Ask question

Ask question

All categories

2 years ago

12

What is the largest car engine in the world?

2 answers:

leonid [27]2 years ago

7 0

Answer:

Congratulations Chrysler!

Step-by-step explanation:

Umnica [9.8K]2 years ago

3 0

The Largest Current Production Car Engine. Congratulations Chrysler! Nobody else is crazy enough to put a 8.4-liter V10 into a production vehicle, but you've done it again. Naturally, General Motors comes to the rescue if you need even more, with the Chevrolet "572" 9.2 V8 crate engine.

You might be interested in

Find the sum: (6a + 6) + (3x - 2) + (2a + 4)

Cerrena [4.2K]

Answer:

8a + 3x + 8

All you have to do is add numbers around to make the process easier.

7 0

2 years ago

Evaluate the expression<br><br> (3+2) x (5-1+2) x 4

Elza [17]

PEMDAS

(3+2)=5
(5-1+2)=6
6*5*4=120

3 0

2 years ago

Find the value of each variable. Lines that appear tangent are tangent, and the dot is the center. (Answer in the form a=? b=? c

Anni [7]

Answer:

a = 60°/2 = 30°

b = 84/2 = 42°

c = (100+60)/2 = 80°

d = 360-100-60-84 = 116°

Answered by GAUTHMATH

3 0

2 years ago

Manuel earns $7.25 per hour helping his neighbor. Last month he earned $348. Choose the correct equation and solution which dete

Otrada [13]

Answer:

7.25h=348

Step-by-step explanation:

4 0

2 years ago

A cable that weighs 2 lbyft is used to lift 800 lb of coal up a mine shaft 500 ft deep. Find the work done.

mr Goodwill [35]

Answer:

W = 650000 ftlb

Step-by-step explanation:

Parameters given:

Weight of the cable = 2 lb/ft

Weight of the coal = 800 lb

Depth of the shaft, x = 500 ft

The total work done in this system is given by the sum of the work done on the coal and the work done on the shaft.

WORK DONE ON THE COAL

The work done on the coal is given as:

$W_{coal}$ = Weight of coal * distance

$W_{coal}$ = 800 * 500

$W_{coal}$ = 400000 ftlb

WORK DONE ON THE SHAFT

If we partition the cable into several segments of depth Δx, we can find the work done on the cable.

The work done on the shaft is given as:

$W_{shaft}$ = 2Δx * x

$W_{shaft}$ = 2xΔx

=> $W_{shaft}$ = $\int\limits^{500}_{0} {2x} \, dx$

$W_{shaft}$ = $\frac{2x^{2}}{2} \left \{ {{500} \atop {0}} \right.$

$W_{shaft}$ = $500^{2} - 0^{2}$

$W_{shaft}$ = 250000 ftlb

Therefore, total work done is:

$W_{total} = 250000 + 400000$

$W_{total} = 650000 ftlb$

4 0

2 years ago