contestada

1134567 A jet with mass m = 1.1 Ă— 105 kg jet accelerates down the runway for takeoff at 2 m/s2. 1) What is the net horizontal force on the airplane as it accelerates for takeoff? 2.2*10^5 N Your submissions: 2.2*10^5 Computed value:220000Submitted:Saturday, January 26 at 2:41 PM Feedback:Correct! 2) What is the net vertical force on the airplane as it accelerates for takeoff? 0 N Your submissions: 0 Computed value:0Submitted:Saturday, January 26 at 2:41 PM Feedback:Correct! 3) Once off the ground, the plane climbs upward for 20 seconds. During this time, the vertical speed increases from zero to 21 m/s, while the horizontal speed increases from 80 m/s to 95 m/s. What is the net horizontal force on the airplane as it climbs upward? N 4) What is the net vertical force on the airplane as it climbs upward? N 5) After reaching cruising altitude, the plane levels off, keeping the horizontal speed constant, but smoothly reducing the vertical speed to zero, in 13 seconds. What is the net horizontal force on the airplane as it levels off? N 6) What is the net vertical force on the airplane as it levels off?

Respuesta :

Initially jet is moving horizontally

so we know that

[tex]m = 1.1 \times 10^5 kg[/tex]

[tex]a = 2 m/s^2[/tex]

now we have

PART a)

Net horizontal force is given as

[tex]F_x = ma_x[/tex]

[tex]F_x = (1.1\times 10^5)(2) [/tex]

[tex]F_x = 2.2 \times 10^5 N[/tex]

Part b)

since the acceleration in vertical direction is zero

so as per Newton's law we know

[tex]F = ma[/tex]

[tex]F = 0 N[/tex]

PART c)

speed of the jet increases in vertical direction from ZERO to 21 m/s in 20 seconds and speed increases in horizontal direction from 95 m/s from 80 m/s in same time

vertical acceleration

[tex]a_y = \frac{21 - 0}{20} = 1.05 m/s^2[/tex]

horizontal acceleration

[tex]a_x = \frac{95 - 80}{20} = 0.75 m/s^2[/tex]

now net horizontal force will be

[tex]F_x = ma_x[/tex]

[tex]F_x = (1.1 \times 10^5)(0.75) = 8.25 \times 10^4 N[/tex]

PART d)

Now for vertical force we have

[tex]F_y = ma_y[/tex]

[tex]F_y = (1.1 \times 10^5)(1.05) = 1.155 \times 10^5 N[/tex]

PART e)

Now after reaching the cruising level the horizontal speed becomes constant and in vertical direction speed decrease to ZERO from 21 m/s in 13 s

So we have

[tex]a_x = 0[/tex]

[tex]a_y = \frac{0 - 21}{13} = -1.62 m/s^2[/tex]

now we have horizontal force as

[tex]F_x = 0[/tex]

Part f)

[tex]F_y = ma_y[/tex]

[tex]F_y = (1.1 \times 10^5)(-1.62) = - 1.78 \times 10^5 N[/tex]

Otras preguntas