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dusya [7]
4 years ago
9

Colonial Adventure Tours calculates the total price of a trip by adding the trip price plus other fees and multiplying the resul

t by the number of persons included in the reservation. List the reservation ID, trip name, customer’s last name, and total cost for all reservations where the number of persons is greater than four. Use the column name TotalCost for the calculated field.
Engineering
1 answer:
Vilka [71]4 years ago
3 0

Answer:

The following query is used to display TripName. Reservationld, FirstName. LastName and TotalCost of trip by adding the trip price plus other fees and multiplying the result by the number of persons which have number of persons >4.

Query:

SELECT ReservationlD, Trip.TripName. Customer.LastName. Customer.FirstName. (TripPrice+OtherFees) 'NumPersons as TotalCost FROM Reservation, Trip, Customer WHERE NumPersons>4 AND Reservation.TriplD=Trip.TriplD AND

Customer. CustomerNum=Reservation.CustomerNum:

Explanation:

  • Select clause is used to retrieve data from specified database table or relation and returns the data in the form of table.
  • ReservationID. Trip.TripName. Customer.LastName. Customer.FirstName are the column name of table.
  • (TripPrice+OtherFees) 'NumPersons will calculate the total cost of the Trip and stored it into TotalCost column.
  • As clause is used to give new name TotalCost to resultant column.
  • FROM clause specifies one or more table from where records to be retrieved. o Reservation. Trip and Customer are the table name.
  • WHERE clause is used in SQL query to retrieve only those records that satisfy the specified condition

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An alloy is evaluated for potential creep deformation in a short-term laboratory experiment. The creep rate (ϵ˙) is found to be
cupoosta [38]

Answer:

Activation energy for creep in this temperature range is Q = 252.2 kJ/mol

Explanation:

To calculate the creep rate at a particular temperature

creep rate, \zeta_{\theta} = C \exp(\frac{-Q}{R \theta} )

Creep rate at 800⁰C, \zeta_{800} = C \exp(\frac{-Q}{R (800+273)} )

\zeta_{800} = C \exp(\frac{-Q}{1073R} )\\\zeta_{800} = 1 \% per hour =0.01\\

0.01 = C \exp(\frac{-Q}{1073R} ).........................(1)

Creep rate at 700⁰C

\zeta_{700} = C \exp(\frac{-Q}{R (700+273)} )

\zeta_{800} = C \exp(\frac{-Q}{973R} )\\\zeta_{800} = 5.5 * 10^{-2}  \% per hour =5.5 * 10^{-4}

5.5 * 10^{-4}  = C \exp(\frac{-Q}{1073R} ).................(2)

Divide equation (1) by equation (2)

\frac{0.01}{5.5 * 10^{-4} } = \exp[\frac{-Q}{1073R} -\frac{-Q}{973R} ]\\18.182= \exp[\frac{-Q}{1073R} +\frac{Q}{973R} ]\\R = 8.314\\18.182= \exp[\frac{-Q}{1073*8.314} +\frac{Q}{973*8.314} ]\\18.182= \exp[0.0000115 Q]\\

Take the natural log of both sides

ln 18.182= 0.0000115Q\\2.9004 = 0.0000115Q\\Q = 2.9004/0.0000115\\Q = 252211.49 J/mol\\Q = 252.2 kJ/mol

3 0
3 years ago
An 800-kg drag racer accelerates from rest to 390 km/hr in 5.8 s. What is the net impulse applied to the racer in the first 5.8
marissa [1.9K]

Answer:

Impulse =14937.9 N

tangential force =14937.9 N

Explanation:

Given that

Mass of car m= 800 kg

initial velocity u=0

Final velocity v=390 km/hr

Final velocity v=108.3 m/s

So change in linear momentum P= m x v

           P= 800 x 108.3

 P=86640 kg.m/s

We know that impulse force F= P/t

So F= 86640/5.8 N

F=14937.9 N

Impulse force F= 14937.9 N

We know that

v=u + at

108.3 = 0 + a x 5.8

a=18.66\ m/s^2

So tangential force F= m x a

F=18.66 x 800

F=14937.9 N

6 0
4 years ago
A missile flying at high speed has a stagnation pressure and temperature of 5 atm and 598.59 °R respectively. What is the densit
alexdok [17]

Answer:

5.31\frac{kg}{m^3}

Explanation:

Approximately, we can use the ideal gas law, below we see how we can deduce the density from general gas equation. To do this, remember that the number of moles n is equal to \frac{m}{M}, where m is the mass and M the molar mass of the gas, and the density is \frac{m}{V}.

For air M=28.66*10^{-3}\frac{kg}{mol} and \frac{5}{9}R=K

So, 598.59 R*\frac{5}{9}=332.55K

pV=nRT\\pV=\frac{m}{M}RT\\\frac{m}{V}=\frac{pM}{RT}\\\rho=\frac{pM}{RT}\\\rho=\frac{(5atm)28.66*10^{-3}\frac{kg}{mol}}{(8.20*10^{-5}\frac{m^3*atm}{K*mol})332.55K}=5.31\frac{kg}{m^3}

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Answer:

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Answer:

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