A Boeing 737—a small, short-range jet with a mass of 51,000 kg— sits at rest at the start of a runway. The pilot turns the pair of jet engines to full throttle, and the thrust accelerates the plane down the runway. After traveling 940 m, the plane reaches its takeoff speed of 70 m/s and leaves the ground. What is the thrust of each engine?

67000N

Explanation:

We solve for the acceleration using the the 3rd constant-acceleration equation.

(Vx)f² = (Vx)i² + 2ax∆x

We have the displacement to be

∆x = Xf - Xi = 940m

Vx = 70m/s

The acceleration = (70m/s)²/2(940m)

= 4900/1880

= 2.61m/s²

From isaac newton's second law,

51000kg x 2.61m/s²

= 133,000N

The engines thrust is half of this value

Therefore thrust = 67000N or 67kN

c. 7/3

step-by-step explanation:

a rational number is like pi- it is a repeating decimal.

a. is -1.8

b. 2.236068

c. 2.3333333333333333333

d.   is 3

i hope that ! =)

yup its correct its: c) b b