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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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3 years ago
You obtain a 100-W light bulb and a 50-W light bulb. Instead of connecting them in the normal way, you devise a circuit that pla
lesantik [10]

Answer:

When they are connected in series

     The  50 W bulb glow more than the 100 W bulb

Explanation:

From the question we are told that

     The power rating  of the first bulb is P_1  = 100 \ W

      The power rating of the second bulb is  P_2  =  50 \ W

     

Generally the power rating of the first bulb is mathematically represented as

      P_1  =  V^2 R

Where  V is the normal household voltage which is constant for both bulbs

  So  

        R_1  =  \frac{V^2}{P_1 }

substituting values

        R_1  =  \frac{V^2}{100}

Thus the resistance of the second bulb would be evaluated as

       R_2  =  \frac{V^2}{50}

From the above calculation we see that

        R_2  >  R_1

This power rating of the first bulb can also be represented mathematically as  

        P_  1  =  I^2_1  R_1

This power rating of the first bulb can also be represented mathematically as    

       P_  2  =  I^2_2 R_2

Now given that they are connected in series which implies that the same current flow through them so

       I_1^2 =  I_2^2

This means  that

       P \ \alpha  \  R

So  when they are connected in series

     P_2  >  P_1

This means that the 50 W bulb glows more than the 100 \ W bulb

3 0
3 years ago
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