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zysi [14]
3 years ago
13

Construction of a reservoir behind a dam Group of answer choices

Engineering
1 answer:
harkovskaia [24]3 years ago
8 0

Answer:

c. can destablize the slopes by increasing pore fluid pressure.

Explanation: A reservoir is a term used in Agriculture or geography to describe a location where water is either collected artificially or naturally for later use in farming or to act as dam for the supply of water in large communities. Most countries of the world make use of reservoirs to store water for future use.

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Reverse masking forms a soft edge on the panel.
valina [46]
True I think I’m not sure?
7 0
2 years ago
Consider the following relational database that Best Airlines uses to keep track of its mechanics, their skills, and their airpo
Musya8 [376]

Answer:

Explanation:

A)

SELECT MECHNAME,AGE FROM MECHANIC;

B)

SELECT AIRNAME,SIZE FROM AIRPORT WHERE SIZE>=20 AND STATE='CALIFORNIA' AND YEAROPENED >=1935 ORDER BY SIZE ASC;

C)

SELECT AIRNAME,SIZE FROM AIRPORT WHERE (SIZE>=20 OR YEAROPENED >=1935) AND STATE='CALIFORNIA';

D)

SELECT AVG(SIZE) FROM AIRPORT WHERE STATE='CALIFORNIA' AND YEAROPENED >=1935;

E)

SELECT COUNT(AIRNAME) FROM AIRPORT WHERE STATE='CALIFORNIA' AND YEAROPENED >=1935;

F)

SELECT COUNT(AIRNAME),STATE FROM AIRPORT WHERE YEAROPENED>=1935 GROUP BY STATE;

G)

SELECT COUNT(AIRNAME),STATE FROM AIRPORT WHERE YEAR OPENED>=1935 GROUP BY STATE HAVING COUNT(*)>=5;

H)

SELECT MECHNAME FROM MECHANIC A JOIN AIRPORT B

ON A.AIRNAME=B.AIRNAME AND B.STATE='CALIFORNIA';

I)  

SELECT MECHNAME FROM MECHANIC A

JOIN QUALIFICATION B

ON A.MECHNUM=B.MECHNUM

AND B.PROFRATE=4

JOIN SKILL C

ON B.SKILLNUM=C.SKILLNUM

AND SKILLNAME='FAN BLADE RELACEMENT';

J)  SELECT MECHNAME FROM MECHANIC A

JOIN QUALIFICATION B

ON A.MECHNUM=B.MECHNUM

AND B.PROFRATE=4

JOIN SKILL C

ON B.SKILLNUM=C.SKILLNUM

AND SKILL NAME='FAN BLADE REPLACEMENT'

JOIN AIRPORT D

ON A.AIRNAME=D.AIRNAME

AND STATE='CALIFORNIA';

K)   SELECT SUM(SALARY),CITY FROM MECHANIC A

JOIN AIRPORT B

ON A.AIRNAME=B.AIRNAME

AND STATE='CALIFORNIA'

GROUP BY CITY;

L)   SELECT MAX(SIZE) FROM AIRPORT ;

M)  SELECT MAX(SIZE) FROM AIRPORT WHERE STATE='CALIFORNIA';

6 0
3 years ago
1. Fatigue equations are based solely on theoretical assumptions. Experimental data is only used to verify the theory. a. True.b
Rainbow [258]

Answer:

1.  b. False

2. b. False

3.  b. False

4.  b. False

5. a. True

6. a. True

7.  b. False

8.  b. False

9. a. True

Explanation:

1. The fatigue properties of a material  are determined by series of test.

2. For most steels there is a level of fatigue limit below which a component will survive an infinite number of cycles, for aluminum and titanium a fatigue limit can not be defined, as failure will eventually occur after enough experienced cycles.

3. Although there is a cyclic stress, there are also stresses complex circumstances involving tensile to compresive and constant stress, where the solution is given into the mean stress and the stress amplitude or stress range, which is double the stress amplitude.

4. Low‐cycle fatigue is defined as few thousand cycles and high cycle fatigue is around more than 10,000 cycles.

5. The number of cycles for failure on brittle materials are less and determined compared with the ductile materials.

6.  The bending fatigue could be handled with specific load requirements  for uniform bending or axial fatigue of the same section size where the material near the surface is subjected to the  maximum stress, as in torsional fatigue, which can be performed on  axial-type specially designed machines also, using the proper fixtures if  the maximum twist required is small, in which linear motion is changed to rotational motion.

7.  A SN-Curve for a given material, is a plot displayed on logarithmic scales of the magnitude of an alternating stress in relation to the number of cycles to failure

8. The strain life method measures the strain resistance of local stresses and strains around stress concentration that controls the fatigue life of the material. It is more accurate than determining fatigue performance as the stress-life method is for long life millions of cycles in elastic stresses, but an it gets an effective stress concentration in fatigue loading.

9. Linear Elastic Fracture Mechanics (LEFM) states that the material is isotropic and linear elastic so, when the stresses near the crack surpasses the material fracture toughness, the crack grows.

7 0
3 years ago
A heat pump cycle whose coefficient of performance is 2.5 delivers energy by heat transfer to a dwelling at a rate of 20kW.
12345 [234]

Answer:

a) 8kW

b) $128

Explanation:

Given the coefficient of performance of the heat pump cycle to be 2.5

Energy delivered by the heat pump = 20kW

a) net power required to operate the heat pump = Energy delivered / coefficient of performance

Net power required = 20/2.5

= 8kW

b) Given the cost of electricity is $0.08 for 1kWhour

Since net power required to operate heat pump = 8kW

If the heat pump operate for 200hours, total power required for a month = 8kW×200hours = 1600kWhour

since 1kWh of electricity costs $0.08, cost of electricity used in a month when the pump operates for 200hour will be 1600kWh×$0.08 which is equivalent to $128

8 0
3 years ago
Befor operating any plant,machinery or equipment,you should​
SIZIF [17.4K]

Answer:

Always make sure that you are properly protected and that it is all clear to be operated on.

3 0
3 years ago
Read 2 more answers
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