To practice Problem-Solving Strategy 3.2 Relative velocity. An airplane pilot wishes to fly directly westward. According to the weather bureau, a wind of 75.0 km/hour is blowing southward. The speed of the plane relative to the air (called the "air speed") as measured by instruments aboard the plane is 310 km/hour . In which direction should the pilot head?

Answers

Answer 1

Answer:

The pilot should head 13.6° north of west.

Why?

We can solve the problem by using trigonometric relations. Since there is a right triangle formed between the direction that the pilot wants to fly to and the wind's speed, we can use the following formula:

$Tan(\alpha )=(WindSpeed)/(PlaneSpeed)\n\n\alpha =Tan((WindSpeed)/(PlaneSpeed))^(-1)=ArcTan((WindSpeed)/(PlaneSpeed))\n\n\alpha =ArcTan((WindSpeed)/(PlaneSpeed))$

Now, substituting the given information and calculating, we have:

$Wind=75(km)/(m) (southward)\n\nAirplane=310(km)/(h)$

$\alpha =ArcTan((WindSpeed)/(PlaneSpeed))\n\n\alpha =(75(km)/(h) )/(310(km)/(h) )=13.6\°$

Hence, we have that the pilot should head to 13.6° north of west.

Have a nice day!

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Answers

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2) 3 N

Explanation:

The relationship between force and elongation in a spring is given by Hooke's law:

$F=kx$

where

F is the force applied

k is the spring constant

x is the elongation

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$x=4 cm$

So the spring constant is

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Therefore, the new elongation is

$x'=(F')/(k)=(6)/(0.5)=12 cm$

In this second problem, we know that the elongation of the spring now is

$x=6 cm$

From part a), we know that the spring constant is

$k=0.5 N/cm$

Therefore, we can use the following equation to find the force:

$F=kx$

And substituting k and x, we find:

$F=(0.5)(6)=3 N$

So, the force to produce an elongation of 6 cm must be 3 N.

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Answers

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