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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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If the input work on a machine is equal to it's output work. the machine has _____ efficiency.
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Explanation:

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An old light bulb draws only 54.3 W, rather than its original 60.0 W, due to evaporative thinning of its filament. By what facto
Lemur [1.5K]

Answer:

The factor of the diameter is 0.95.

Explanation:

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Power = 60 W

We know that,

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R\propto\dfrac{1}{D}

The power is inversely proportional to the resistance.

P\propto\dfrac{1}{R}

P\propto D^2

We need to calculate the factor of the diameter of the filament reduced

Using relation of power and diameter

\dfrac{P_{i}}{P_{f}}=\dfrac{D_{i}^2}{D_{f}^2}

Put the value into the formula

\dfrac{D_{i}^2}{D_{f}^2}=\dfrac{54.3}{60}

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7 0
3 years ago
Read 2 more answers
A ball is kicked at an angle of 35° with the ground.a) What should be the initial velocity of the ball so that it hits a target
stiks02 [169]

Answer:

a.18.5 m/s

b.1.98 s

Explanation:

We are given that

\theta=35^{\circ}

a.Let v_0 be the initial velocity of the ball.

Distance,x=30 m

Height,h=1.8 m

v_x=v_0cos\theta=v_0cos35

v_y=v_0sin\theta=v_0sin35

x=v_0cos\theta\times t=v_0cos35\times t

t=\frac{30}{v_0cos35}

h=v_yt-\frac{1}{2}gt^2

Substitute the values

1.8=v_0sin35\frac{30}{v_0cos35}-\frac{1}{2}(9.8)(\frac{30}{v_0cso35})^2

1.8=30tan35-\frac{6574.6}{v^2_0}

\frac{6574.6}{v^2_0}=21-1.8=19.2

v^2_0=\frac{6574.6}{19.2}

v_0=\sqrt{\frac{6574.6}{19.2}}=18.5 m/s

Initial velocity of the ball=18.5 m/s

b.Substitute the value then we get

t=\frac{30}{18.5cos35}

t=1.98 s

Hence, the time for the ball to reach the target=1.98 s

7 0
3 years ago
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