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drek231 [11]
3 years ago
6

Give four effects of water hammer.​

Engineering
1 answer:
makkiz [27]3 years ago
4 0

Explanation:

The hammer effect (or water hammer) can harm valves, pipes, and gauges in any water, oil, or gas application. It occurs when the liquid pressure is turned from an on position to an off position abruptly. When water or a liquid is flowing at full capacity there is a normal, even sound of the flow.

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A _____ satellite system employs many satellites that are spaced so that, from any point on the Earth at any time, at least one
Wittaler [7]

Answer:

d. low earth orbit (LEO)

Explanation:

This type of satellites form a constellation deployed as a series of “necklaces” in such a way that at any time, at least one satellite is visible by a receiver antenna, compensating the movement due to the earth rotation.

Opposite to that, a geostationary satellite is at an altitude that makes it  like a fixed point over the Earth´s equator, rotating synchronously with the Earth, so it is always visible in a given area.

3 0
3 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
2 years ago
g If the rails are originally laid in contact, what is the stress in them on a summer day when their temperature is 31.0
Anettt [7]

A) The amount of space must be left between adjacent rails if they are just to touch on a summer day when their temperature is 31.0 °C is; 0.6048 cm

B) The stress in the rails on a summer day when their temperature is 31.0 °C is; 86.4 × 10⁶ Pa

<h3>Linear Thermal Expansion</h3>

We are given;

Length; L = 14 m

Initial Temperature; T_i = −5 °C

Final Temperature; T_f = 31 °C

The formula for Linear Thermal Expansion is;

ΔL = L_i * α * ΔT

where;

L_i is initial length

α is thermal expansion

ΔL is change in length

ΔT is change in temperature

Now, the thermal expansion of steel from online tables is α = 1.2 × 10⁻⁵ C⁻¹

Thus;

ΔL = 14 * 1.2 × 10⁻⁵  * (31 - (-5))

ΔL = 6.048 × 10⁻³ m = 0.6048 cm

The formula to get the stress is;

σ = Y * α  * ΔT

where;

Y is young's modulus of steel = 20 × 10¹⁰ Pa

α is thermal expansion

ΔT is change in temperature

Thus;

σ = 20 × 10¹⁰ × 1.2 × 10⁻⁵ × (31 - (-5))

σ = 86.4 × 10⁶ Pa

The complete question is;

Steel train rails are laid in 14.0-m long segments placed end to end. The rails are laid on a winter day when their temperature is −5 °C.

(a) How much space must be left between adjacent rails if they are just to touch on a summer day when their temperature is 31.0 °C?

(b) If the rails are originally laid in contact, what is the stress in them on a summer day when their temperature is 31.0 °C?

Read more about Linear Thermal Expansion at; brainly.com/question/6985348

4 0
2 years ago
Railroad tracks made of 1025 steel are to be laid during the time of year when the temperature averages 4C (40F). Of a joint spa
DENIUS [597]

Answer:

41.5° C

Explanation:

Given data :

1025 steel

Temperature = 4°C

allowed joint space = 5.4 mm

length of rails = 11.9 m

<u>Determine the highest possible temperature </u>

coefficient of thermal expansion ( ∝ ) = 12.1 * 10^-6 /°C

Applying thermal strain ( Δl / l )  = ∝ * ΔT

                                    ( 5.4 * 10^-3 / 11.9 )  = 12.1 * 10^-6 * ( T2 - 4 )

∴  ( T2 - 4 ) =  ( 5.4 * 10^-3 / 11.9 ) / 12.1 * 10^-6

hence : T2 = 41.5°C

8 0
3 years ago
Determine the voltages at all nodes and the currents through all branches. Assume that the transistor B is 100,
iren [92.7K]

Answer:

The voltages of all nodes are, IE = 4.65 mA, IB =46.039μA,  IC=4.6039 mA, VB = 10v, VE =10.7, Vc =4.6039 v

Explanation:

Solution

Given that:

V+ = 20v

Re = 2kΩ

Rc = 1kΩ

Now we will amke use of the method KVL in the loop.

= - Ve + IE . Re + VEB + VB = 0

Thus

IE = V+ -VEB -VB/Re

Which gives us the following:

IE = 20-0.7 - 10/2k

= 9.3/2k

so, IE = 4.65 mA

IB = IE/β +1 = 4.65 m /101

Thus,

IB = 0.046039 mA

IB = 46.039μA

IC =βIB

Now,

IC = 100 * 0.046039

IC is 4.6039 mA

Now,

VB = 10v

VE = VB + VEB

= 10 +0.7 = 10.7 v

So,

Vc =Ic . Rc = 4.6039 * 1k

=4.6039 v

Finally, this is the table summary from calculations carried out.

Summary Table

Parameters          IE       IC           IB            VE       VB         Vc

Unit                     mA     mA          μA            V           V          V

Value                  4.65    4.6039   46.039    10.7      10     4.6039

4 0
2 years ago
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