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Mariana [72]
3 years ago
15

An airplane undergoes the following displacements: First, it flies 40 km in a direction 30° east of north. Next, it flies 56 km

due south. Finally, it flies 100 km 30° north of west. Using analytical methods, determine how far the airplane ends up from its starting point.

Physics
1 answer:
goldfiish [28.3K]3 years ago
4 0

Answer:

Distance from start point is 72.5km

Explanation:

The attached Figure shows the plane trajectories from start point (0,0) to (x1,y1) (d1=40km), then going from (x1,y1) to (x2,y2) (d2=56km), then from (x2,y2) to (x3,y3) (d3=100). Taking into account the angles and triangles formed (shown in the Figure), it can be said:

x1=d1*cos(60), y1=d1*sin(60)\\\\ x2=x1 , y2=y1-d2\\\\ x3=x2-d3*cos(30) , y3=y2+d3*sin(30)

Using the Pitagoras theorem, the distance from (x3,y3) to the start point can be calculated as:

d=\sqrt{x3^{2} +y3^{2} }

Replacing the given values in the equations, the distance is calculated.

You might be interested in
You throw a ball upward with a speed of 14m/s. What is the acceleration of the ball after it leaves your hand? Ignore air resist
omeli [17]

The acceleration of the ball after leaving the hand is 9.8 m/s^2 downward

Explanation:

In order to find the acceleration of the ball during its motion, we have to study which forces are acting on it.

After the ball leaves the hand, if we neglect air resistance, there is only one force acting on the ball: the force of gravity, whose magnitude is

F=mg

where m is the mass of the ball and g is the acceleration of gravity (g=9.8 m/s^2), acting in the downward direction.

According to Newton's second law, the acceleration of the ball is given by

a=\frac{\sum F}{m}

where

\sum F is the net force acting on the ball

After the ball leaves the hand, the only force acting on it is the force of gravity, so we can substitute (mg) into the previous equation:

a=\frac{mg}{m}=g=9.8 m/s^2

This means that the acceleration of the ball remains 9.8 m/s^2 downward for the entire motion, after leaving the hand.

Learn more about Newton's second law:

brainly.com/question/3820012

#LearnwithBrainly

5 0
3 years ago
Xavier is roller skating at 14 km/h and tosses a set of keys forward on the ground at 8 km/h. The speed of the keys relative to
Arturiano [62]

Answer:

22 km/h

Explanation:

Given that,

Speed of Xavier, v = 14 km/h

He tosses a set of keys forward on the ground at 8 km/h, v' = 8 km/h

We need to find the speed of the keys relative to the ground. Let it is V.

As both Xavier and the keys are moving in same diretion. The relative speed wrt ground is given by :

V = v+v'

V= 14 + 8

V = 22 km/h

So, the speed of the keys relative to the ground is 22 km/h.

4 0
2 years ago
PLS ANSWER ASAP!!
Molodets [167]

b) between poles M1 and M2

Explanation:

From the expression, we can deduce that r is the distance between two magnetic poles M1 and M2.

The law of attraction between two magnetic poles states that:

<em>  the force of attraction or repulsion between two magnetic poles is a function of the product of the strength of the magnetic poles and the square of the distance between the pole</em>s

 

    Mathematically:

            FM = K \frac{M1 M2}{r^{2} }

 here r is the distance between the poles

  FM is the magnetic force between the poles

   M1 is the strength of the first magnetic pole

   M2 is the strength of the second pole

   K is the magnetic field constant

learn more:

magnetic pole brainly.com/question/2191993

#learnwithBrainly

8 0
3 years ago
Pls need an answer asap
Andrei [34K]

c] Electrons

Electron gets transferred

3 0
2 years ago
A block moving to the right on a level surface with friction is pulled by an increasing horizontal force also directed to the ri
KengaRu [80]

Answer:

e) True  Yes both are constant

Explanation:

Let us propose the solution of the problem before reviewing the statements, we use Newton's second law

       F - fr = m a

      N- W = 0

      N = mg

The equation for the force of friction is

     fr = μ N

     F - μ mg = m a

     F = m (a- μ g)

Now let's review the claims

.a) False. Normal force and friction are constants.

.b) False. Both are constant.

.c) False. Both are constant.

d.) False

e) True  Yes both are constant

6 0
3 years ago
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