In 2008 502 motorcyclists fire in Florida - an increase from<br> The number killed in 2004

Four race cars are traveling on a 2.5-mile tri-oval track. The four cars are traveling at constant speeds of 195 mi/h, 190 mi/h,

Snezhnost [94]

Answer:

Explanation:

1) The number of times, the car with the speed of 195 mph will cross the given point is equal to 30 minutes divided by the time taken by car to cross the 2.5 miles.

0 .5*195/2.5 = 39

Likewise, the car with the speed of 190 mph crosses the point 38 times; the car with the speed of 185 mph crosses the point 37 times

and car with the speed of 180 mph crosses it 36 times

here, the time-mean speed, vt is given below,

vt = (39*195 +38*190+37*185+36*180)/(39+38+37+38)

= 186.433 mph

and space mean speed is given by,

= (39+38+37+36)/(39/195+38/190+37/1850+36/180)

1) The number of times, the car with the speed of 195 mph will cross the given point is equal to 30 minutes divided by the time taken by car to cross the 2.5 miles.

0 .5*195/2.5 = 39

Likewise, the car with the speed of 190 mph crosses the point 38 times; the car with the speed of 185 mph crosses the point 37 times

and car with the speed of 180 mph crosses it 36 times

here, the time-mean speed, vt is given below,

vt = (39*195 +38*190+37*185+36*180)/(39+38+37+38)

= 186.433 mph

and space mean speed is given by,

= (39+38+37+36)/(39/195+38/190+37/1850+36/180)

=187.5 mph

2) There would be only four number of observations when the aerial photo is given, therefore time mean speed, vt in that condition will be calculated as

Vt = 195+190+185+180/4

= 187.5

Vs= 4/(1/195+1/190+1/185+1/180)

= 188.36 mph

2) There would be only four number of observations when the aerial photo is given, therefore time mean speed, vt, in that condition will be calculated as

Vt = 195+190+185+180/4

= 187.5

Vs= 4/(1/195+1/190+1/185+1/180)

= 188.36 mph

4 0

4 years ago

Write a SELECT statement that returns these four columns:

labwork [276]

A SELECT command pulls zero or more rows from one or more database tables or views. SELECT is the most often used data manipulation language (DML) command in most applications. See the statement required below.

<h3>What is the SELECT Statement that gives the above results?</h3>

SELECT VendorName, InvoiceNumber, InvoiceDate,

InvoiceTotal - PaymentTotal - CreditTotal AS Balance

FROM Vendors JOIN Invoices

ON Vendors.VendorID = Invoices.VendorID

WHERE InvoiceTotal - PaymentTotal - CreditTotal > 0

ORDER BY VendorName;

Learn more about SELECT Statements at;

brainly.com/question/19338967
#SPJ1

3 0

2 years ago

A skilled worker with the ability to operate computer numerically controlled (CNC) machines is qualified to work in which of the

KengaRu [80]

Answer:

Machinist

Explanation:

A skilled worker with the ability to operate computer numerically controlled (CNC) machines is qualified to work in a machinist position.

A machinist is a person who is properly skilled and consists of advanced knowledge regarding the functions of a CNC machine. He can use different mechanisms and complex numerical functions of the machine to carry out different tasks. Any person who lacks the official learning of mechanisms cannot operate such machines effectively.

3 0

3 years ago

Design the necessary circuit using Logisim to implement the situation described above. Use Kmaps for simplification. Be VERY car

Phantasy [73]

Answer:

The circuit is shown in my attachments, please take a good look

3 0

3 years ago

Read 2 more answers

Steam enters an adiabatic turbine at 10MPa and 500 C and leaves at 10 kPa with a quality of 90%. Neglecting the changes in kinet

Amanda [17]

Answer:

flow ( m ) = 4.852 kg/s

Explanation:

Given:

- Inlet of Turbine

P_1 = 10 MPa

T_1 = 500 C

- Outlet of Turbine

P_2 = 10 KPa

x = 0.9

- Power output of Turbine W_out = 5 MW

Find:

Determine the mass ow rate required

Solution:

- Use steam Table A.4 to determine specific enthalpy for inlet conditions:

P_1 = 10 MPa

T_1 = 500 C ---------- > h_1 = 3375.1 KJ/kg

- Use steam Table A.6 to determine specific enthalpy for outlet conditions:

P_2 = 10 KPa -------------> h_f = 191.81 KJ/kg

x = 0.9 -------------> h_fg = 2392.1 KJ/kg

h_2 = h_f + x*h_fg

h_2 = 191.81 + 0.9*2392.1 = 2344.7 KJ/kg

- The work produced by the turbine W_out is given by first Law of thermodynamics:

W_out = flow(m) * ( h_1 - h_2 )

flow ( m ) = W_out / ( h_1 - h_2 )

- Plug in values:

flow ( m ) = 5*10^3 / ( 3375.1 - 2344.7 )

flow ( m ) = 4.852 kg/s

3 0

3 years ago

In 2008 502 motorcyclists fire in Florida - an increase from The number killed in 2004