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sdas [7]
3 years ago
15

A cloud is drifting across the sky at a constant velocity of 724metersperminute to the east. It is at an altitude of 9,200 meter

s. How far will the cloud move in 12.5minutes?
Physics
1 answer:
alisha [4.7K]3 years ago
7 0

Answer:

The cloud moves 9050 meters to the east in 12.5 minutes.

Explanation:

Let suppose that mass of the cloud is negligible. meaning that effects of gravity are negligible and that altitude of the cloud remains constant. If the cloud drifts at constant velocity, travelled distance is defined by following formula:

\Delta s = v\cdot \Delta t (1)

Where:

v - Velocity, in meters per second.

\Delta t - Time, in seconds.

If we know that v = 724\,\frac{m}{min} and \Delta t = 12.5\,min, then the travelled distance after 12.5 minutes is:

\Delta s = 9050\,m

The cloud moves 9050 meters to the east in 12.5 minutes.

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Where the weight w of the block has an x-component and y-component:

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As well as the Normal Force N:

N_{x}=Nsin(\theta)    (3)

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In addition, we know N=w, then \sum F_{y}=0

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Substituting (1) in (5):

wsin(\theta)=m.a    (6)

In addition, we know w=m.g, where m is the mass of the block and g the gravity acceleration, which is equal to 9.8m/{s}^{2}  

So:

m.g.sin(\theta)=m.a   (7)

a=g.sin(\theta)    (8)

a=5.88m/{s}^{2}    (9)   >>>>This is the acceleration of the block

On the other hand, we have the following equation that expresses a <u>relation between</u> the distance d with the acceleration a and time t:

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We already know the value of  d and calculated a, we have to find t:

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t=\sqrt{\frac{2(0.75m)}{5.88m/{s}^{2}}}   (12)

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V_{f}-V_{o}=a.t   (13)

If V_{o}=0

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(a) 5.66 m/s

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