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Ivan
2 years ago
7

Write multiple if statements: If carYear is before 1968, print "Probably has few safety features." (without quotes). If after 19

71, print "Probably has head rests.". If after 1991, print "Probably has anti-lock brakes.". If after 2000, print "Probably has tire-pressure monitor.". End each phrase with period and newline. Ex: carYear
Engineering
1 answer:
Rzqust [24]2 years ago
3 0

Answer:

The solution code is written in Python

  1. if(carYear < 1968):
  2.    print("Probably has few safety features.\n")
  3. if(carYear > 1971):
  4.    print("Probably has head rests\n")
  5. if(carYear > 1991):
  6.    print("Probably has anti-lock brakes.\n")
  7. if(carYear > 2000):
  8.    print("Probably has tire-pressure monitor.\n")

Explanation:

To create if statement, we just use the keyword "if" and joined with an operator either < (before) or > (after). Then we can use the print function to display the message based on the condition that met.

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A non-licensed person may be the SOLE owner of a civil, electrical, or mechanical engineering business under which of the follow
kotegsom [21]

Answer:

(d) None. No provisions exist.

Explanation:

B&P Code § 6738 prohibits a non-licensed person from being the sole proprietor of an engineering business. The non-licensed can be a partner in an engineering business that offers civil, electrical, or mechanical services. It is mandatory that at least one licensed engineer must be a co-owner of the business.

5 0
2 years ago
Movies( title, year, length, genre, studioName, producerC#) StarsIn(movieTitle, movie Year, starName) MovieStar(name, address, g
ehidna [41]

Answer:

a) ALTER TABLE Movies  

Include CONSTRAINT FK_Producer  

Outside KEY (producer#C) REFERENCES MovieExec(cert#);  

b) Here we will make the segment as NOTNULL.  

Modify TABLE Movies  

Modify COLUMN producer#C varchar2 NOT NULL;  

c) The outside key falling choices figure out what activities the database motor should take on the off chance that you attempt to erase or refresh information in the referenced sections in the parent table.  

Modify TABLE Movies  

Include CONSTRAINT FK_Producer  

Remote KEY (producer#C) REFERENCES MovieExec(cert#)  

ON DELETE CASCADE  

ON UPDATE CASCADE;  

d) ALTER TABLE StarsIn  

Include CONSTRAINT FK_mname  

Remote KEY (movieTitle) REFERENCES Movies(title);  

e) ALTER TABLE StarsIn  

Include CONSTRAINT FK_sname  

Remote KEY (starname) REFERENCES MovieStar(name)  

ON DELETE CASCADE;

7 0
2 years ago
Determine the required dimensions of a column with a square cross section to carry an axial compressive load of 6500 lb if its l
exis [7]

Answer: 0.95 inches

Explanation:

A direct load on a column is considered or referred to as an axial compressive load. A direct concentric load is considered axial. If the load is off center it is termed eccentric and is no longer axially applied.

The length= 64 inches

Ends are fixed Le= 64/2 = 32 inches

Factor Of Safety (FOS) = 3. 0

E= 10.6× 10^6 ps

σy= 4000ps

The square cross-section= ia^4/12

PE= π^2EI/Le^2

6500= 3.142^2 × 10^6 × a^4/12×32^2

a^4= 0.81 => a=0.81 inches => a=0.95 inches

Given σy= 4000ps

σallowable= σy/3= 40000/3= 13333. 33psi

Load acting= 6500

Area= a^2= 0.95 ×0.95= 0.9025

σactual=6500/0.9025

σ actual < σallowable

The dimension a= 0.95 inches

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