Ask question

Ask question

All categories

2 years ago

5

If an airplane were traveling eastward with a thrust force 450 N and there was a tailwind of 200 N, what would the resulting net

force be on the airplane?
*20 points*

2 answers:

Vsevolod [243]2 years ago

4 0

Answer: The resulting net force is 650N

Explanation:

I jus know stuff HAVE A GOOD DAY MERRY CHRISTMAS AND HAPPY NEW YEAR ECT.

Finger [1]2 years ago

3 0

Answer:

650N

Explanation:

450+200=650

You might be interested in

How long does it take for the speed of light and sound to travel around the world

irina [24]

-- Light travels straight, not around in a circle. But if it did, it would cover
a distance equal to the length of the equator in about <em>0.13 second</em>.

-- At the speed of sound (in air at standard temperature and pressure),
it would take about <em>32.6 hours </em>to cover the same distance.

6 0

3 years ago

A truck is moving with a certain uniform velocity. It is accelerated uniformly by 0.75 m/s^2. After 20 seconds , the velocity be

scoundrel [369]

Answer:

Vi = 5 m/s

Explanation:

let (a) acceleration = 0.75 m/s²

(t) time = 20 seconds

Vf = final velocity = 72 km/hr (convert to m/s to units consistency = 20 m/s)

find Initial velocity (Vi)

Vf - Vi

a = -----------

t

Vi = Vf - (a * t) = 20 - (0.75 * 20)

Vi = 5 m/s

5 0

3 years ago

What human process are carbon sources that are upsetting the balance between CO to update via planets and CO2 released by living

Serhud [2]

The combustion of fossil fuels is releasing more co2 into the atmosphere then what would occur naturally

5 0

3 years ago

A geosynchronous Earth satellite is one that has an orbital period of precisely 1 day. Such orbits are useful for communication

koban [17]

Answer:

r = 4.24x10⁴ km.

Explanation:

To find the radius of such an orbit we need to use Kepler's third law:

$\frac{T_{1}^{2}}{T_{2}^{2}} = \frac{r_{1}^{3}}{r_{2}^{3}}$

<em>where T₁: is the orbital period of the geosynchronous Earth satellite = 1 d, T₂: is the orbital period of the moon = 0.07481 y, r₁: is the radius of such an orbit and r₂: is the orbital radius of the moon = 3.84x10⁵ km. </em>

From equation (1), r₁ is:

$r_{1} = r_{2} \sqrt[3] {(\frac{T_{1}}{T_{2}})^{2}}$

$r_{1} = 3.84\cdot 10^{5} km \sqrt[3] {(\frac{1 d}{0.07481 y \cdot \frac{365 d}{1 y}})^{2}}$

$r_{1} = 4.24 \cdot 10^{4} km$

Therefore, the radius of such an orbit is 4.24x10⁴ km.

I hope it helps you!

3 0

3 years ago

A yacht is pushed along at a constant velocity by the wind. The wind force is 180N at an angle of 25 degrees to the direction of

damaskus [11]

The parallel component is given by
F=180cos(25)=163.14N

4 0

3 years ago