(Lb) + 0.5²(2)Cov (La, Lb)

but Cov (La, Lb) = Corr(La, Lb) s(La) s(Lb) = 0.4s (La) s(Lb)

so we substitute

s²(Lt) = 0.25s² (La) + 0.25s² (Lb) + 0.4s (La) s(Lb)

0.4² = 0.25 (0.5²) + 0.25s² (Lb) + (0.5)0.4(0.5) s(Lb)

0.64 = 0.25 + s²(Lb) + 0.4s(Lb)

s²(Lb) + 0.4s(Lb) - 0.39 = 0

s(Lb) = { -0.4 ± √(0.16 + (4*0.39)) } / 2

s (Lb) = 0.4557

therefore the standard deviation STD of machine B is s (Lb) = 0.4557

8 0
3 years ago
Transformation is the change of one energy form to another.<br> O True<br> O False
Doss [256]
I think is true try that
3 0
2 years ago
Read 2 more answers
Compute the atomic density (the number of atoms per cm3 rather than mass density g/cm3) for a perfect crystal of silicon at room
Ann [662]

Answer:

5E22 atoms/cm³

Explanation:

We need to find the number of moles of silicon per cm³

number of moles per cm³ = density/atomic weight = 2.33 g/cm ÷ 28.09 g/mol = 0.083 mol/cm³.

Since there are 6.022 × 10²³ atoms/mol, then the number of atoms of silicon per cm³ = number of atoms per mol × number of moles per cm³

= 6.022 × 10²³ atoms/mol × 0.083 mol/cm³

= 0.4995 × 10²³ atoms/cm³

= 4.995 × 10²² atoms/cm³

≅ 5 × 10²² atoms/cm³

= 5E22 atoms/cm³

5 0
3 years ago
Air is compressed by a 40-kW compressor from P1 to P2. The air temperature is maintained constant at 25°C during this process a
AlexFokin [52]

Answer:

the rate of entropy change of the air is -0.1342 kW/K

the assumptions made in solving this problem

- Air is an ideal gas.

- the process is isothermal ( internally reversible process ). the change in internal energy is 0.

- It is a steady flow process

- Potential and Kinetic energy changes are negligible.

Explanation:

Given the data in the question;

From the first law of thermodynamics;

dQ = dU + dW ------ let this be equation 1

where dQ is the heat transfer, dU is internal energy and dW is the work done.

from the question, the process is isothermal ( internally reversible process )

Thus, the change in internal energy is 0

dU = 0

given that; Air is compressed by a 40-kW compressor from P1 to P2

since it is compressed, dW = -40 kW

we substitute into equation 1

dQ = 0 + ( -40 kW )

dQ = -40 kW

Now, change in entropy of air is;

ΔS_{air = dQ / T

given that T = 25 °C = ( 25 + 273.15 ) K = 298.15 K

so we substitute

ΔS_{air =  -40 kW / 298.15 K

ΔS_{air =  -0.13416 ≈ -0.1342 kW/K

Therefore, the rate of entropy change of the air is -0.1342 kW/K

the assumptions made in solving this problem

- Air is an ideal gas.

- the process is isothermal ( internally reversible process ). the change in internal energy is 0.

- It is a steady flow process

- Potential and Kinetic energy changes are negligible.

7 0
3 years ago
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