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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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Answer:

q = 2.067 \times 10^{-5}\ C

Explanation:

Given,

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Electric field,E = 670 N/C.

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Equating both the equation of motion

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3 years ago
In a physics laboratory experiment, a coil with 200 turns enclosing an area of 13.1 cm2 is rotated during the time interval 3.10
sergij07 [2.7K]

Answer:

A)\Phi=83.84\times 10^{-9}

B)\Phi=0 Wb

C)emf=5.4090\times 10^{-4}V

Explanation:

Given that:

  • no. of turns i the coil, n=200
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(A)

Initially the coil area is perpendicular to the magnetic field.

So, magnetic flux is given as:

\Phi=B.a\,cos \theta..................................(1)

\theta is the angle between the area vector and the magnetic field lines. Area vector is always perpendicular to the area given. In this case area vector is parallel to the magnetic field.

\Phi=6.4\times 10^{-5}\times 13.1 \times 10^{-4}\, cos 0^{\circ}

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(B)

In this case the plane area is parallel to the magnetic field i.e. the area vector is perpendicular to the magnetic field.

∴  \theta=90^{\circ}

From eq. (1)

\Phi=6.4\times 10^{-5}\times 13.1 \times 10^{-4}\, cos 90^{\circ}

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According to the Faraday's Law we have:

emf=n\frac{B.a}{t}

emf=\frac{200\times 6.4\times 10^{-5}\times 13.1 \times 10^{-4}}{3.1\times 10^{-2}}

emf=5.4090\times 10^{-4}V

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