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Whitepunk [10]
3 years ago
14

Draw perpendicular lines based on the info in the image.

Engineering
1 answer:
Over [174]3 years ago
5 0
Wait let me try it…..
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An engineer measures a sample of 1200 shafts out of a certain shipment. He finds the shafts have an average diameter of 2.45 inc
Vadim26 [7]

Answer: 78.89%

Explanation:

Given : Sample size : n=  1200

Sample mean : \overline{x}=2.45

Standard deviation : \sigma=0.07

We assume that it follows Gaussian distribution (Normal distribution).

Let x be a random variable that represents the shaft diameter.

Using formula, z=\dfrac{x-\mu}{\sigma}, the z-value corresponds to 2.39 will be :-

z=\dfrac{2.39-2.45}{0.07}\approx-0.86

z-value corresponds to 2.60 will be :-

z=\dfrac{2.60-2.45}{0.07}\approx2.14

Using the standard normal table for z, we have

P-value = P(-0.86

=P(z

Hence, the percentage of the diameter of the total shipment of shafts will fall between 2.39 inch and 2.60 inch = 78.89%

7 0
3 years ago
A distribution center is used in which of the following applications?
FrozenT [24]

both b and c are the right

5 0
3 years ago
Explain how feedback control is used to<br> adjust robotic movements.
LuckyWell [14K]

Answer:

Feedback control of arm movements using Neuro-Muscular Electrical Stimulation (NMES) combined with a lockable, passive exoskeleton for gravity compensation

6 0
2 years ago
What mass of LP gas is necessary to heat 1.4 L of water from room temperature (25.0 ∘C) to boiling (100.0 ∘C)? Assume that durin
DochEvi [55]

Answer:

m_{LP}=0.45\,kg

Explanation:

Let assume that heating and boiling process occurs under an athmospheric pressure of 101.325 kPa. The heat needed to boil water is:

Q_{water} = (1.4\,L)\cdot(\frac{1\,m^{3}}{1000\,L} )\cdot (1000\,\frac{kg}{m^{3}} )\cdot [(4.187\,\frac{kJ}{kg\cdot ^{\textdegree}C} )\cdot (100^{\textdegree}C-25^{\textdegree}C)+2257\,\frac{kJ}{kg}]

Q_{water} = 3599.435\,kJ

The heat liberated by the LP gas is:

Q_{LP} = \frac{3599.435\,kJ}{0.16}

Q_{LP} = 22496.469\,kJ

A kilogram of LP gas has a minimum combustion power of 50028\,kJ. Then, the required mass is:  

m_{LP} = \frac{22496.469\,kJ}{50028\,\frac{kJ}{kg} }

m_{LP}=0.45\,kg

6 0
3 years ago
Consider a dip-coating process where a very long (assume infinitely long) wire(solid) with radius, ri, is being pulled verticall
Gekata [30.6K]

Answer:

See explaination and attachment.

Explanation:

Navier-Stokes equation is to momentum what the continuity equation is to conservation of mass. It simply enforces F=ma in an Eulerian frame.

The starting point of the Navier-Stokes equations is the equilibrium equation.

The first key step is to partition the stress in the equations into hydrostatic (pressure) and deviatoric constituents.

The second step is to relate the deviatoric stress to viscosity in the fluid.

The final step is to impose any special cases of interest, usually incompressibility.

Please kindly check attachment for step by step solution.

6 0
3 years ago
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