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3 years ago

How is the angle between power strokes determined?

1 answer:

5 0

And there's an equation to help determine which configurations will work best. In a four-stroke engine, an individual piston fires every 720 degrees (two crankshaft rotations). ... The flat-four fires at 180-degree intervals, and its V angle is 180 degrees, which leads to a balance of firing forces.Jan 14, 2011

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A motor driven water pump operates with an inlet pressure of 96 kPa (absolute) and mass flow rate of 120 kg/min. The motor consu

NeX [460]

Answer:

The maximum water pressure at the discharge of the pump (exit) = 496 kPa

Explanation:

The equation expressing the relationship of the power input of a pump can be computed as:

$E _{pump,u} = \dfrac{m(P_2-P_1)}{\rho}$

where;

m = mass flow rate = 120 kg/min

the pressure at the inlet $P_1$ = 96 kPa

the pressure at the exit $P_2$ = ???

the pressure $\rho$ = 1000 kg/m³

∴

$0.8 \times 10^{3} \ W = \dfrac{(120 \ kg/min * 1min/60 s)(P_2-96000)}{1000}$

$0.8 \times 10^{3}\times 1000 = {(120 \ kg/min * 1min/60 s)(P_2-96000)}$

$800000 = {(120 \ kg/min * 1min/60 s)(P_2-96000)}$

$\dfrac{800000}{2} = P_2-96000$

400000 = P₂ - 96000

400000 + 96000 = P₂

P₂ = 496000 Pa

P₂ = 496 kPa

Thus, the maximum water pressure at the discharge of the pump (exit) = 496 kPa

8 0

2 years ago

PLEASE HELP!!! ILL GIVE BRANLIEST *EXTRA POINTS* dont skip :((

Oksi-84 [34.3K]

The first one is the answer

8 0

3 years ago

Read 2 more answers

Write a SELECT statement that returns these column names and data from the Products table: product_name The product_name column

I am Lyosha [343]

Answer:

SELECT

product_name, list_price, discount_percent,

ROUND(list_price * (discount_percent / 100), 2) AS discount_amount,

ROUND(list_price - (discount_percent / 100 * list_price), 2) AS discount_price

FROM

Products

ORDER BY (list_price - (discount_percent / 100 * list_price)) DESC

LIMIT 5;

Explanation:

In the above SELECT statement is used to select columns from Products table.

The columns are product_name,list_price and discount_percent

Next a column discount_amount is selected which is the calculated from previous two columns that are list_price and discount_percent. The discount amount is basically calculated by multiplying list_price with discount_percent/100 and the resultant column is named as discount_amount using Alias which is used to give a temporary name to set of columns or a table.

Next another column discount_price is obtained from previous three columns that are list_price , discount_percent and discount_amount as: list_price - (discount_percent / 100 * list_price) This as a whole is given a column name discount_price.

FROM is used to refer to a table from which these columns are to be selected. The table name is Product.

The result set is sorted in descending order by discount_price so ORDER BY is used to order the resultant records and DESC is used to sort these records according to the discount_price in descending order.

LIMIT statement is used to extract the records from Product and limit the number of rows returned as a result based on a limit value which is 5 here.

5 0

3 years ago

The following laboratory test results for Atterberg limits and sieve-analysis were obtained for an inorganic soil. [6 points] Si

alexira [117]

Answer: hello the complete question is attached below

answer:

A) Group symbol = SW

B) Group name = well graded sand , fine to coarse sand

C) It is not a clean sand given that ≤ 50% particles are retained on No 200

Explanation:

A) Classifying the soil according to USCS system

( using 2nd image attached below )

description of sand :

The soil is a coarse sand since ≤ 50% particles are retained on No 200 sieve, also

The soil is a sand given that more than 50% particles passed from No 4 sieve

The soil can be a clean sand given that fines ≤ 12%

The soil can be said to be a well graded sand because the percentage of particles passing through decreases gradually over time

Group symbol as per the 2nd image attached below = SW

B) Group name = well graded sand , fine to coarse sand

C) It is not a clean sand given that ≤ 50% particles are retained on No 200

5 0

3 years ago

The human eye, as well as the light-sensitive chemicals on color photographic film, respond differently to light sources with di

jeka57 [31]

Answer:

a) at T = 5800 k

band emission = 0.2261

at T = 2900 k

band emission = 0.0442

b) daylight (d) = 0.50 μm

Incandescent ( i ) = 1 μm

Explanation:

To Calculate the band emission fractions we will apply the Wien's displacement Law

The ban emission fraction in spectral range λ1 to λ2 at a blackbody temperature T can be expressed as

F ( λ1 - λ2, T ) = F( 0 ----> λ2,T) - F( 0 ----> λ1,T )

Values are gotten from the table named: blackbody radiation functions

a) Calculate the band emission fractions for the visible region

at T = 5800 k

band emission = 0.2261

at T = 2900 k

band emission = 0.0442

attached below is a detailed solution to the problem

b)calculate wavelength corresponding to the maximum spectral intensity

For daylight ( d ) = 2898 μm *k / 5800 k = 0.50 μm

For Incandescent ( i ) = 2898 μm *k / 2900 k = 1 μm

3 0

3 years ago