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dimaraw [331]
3 years ago
8

Introduce JTA and JT

Engineering
1 answer:
Vinil7 [7]3 years ago
7 0
JT2 oaoaosnforeneomdocdmsnsksmsmsodnfnfj
You might be interested in
W
malfutka [58]

Answer:

c

Explanation:

because I took it in my state test

7 0
3 years ago
Which conditions are required for nuclear fusion to begin
Thepotemich [5.8K]
Fusion processes require fuel and a confined environment with sufficient temp, pressure, and confinement time to create a plasma in which fusion can occur.
6 0
3 years ago
Researchers at the University of__________modified the iPhone to allow it to create medical images.
Morgarella [4.7K]

Answer:

B

Explanation:

I'm so sorry if this is wrong. This is what I think it is though, I had a class like this and I think this is what I said.

3 0
3 years ago
Read 2 more answers
In the following code, determine the values of the symbols here and there. Write the object code in hexadecimal. (Do not predict
allsm [11]

Answer:

Answer explained below

Explanation:

The value of here is 9

The value of there is hexadecimal value of DECO here, d = 0x39 aaaa (aaaa is the memory address of here )

We have the object code :-

let's take there address is 0x0007

0x0005 BR there :- 0x120020

0x0007 here: .WORD 9

310003 there: DECO here,d - 0x390007

310005 STOP

.END

4 0
3 years ago
On July 23, 1983, Air Canada Flight 143 required 22,300 kg of jet fuel to fly from Montreal to Edmonton. The density of jet fuel
Natasha2012 [34]

Answer:

20, 083 L

Explanation:

The mistake was the result of not using units when converting the 7862 l to Kg. They used the density in pounds hence they multiplied by 1.77 Lb/L and obtained 13597 Lb not Kg as they assumed.

To obtain the amount needed to refuel they subtracted this quantity from the 22,300 Kg required for the trip again obtaining the wrong quantity of 8703 Kg and they converted this to liters by dividing the density to get 4916 L and then placed then 5000 L of fuel

The quantity required was

7862 L * 1.77 Lb/L = 13915.74 Lb (pounds not kilos)

then converting this pounds to Kg by multiplying by  0.454 Kg/L one gets

6173 Kg on board

Amount Required

( 22,300 -6173)  :  16127 Kg

16127 Kg/ 0.803 Kg/L =  20083 L

5 0
3 years ago
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