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WITCHER [35]
2 years ago
14

A plane flies along a straight line path after taking off, and it ends up 220 km farther east and 100.0 km farther north, relati

ve to where it started. In what direction did it fly on the straight line path?
Physics
1 answer:
love history [14]2 years ago
3 0

Answer:

North east

Explanation:

The distance travelled by a plane from one point to another is the shortest distance between the two points. This distance travelled is usually a straight line path from point 1 to point 2.

Since the plane ends up 220 km farther east and 100.0 km farther north, the direction of flight for the plane is in the North-East direction.

Let x represent the distance travelled by the plane. Hence:

x² = 100² + 220²

x² = 58400

x = 241.66 km

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La cinemática permite encontrar la respuesta para la aceleracion del cuerpo en el cañón es:

            a = 1,8 10⁶ m/s²  

 

La cinemática estudia el movimiento de los cuerpos, buscando relaciones entre la posicion, la velocidad y la aceleración.

       v² = v₀² + 2 a x

Donde v y v₀ son la velocidad actual e inicial, respectivamente, a es la aceleracion y x la distancia recorrida.

Indica que la longitud de cañon es x= 18 m la velocidad de  salida es  

            v= 29000 km/h ( \frac{1000m}{1 km} ) ( \frac{1h}{3600s} s) = 8,055,56 m/s.

La velocidad inicial del proyectil es cero.

            a = \frac{v^2}{2x}  

            a = \frac{ 8055.56^2}{2 \ 18}  

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En conclusión usando la cinemática podemos encontrarla respuesta para la aceleracion del cuerpo en el cañón es:

            a = 1,8 10⁶ m/s²  

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3 years ago
Jonathan answered 28 out of 35 questions correctly on his chemistry test. What is his percent score?
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Seriously? It’s that hard to use a calculator??


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Imagine that person B is more massive than person A in the picture above.
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3 years ago
A solid cylinder of mass 7 kg and radius 0.9 m starts from rest at the top of a 20º incline. It is released and rolls without sl
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Answer:

157.8 J

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m = mass of the cylinder = 7 kg

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g = acceleration due to gravity = 9.8 m/s²

TE = Total Energy at the bottom

PE = Gravitational potential energy at the top

Using conservation of energy

Total Energy at the bottom = Gravitational potential energy at the top  

TE = PE

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TE = 157.8 J

7 0
3 years ago
A projectile is launched at an angle of 29 degrees above the horizontal with an initial velocity of 36.6 at an unknown height.
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The magnitude of the unknown height of the projectile is determined as 16.1 m.

<h3>Magnitude of the height</h3>

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