I need help, worth 40 Points!

A plane flies due north (90° from east) with a velocity of 100 km/h for 3 hours. During this time, a steady wind blows southeast at 30 km/h at an angle of 315° from due east. After 3 hours, where will the plane’s position be relative to its starting point?

*Please try to fill the data table.

I Need Help, Worth 40 Points!A Plane Flies Due North (90 From East) With A Velocity Of 100 Km/h For 3
I Need Help, Worth 40 Points!A Plane Flies Due North (90 From East) With A Velocity Of 100 Km/h For 3

Answers

Answer 1

Answer:

thank you

Explanation:

Answer 2

Split up the two given velocity vectors into horizontal and vertical components.

Plane:

p = (100 km/h) (cos(90°) i + sin(90°) j ) = (100 km/h) j

Its direction is given, 90° relative to east.

Traveling for 3 hours with this velocity results in a displacement of

p * (3 h) = (300 km/h) (3h) j = (900 km) j

meaning that this velocity contributes to a north-facing displacement of 900 km.

Wind:

w = (30 km/h) (cos(315°) i + sin(315°) j ) ≈ (21.2 km/h) i + (-21.2 km/h) j

Its direction is also given, 315° relative to east.

After 3 h, this velocity contributes to a displacement of

w * (3h) ≈ (63.6 km) i + (-63.6 km) j

meaning if the plane had no velocity of its own but somehow stayed in the air, the wind would have pushed it about 63.6 km south and 63.6 km east, which translates to a net displacement of

√((63.6 km)² + (-63.6 km)²) = 90 km

due southeast.

Putting everything together, your table displacement table should read:

.                       |   wind   |   plane

velocity           |   30      |   100       ...   km/h

direction         |   90°     |   315°      ...   relative to east

x                      |   63.6   |   0           ...   km

y                      |  -63.6   |   900      ...   km

The resultant vector is the sum of these, r = p + w :

r ≈ (21.2 km/h) i + (78.8 km/h) j

After 3 h, the resultant displacement is

r * (3h) ≈ (63.6 km) i + (236.4 km) j

with magnitude

√((63.6 km)² + (236.4 km)²) ≈ 244.8 km

and direction θ such that

tan(θ) ≈ (236.4 km) / (63.6 km)

θ ≈ arctan(3.714) ≈ 74.9°

So the resultant table should read

xnet             |   63.6 km

ynet             |   236.4 km

magnitude   |   244.8 km

theta            |   74.9° relative to east


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Answer:

Explanation:

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Answer:

[tex]F=8.64\times 10^{-9}\ N[/tex]

Explanation:

Given that,

Mass of student 1, m₁ = 120 kg

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Let they are at a distance of 10 m.

We need to find the gravitational pull they have on each other. The gravitational force acting between two objects is given by :

[tex]F=G\dfrac{m_1m_2}{r^2}\\\\F=6.67\times 10^{-11}\times \dfrac{120\times 108}{(10)^2}\\\\=8.64\times 10^{-9}\ N[/tex]

So, the gravitational pull between students is [tex]8.64\times 10^{-9}\ N[/tex].

An astronaut is working outside the International Space Station where the atmospheric pressure is essentially zero. The pressure gauge on her air tank reads 690000 Pa. What force does the air inside the tank exert on the flat end of the cylindrical tank, a disk 0.150 m in diameter?

Answers

Answer:

Explanation:

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Given

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Area will be the area of the circular section of the cylindrical tank

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Hence the force applied by the air inside the tank is 12,187.125N

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Which part of a DC generator keeps the electric current flowing in only one
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D. The north pole of one magnet attracts the north pole of another
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Answer:

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Answer:

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Answer:

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Se llama permutación de m elementos a cada una de las diferentes ordenaciones  que se pueden hacer con esos elementos. En otras palabras, se llama permutaciones de  m elementos en n posiciones a las distintas formas en que se pueden ordenar los m elementos ocupando únicamente las n  posiciones.

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La selección de los alumnos se puede realizar de 2300 formas.

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25C3 = 25! / 3! 22!

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Answers

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Velocity is the answer

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Answers

Given :

A farmer is pushing a 75 kg plow across a field with 700 N at an angle of 35 degrees above the ground.

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The acceleration of the plow.

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Normal force = mg + F sin Ф

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= 1151.50 N

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Answer:

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See the comments.

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True

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Explanation:

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