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frozen [14]
3 years ago
7

A car transports its passengers between 3 buildings. It moves from the first building to the second building, 4.76km away, in a

direction 37° north of east. It then moves from second building to the third building in a direction 69° west of north. Finally, it returns to the first building, sailing in a direction 28° east of south. Calculate the distance between (a) the second and third buildings and (b) the first and third buildings.​

Physics
1 answer:
BigorU [14]3 years ago
8 0

Answer:

1) a = 6.14 km

2) b = 4.69 km

Explanation:

Let the first building be A, second building be B and third building be C.

Now, bearing of A = 4.76 km in a direction 37° north of east

Bearing of B = 69° west of north

Bearing of C = 28° east of south

Thus if this 3 points form a triangle, we will have the following angles;

Angle at point A = 28 + (90 - 37) = 81°

Angle at point B = 28 + (90 - 69) = 49°

Angle at point C = 180 - (81 + 49) = 50°

Now, the distance between second and third building is "a" which is represented by BC in the triangle attached while the given distance of 4.76 represents side AB. Thus;

Using sine rule, we can find "a".

a/sin 81 = 4.76/sin 50

a = 6.14 km

B) distance between first and third building is AB in the triangle depicted by "b".

Similar to the first problem, we will use sine rule again.

b/sin49 = 4.76/sin 50

b = 4.69 km

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Answer:

Magnitude = 4.056 m

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Explanation:

The vector is resolved in terms of the vertical and horizontal components. Let's look each of these separately.

The vector 4.40 is directed East. This automatically becomes a horizontal component.

But we know that there is a vector 3.40 North West. The angle the vector makes with the horizontal is 61⁰.

Resolving the vectors should yield the horizontal and vertical components:

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The first component is 4.40 m

The second one is derived by resolving 3.40 to the horizontal like this 3.40 × - cos 61⁰ = -1.648 m

Adding the horizontal component gives 4.40 m + ( -1.648 m) = 2.752 m

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<h2>Answer:</h2>

He saves 13.2 minutes

<h2>Explanation:</h2>

Hey! The question is incomplete, but it can be found on the internet. The question is:

How many minutes did he save?

Let's call:

t_{1}:Time \ at \ speed \ 65mph \\ \\ t_{2}:Time \ at \ speed \ 73mph \\ \\ v_{1}=65mph \\ \\ v_{2}=73mph

We know that the 135 miles are on the interstate highway where the speed limit is 65 mph. From this, we can calculate the time it takes to drive on this highway. Assuming Richard maintains constant the speed:

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t_{2}=\frac{135}{73} \\ \\ t_{2}=1.85 \ hours

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t_{s}=0.22h\left(\frac{60min}{1h}\right) \\ \\ \boxed{t_{s}=13.2min}

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