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Hypersonic scramjet. On March 27, 2004, the United States
successfully tested the hypersonic X-43A scramjet, which flew at
Mach 7 (seven times the speed of sound) for 11 seconds. (b) How many kilometers
did the scramjet travel during its 11 second test?


Sagot :

25.487 km did the scramjet travel during its 11 second test.

Given that the scramjet travels at a speed seven times that of sound, speed v is equal to 7 x 331 m/s or 2317 m/s.

(A) We require the scramjet's travel time over a distance of

       d = 5000 km (5x10^6 m).

Using the formula

time taken = distance traveled/speed,

we can calculate the following:

(5x10^6)/2317 sec

= (5x10^6)/(23175x60) minute

= 5000000/139020 minutes

= 35.97 seconds

As 1 second equals 1/60 of a minute, Therefore, the travel time on Scramjet from San Francisco to New York will be 35.97 minutes.

(B) The distance covered by the Scramjet in 11 seconds must now be determined. as a result,

using speed, distance, and time relation.

Distance = Speed x Time

               = 7  x 331  x 11

                =  25487 m

Since 1 metre equals 1/1000 of a kilometer, 25487 m of 25.487 km

Learn more about Distance here:

https://brainly.com/question/15172156

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