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Rasek [7]
2 years ago
5

A plane takes off in san francisco at noon and flies toward the southeast. an hour later, it is 400 kilometers east and 300 kilo

meters south of its starting location. assuming the plane flew in a straight line, how far did it travel? how many degrees south of east did the plane fly?

Physics
1 answer:
eduard2 years ago
5 0
This first step involves a right triangle. If the plane is 400 km east and 300 km south of the origin, and it flew in a straight line, then you can construct a right triangle with side lengths 300, 400, and c. You may recognize that these are multiples of the Pythagorean triple 3, 4, 5, so the side length c is 500 km. Otherwise, you would write 
c^2 = 300^2+400^2 \\ c = 500. 

This second step is, if I am correctly interpreting "degrees south of east," to find the angle formed by the horizontal line representing the east and the path of the plane. I made a diagram that does just that (see attached). You can use a trig function of one of the angles to solve. I chose
tan(C)= \frac{3}{4}  \\  arctan( \frac{3}{4})=36.870. Thus, I believe it is 37° south of east.

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A spring on a horizontal surface can be stretched and held 0.5 m from its equilibrium position with a force of 60 N. a. How much
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Answer:

a)1815Joules b) 185Joules

Explanation:

Hooke's law states that the extension of a material is directly proportional to the applied force provided that the elastic limit is not exceeded. Mathematically;

F = ke where;

F is the applied force

k is the elastic constant

e is the extension of the material

From the formula, k = F/e

F1/e1 = F2/e2

If a force of 60N causes an extension of 0.5m of the string from its equilibrium position, the elastic constant of the spring will be ;

k = 60/0.5

k = 120N/m

a) To get the work done in stretching the spring 5.5m from its position,

Work done by the spring = 1/2ke²

Given k = 120N/m, e = 5.5m

Work done = 1/2×120×5.5²

Work done = 60× 5.5²

Work done = 1815Joules

b) work done in compressing the spring 1.5m from its equilibrium position will be gotten using the same formula;

Work done = 1/2ke²

Work done =1/2× 120×1.5²

Works done = 60×1.5²

Work done = 135Joules

8 0
2 years ago
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2 years ago
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3 years ago
A delivery truck leaves a warehouse and travels 3.20 km east. The truck makes a right turn and travels 2.45 km south to arrive a
boyakko [2]

Explanation:

It is given that,

Displacement of the delivery truck, d_1=3.2\ km (due east)

Then the truck moves, d_2=2.45\ km (due south)

Let d is the magnitude of the truck’s displacement from the warehouse. The net displacement is given by :

d=\sqrt{d_1^2+d_2^2}

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Let \theta is the direction of the truck’s displacement from the warehouse from south of east.

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So, the magnitude and direction of the truck’s displacement from the warehouse is 4.03 km, 37.4° south of east.

8 0
3 years ago
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