Which statement below can be used to read data from a file one character at a time?

Given a two-dimensional steady inviscid air flow field with no body forces described by the velocity field given below. Assuming

kolbaska11 [484]

Answer:

the pressure gradient in the x direction = -15.48Pa/m

Explanation:

The concept of partial differentiation was used in the determination of the expression for u and v.
each is partially differentiated with respect to x and the appropriate substitution was done to get the value of the pressure gradient as shown in the attached file.

4 0

3 years ago

What is the resultant force on one side of a 25cm diameter circular plate standing at the bottom of 3m of pool water?

Tom [10]

Answer:

F=1.47 KN

Explanation:

Given that

Diameter of plate = 25 cm

Height of pool h = 3 m

We know that force can be given as

F= P x A

P=ρ x g x h

Now by putting the values

P=1000 x 10 x 3

P= 30 KPa

$A=\dfrac{\pi}{4}\times 0.25^2\ m^2$

$A=0.049\ m^2$

F= 30 x 0.049 KN

F=1.47 KN

So the force on the plate will be 1.47 KN.

4 0

3 years ago

Which career best fits the group of the words below? High voltage, lines

Llana [10]

Answer:

I forget the word for it, but probably the guys who set up the power lines in the city.

Explanation:

5 0

3 years ago

Read 2 more answers

-Mn has a cubic structure with a0 = 0.8931 nm and a density of 7.47 g/cm3. -Mn has a different cubic structure, with a0 = 0.63

Fudgin [204]

Answer:

The percentage volume change is -3.0%

Explanation: We are to determine the percentage change that will occurs is alpha-Mn is transformed to beta-Mn

Value are defined as;

Cubic structure (a0) for alpha-Mn = 0.8931nm = 0.8931e-9m = 7.1236e-28cm3

Cubic structure (a0) for beta-Mn = 0.6326nm = 0.6326e-9m = 2.5316e-28cm3

Density of alpha-Mn = 7.47g/cm3

Density of beta-Mn = 7.26g/cm3

Atomic weight of Mn = 54.938g/mol

Atomic radius of Mn = 0.112nm

STEP1: CALCULATE THE ATOM NUMBER PER CELL IN THE ALPHA-Mn;

Atom per cell = (density × cubic structure × Avogadro's constant) ÷ (atomic weight ) × 100000

(7.47× 7.1236e-28 × 6.02e23) ÷ 54.938 = 58.31

Therefore the number of Atom in alpha-Mn is 58.31 atom per cell

STEP2: CALCULATE THE NUMBER OF ATOM PER CELL IN THE BETA-Mn

Atom per cell = (density × cubic structure × Avogadro's constant) ÷ (atomic weight) × 1000000

(7.26 × 2.5316e-28 × 6.02e23) ÷ 54.938 = 20.14

Therefore the number of Atom in beta-Mn is 20.14 atom per cell

STEP3: CALCULATE THE PERCENTAGE VOLUME OF ALPHA-Mn AND BETA-Mn

V% = [(volume of atom × number of atom per cell) ÷ volume of unit cell] × 1000

For Alpha-Mn:

[(1.4049e-30 × 58.31) ÷ 7.1236e-28] × 1000 = 114.998%

For Beta-Mn:

[(1.4049e-30 × 20.14) ÷ 2.5316e-28] × 1000 = 111.766%

STEP4: CALCULATE THE CHANGE IN PERCENTAGE VOLUME FOR ALPHA TO TRANSFORM TO BETA

change = final state - initial state

Therefore;

Change = 111.766 - 114.998 = -3.23%

Therefore for a transformation of Alpha-Mn to Beta-Mn they will be a decrease in volume

3 0

3 years ago

Ann’s Retail, a women’s clothing store, hires female attendants to assist clients in the store’s dressing rooms. Larry, a male,

mojhsa [17]

Answer:

A bona fide occupational qualification defense

Explanation:

Since the store is for women clothing, the retail may prefer to employ only female to assist the customers. Under a bona fide occupational qualification defense, an employer is allowed to discriminate if a characteristic is a necessity for the performance of the job and for the business. Therefore, the store has a bona fide occupational qualification defense.

3 0

3 years ago