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serg [7]
3 years ago
6

An airplane pilot flies due west at a speed of 216 km/hr with respect to the air. After flying for a half an hour, the pilot fin

ds themselves over a town that is 119 km west and 27 km south of their starting point.
a- Determine the magnitude of the velocity of the wind with respect to the ground.
(include units with the answer)

b-Determine the direction of the velocity of the wind with respect to the ground measured as west from the south.
Physics
1 answer:
Yuki888 [10]3 years ago
5 0

Answer:

speed wind  Vw = 54.04 km / h   θ = 87.9º

Explanation:

We have a speed vector composition exercise

In the half hour the airplane has traveled X = 108 km to the west, but is located at coordinated 119 km west and 27 km south

Let's add the vectors in each coordinate axis

   

X axis (East-West)

      -Xvion - Xw = -119

      Xw = -Xavion + 119

      Xw = 119 -108

      Xwi = 1 km

Calculate the speed for time of  t = 0.5 h

     Vwx = Xw / t

     Vwx= 1 /0.5

     Vwx = - 2 km / h

Y Axis (North-South)

    Y plane - Yi = -27

    Y plane = 0

    Yw = 27 km

    Vwy = 27 /0.5

    Vwy = 54 km / h

Let's use the Pythagorean theorem and trigonometry to compose the answer

 Vw = √ (Vwx² + Vwy²)

  Vw = R 2² + 54²

  Vw = 54.04 km / h

  tan θ = Vwy / Vwx

  tan θ = 54/2 = 27

  θ = Tan⁻¹ 1 27

  θ = 87.9º

The speed direction is 87. 9th measure In the third quadrant of the X axis in the direction 90-87.9 = 2.1º  west from the south

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zavuch27 [327]

#2

As it is given here

initial speed is

v_i = 6 m/s

After 4 seconds the final speed is

v_f = 14 m/s

so here we can use the formula of acceleration using kinematics

a = \frac{v_f - v_i}{t}

a = \frac{14 - 6}{4}

a = 2 m/s^2

so here it will accelerate at 2 m/s^2 rate.

#3

As it is given here

initial it starts from rest

v_i = 0 m/s

After 2.5 seconds the final speed is

v_f = 15 m/s

so here we can use the formula of acceleration using kinematics

a = \frac{v_f - v_i}{t}

a = \frac{15 - 0}{2.5}

a = 6 m/s^2

so here it will accelerate at 6 m/s^2 rate.


#4

i think question is not correct as in first line it is saying about a bag of trash and then in next line it is asking for the position of Jumper and bridge.

7 0
2 years ago
An airplane is flying at a constant speed in a positive direction. It slows down when it approaches the airport where it's going
KatRina [158]
3.negative acceleration....
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3 years ago
4) Write down the transformation of energy in torch light.<br>​
Elenna [48]

Answer:

<u>The transformation of energy in a torch light is as follows:</u>

1) When the torch is turned ON, the chemical energy in the batteries is converted into electrical energy.

2) The electrical energy is converted into heat and light energy. (We feel the torch to be hot after some time and we can see the light energy)

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<h2>~AnonymousHelper1807</h2>
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For a projectile launched horizontally, which of the following best describes the downward component of a projectile's velocity?
Angelina_Jolie [31]

C. The downward component of the projectile's velocity continually increases

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The motion of a projectile consists of two independent motions:  

- A uniform motion (with constant velocity) along the horizontal direction  

- A uniformly accelerated motion, with constant acceleration (equal to the acceleration of gravity) in the downward direction  

Here we want to study the downward component of the projectile's velocity. Since the vertical motion is a uniformly accelerated motion, the vertical velocity is given by:

v=u+at

where

u = 0 is the initial vertical velocity (zero since the projectile is fired horizontally)

a=g=9.8 m/s^2 downward is the acceleration of gravity

t is the time

So the equation becomes

v=gt

This means that

C. The downward component of the projectile's velocity continually increases

Because every second, it increases by 9.8 m/s in the downward direction.

Learn more about projectile motion:

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A point charge of 5.7 µc moves at 4.5 × 105 m/s in a magnetic field that has a field strength of 3.2 mt, as shown in the diagram
Sauron [17]

The magnitude of the force on the charge by the influence of the magnetic field will be 6.6*10^-3 N

<h3>What is magnetic force?</h3>

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where

The magnitude of the charge   q=5.7\ \mu C =5.7\times 10^{-6} \ C

The velocity of the charge        v=4.5\times 10^5\ \frac{m}{s}

The magnitude of the magnetic field   3.2\ mT=0.0032\ T

The angle between the directions of v and B  \theta =90^o-37^o=53^o

By substituting the values we will get:

F=(5.7\times 10^}-6})(4.5\times 10^5)(0.0032)(Sin53^o)

F=6.6\times 10^{-3}\ N

Hence the magnitude of the force on the charge by the influence of the magnetic field will be 6.6*10^-3 N

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