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Luba_88 [7]
3 years ago
6

To a motorist travelling due North at 50km/hr, the wind appears to come from North West at 60mk/hr. Find the true velocity of th

e wind.​
Physics
1 answer:
kvv77 [185]3 years ago
7 0

Answer:

The true velocity of wind will be 43.1 km/h.

Explanation:

Given that,

Velocity of motor \vec{v_{m}}= 50\hat{j}\ km/h

The resultant velocity of wind

\ver{v_{r}}=60\hat{i}\times\dfrac{1}{\sqrt{2}}+60\hat{j}}\times\dfrac{1}{\sqrt{2}}

Suppose, the true velocity of wind is \vec{v_{w}}.

We need to calculate the true velocity of wind

Using formula of resultant velocity

\vec{v_{m}}+\vec{v_{w}}=\vec{v_{r}}

\vec{v_{w}}=\vec{v_{r}}-\vec{v_{m}}

Where, \vec{v_{m}} = velocity of motor

\vec{v_{w}} = velocity of wind

\vec{v_{r}} = resultant velocity

Put the value into the formula

\vec{v_{w}}=\dfrac{60}{\sqrt{2}}\hat{i}+(\dfrac{60}{\sqrt{2}}-50)\hat{j}

\vec{v_{w}}=42.43\hat{i}-7.57\hat{j}

The magnitude of true velocity is,

v_{m}=\sqrt{(42.43)^2+(-7.57)^2}

v_{m}=43.0.9\approx 43.1\ km/h

Hence,  The true velocity of wind will be 43.1 km/h.

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An object is dropped from a height of 75.0 m above ground level. (a) Determine the distance traveled during the first second. (b
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Answer:

a)Distance traveled during the first second = 4.905 m.

b)Final velocity at which the object hits the ground = 38.36 m/s

c)Distance traveled during the last second of motion before hitting the ground = 33.45 m

Explanation:

a) We have equation of motion

             S = ut + 0.5at²

     Here u = 0, and a = g

              S = 0.5gt²

    Distance traveled during the first second ( t =1 )

              S = 0.5 x 9.81 x 1² = 4.905 m

   Distance traveled during the first second = 4.905 m.

b)  We have equation of motion

            v² = u² + 2as

      Here u = 0, s= 75 m and a = g

           v² = 0² + 2 x g x 75 = 150 x 9.81

           v = 38.36 m/s

      Final velocity at which the object hits the ground = 38.36 m/s

c) We have S = 0.5gt²

                   75 = 0.5 x 9.81 x t²

                    t = 3.91 s

   We need to find distance traveled last second

   That is

          S = 0.5 x 9.81 x 3.91² - 0.5 x 9.81 x 2.91² = 33.45 m

   Distance traveled during the last second of motion before hitting the ground = 33.45 m

       

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