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serg [7]
3 years ago
7

A bouncing belt tensioner indicates what

Engineering
1 answer:
Anvisha [2.4K]3 years ago
4 0

Answer:

worn down tensioner

Explanation:

when your belt tensioner is bouncing it is worn out the tensioner belt pulley arm is worn out due to a spring that has gone bad this will cause the tensioner to bounce you will want to replace the belt tensioner

You might be interested in
The Clausius inequality expresses which of the following laws? i. Law of Conservation of Mass ii. Law of Conservation of Energy
DanielleElmas [232]

Answer:

(iv) second law of thermodynamics

Explanation:

The Clausius  inequality expresses the second law of thermodynamics it applies to the real engine cycle.It is defined as the cycle integral of change in entropy of a reversible system is zero. It is nothing but mathematical form of second law of thermodynamics . It also states that for irreversible process the cyclic integral of change in entropy is less than zero

3 0
3 years ago
When preparing a foundation for a heavy duty machine tool, discuss any four (4) statics machine characteristics to be considered
kozerog [31]

Answer:

1 ) Accuracy of the Machine Tool

2) Load bearing capacity

3) Linearity in the product line

4) Torque of the machine

Explanation:

we know that machine tool is the permanent essential in manufacturing industries

it is a machine use for different form like cutting , grinding and boring etc

so 1st is

1 ) Accuracy of the Machine Tool

we know it is very important Characteristic of the machine tool because when we use it in manufacturing unit Accuracy of the Machine Tool should be higher concern

2) Load bearing capacity

we should very careful about Load bearing capacity because how much amount of load tool will bear check by some parameter like creep , shear stress and strength etc

3) Linearity in the product line

Linearity in the product line mean that it should be group of related product produced by the any one of the manufacturer otherwise it will take time or it may be intermixing

4) Torque of the machine

we know that Torque is a rotational force or a turning force so amount of force multiplied by the distance of the operation

and we know torque per second give the power rating of machine tool

5 0
4 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
Bằng lý luận và thực tiễn, bạn hãy trình bày ý kiến của mình về quan điểm sau đây: "Quá trình dạy học là quá trình truyền thụ tr
mrs_skeptik [129]

Answer:

can you please ask in English I can't understand this language

7 0
3 years ago
A three-point bending test is performed on a glass specimen having a rectangular cross section of height d 5 mm (0.2 in.) and wi
Anon25 [30]

Answer:

The flexural strength of a specimen is = 78.3 M pa

Explanation:

Given data

Height = depth = 5 mm

Width = 10 mm

Length L = 45 mm

Load = 290 N

The flexural strength of a specimen is given by

\sigma = \frac{3 F L}{2 bd^{2} }

\sigma = \frac{3(290)(45)}{2 (10)(5)^{2} }

\sigma = 78.3 M pa

Therefore the flexural strength of a specimen is = 78.3 M pa

4 0
3 years ago
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