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il63 [147K]
3 years ago
14

A plane travels down a runway 2750 m before it lifts off at an angle of 37 degrees from the horizontal. The plane has traveled 1

.8 km since its wheels left the ground. What is the displacement of the plane?
Physics
1 answer:
ss7ja [257]3 years ago
5 0

Answer: 4.236km

Explanation:

Let's define the point (x, y) as:

x = horizontal distance moved.

y = vertical distance moved.

If the plane starts in the point (0, 0) then:

"A plane travels down a runway 2750 m before it lifts off..."

At this time, the position will be:

P = (0 + 2750m, 0) = (2750m, 0).

"it lifts off at an angle of 37 degrees from the horizontal. The plane has traveled 1.8 km since its wheels left the ground."

In this case, as the angle is measured from the horizontal, the components will be:

x = 1.8km*cos(37°) = 1.438km

y = 1.8km*sin(37°) =  1.083 km

Then the new position is:

P = (2750m + 1.438 km, 0 + 1.083 km)

Let's write it using the same units for all the quantities:

we know that

1km = 1000m

Then:

2750m = (2750/1000) km = 2.750 km.

Then we can write the new position as:

P = (2.750 km + 1.438km, 1.083km) = (4.188km, 1.083km)

Now, we define the displacement as the distance between the final position and the initial position.

The distance between two points (a, b) and (c, d) is:

D = √( (a  c)^2 + (b - d)^2)

In this case the points are:

(0, 0) for the initial position

(4.188km, 1.083km) for the final position.

And the displacement will be:

D = √( (4.188km - 0)^2 + (1.083 - 0)^2) = 4.236km

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Answer:

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Explanation:

The problem is asking for the Force of pushing off the ground.

  • The formula of Force is: F = mass x acceleration

Given = <em>Mass</em>: 600 newtons (N)

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We have to convert the mass into kg first. Remember that <u>1 kg is equal to 9.80665 newtons.</u>

Let x be the<em> mass in newtons</em>.

Let's convert: \frac{1 kg}{9.80665 N} x \frac{x}{600 N} = \frac{600}{9.80665} = 61.18 kg

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Let's go back to finding the force.

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3 years ago
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Explanation:

Explanation:

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We know that,

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The balance equation is along y-axis

The box will not move in y-axis therefore, the net force in the y-axis will be zero.

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