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padilas [110]
2 years ago
14

Holmes owns two suits: one black and one tweed. He always wears either a tweed suit or sandals. Whenever he wears his tweed suit

and a purple shirt, he chooses to not wear a tie. He never wears the tweed suit unless he is also wearing either a purple shirt or sandals. Whenever he wears sandals, he also wears a purple shirt. Yesterday, Holmes wore a bow tie. What else did he wear?
Engineering
1 answer:
vazorg [7]2 years ago
4 0

Answer:

He wore his black suit, another color of shirt (not purple) and shoes

Explanation:

Holmes owns two suits: one black and one tweed.

Whenever he wears his tweed suit and a purple shirt, he chooses not to wear a tie and whenever he wears sandals, he always wears a purple shirt.

So, if he wore a bow tie yesterday, it means he wore his black suit, another color of shirt (not purple) and shoes because the shirt color is not purple

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The pressure drop across a valve through which air flows is expected to be 10 kPa. If this differential were applied to the two
ozzi

Answer:hit that soulja boy

Explanation:

6 0
2 years ago
Suppose we store a relation R (x,y) in a grid file. Both attributes have a range of values from 0 to 1000. The partitions of thi
leva [86]

Answer:

For (a) The total number of buckets from the given query for the relation is 25 buckets (b) the nearest neighboring query is (80, 200) (80, 150), (100, 150), (120,150) and (120, 200)

Explanation:

From the question stated, we need to define what a Grid file is

Grid File it is a structure of data that are used to divide the total space into a grid non-periodic, where set of point (small) are defined by more than one cells of the grid.

(a)Finding buckets for the query

The relation is divided into two parts which ranges from 0 to 1000, the first part is partitioned in every 20 units, at 20, 40, 60 etc; a second part is partitioned into every 50 units at 50, 100, 150 etc.

The total number of buckets from the given query for the relation is 25 buckets

(b)Finding the closest point or nearest point

The closest point discovered in the distance is little above 15

These points are are the points closer to the point target (110, 205) which can be found in five neighboring rectangles with left corners lower is stated as follows:

(80, 200) (80, 150), (100, 150), (120,150) and (120, 200)

3 0
2 years ago
A shaft consisting of a steel tube of 50-mm outer diameter is to transmit 100 kW of power while rotating at a frequency of 34 Hz
nikitadnepr [17]

Answer:

25 - \sqrt[4]{26.66*10^{-8} }  mm

Explanation:

Given data

steel tube : outer diameter = 50-mm

power transmitted = 100 KW

frequency(f) = 34 Hz

shearing stress ≤ 60 MPa

Determine tube thickness

firstly we calculate the ; power, angular velocity and torque of the tube

power = T(torque) * w (angular velocity)

angular velocity ( w ) = 2\pif = 2 * \pi * 34 = 213.71

Torque (T) = power / angular velocity = 100000 / 213.71 = 467.92 N.m/s

next we calculate the inner diameter  using the relation

  \frac{J}{c_{2}  } = \frac{T}{t_{max} }  = 467.92 / (60 * 10^6) =  7.8 * 10^-6 m^3

also

c2 = (50/2) = 25 mm

\frac{J}{c_{2} } = \frac{\pi }{2c_{2} } ( c^{4} _{2} - c^{4} _{1} ) =  \frac{\pi }{0.050} [ ( 0.025^{4} - c^{4} _{1}  ) ]

therefore; 0.025^4 - c^{4} _{1} = 0.050 / \pi (7.8 *10^-6)

c^{4} _{1} = 39.06 * 10 ^-8 - ( 1.59*10^-2 * 7.8*10^-6)

    39.06 * 10^-8 - 12.402 * 10^-8 =26.66 *10^-8

c_{1} = \sqrt[4]{26.66 * 10^{-8} }  =

THE TUBE THICKNESS

c_{2} - c_{1} = 25 - \sqrt[4]{26.66*10^{-8} }  mm

4 0
2 years ago
A horizontal, cylindrical, tank, with hemispherical ends, is used to store liquid chlorine at 10 bar. The vessel is 4 m internal
sp2606 [1]

Answer: 0.021818m =2.18x10^-²m

Explanation: Using Laplace principle

Design pressure is =12 bar= 1200000N/m²

T = Wall thickness

Design stress=110Mn/m²= 110,000,000N/m²

Radius= diameter/2 =4/2=2m

Design stress=Hoop stress =Pr/t where p=internal pressure or internal pressure, r=radius and t= wall thickness.

As Laplace equation stated.

110,000,000= 1,2000,00 x 2/t

t= 2,400,000/110,000,000.

t= 0.021818m

t=2.18x10^-2m.

3 0
2 years ago
If water molecules pass through a membrane with a steady state flux of 220 mole/(m2 day), how long will it take, in hours, for 0
goblinko [34]

Answer:

<em>0.0386 hr</em>

<em></em>

Explanation:

Area = 565 cm^2 = 0.0565 m^2  (1 cm^2 = 0.0001 m^2)

flux state rate = 220 mole/m^2-day

<em>There are 24 hrs in a day,</em> therefore rate in hrs will be

220/24 = 9.17 mole/m^2-hr

mass of water = 0.4 kg

molar mass of water = (1 x 2) + 16 = 18 kg/mole

therefore,

<em>mole of water = mass of water/molar mass of water</em>

mole of water = 0.4/18 = 0.02 mole

<em>mole flux = mole/area</em> = 0.02/0.0565 = 0.354 mol/m^2

<em>time that will be taken will be for water to pass = mole flux/mole flux rate</em>

time = 0.354/9.17 = <em>0.0386 hr</em>

7 0
2 years ago
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