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Sphinxa [80]
3 years ago
10

The latitude of any location on earth is the angle formed by the two rays drawn from the center of earth to the location and to

the equator. the ray through the location is the initial ray. use 3960 miles as the radius of the earth. suppose city a is due north of city
b. find the distance between city a​ (north latitude 45 degrees 28 prime upper n45°28′ n​) and city b​ (latitude 38 degrees 52 prime upper n38°52′ n​).

Physics
1 answer:
Alina [70]3 years ago
5 0

The distance between city a and city b is 833.345 miles.

We know that

1°=60'

The distance of city a from the initial ray is  calculated as

x_a=3960*tan45.46°=4024.101 miles

The distance of city b from the initial ray is calculated as

x_b=3960*tan 38.86°=3190.75 miles

Now the distance between city a and b is equal to

4024.101-3190.75=833.345 miles

This is the vertical distance between the cities.

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7 0
2 years ago
A miniature quadcopter is located at x = -2.25 m and y, - 5.70 matt - 0 and moves with an average velocity having components Vv,
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\dfrac{y_2 - 5.70\,\mathrm m}{1.60\,\mathrm s} = -2.50\dfrac{\rm m}{\rm s}

Solve for x_2 and y_2, which are the <em>x</em>- and <em>y</em>-components of the copter's position vector after <em>t</em> = 1.60 s.

x_2 = -2.25\,\mathrm m + \left(2.70\dfrac{\rm m}{\rm s}\right)(1.60\,\mathrm s) \implies \boxed{x_2 = 2.07\,\mathrm m}

y_2 = 5.70\,\mathrm m + \left(-2.50\dfrac{\rm m}{\rm s}\right)(1.60\,\mathrm s) \implies \boxed{y_2 = 1.70\,\mathrm m}

Note that I'm reading the given details as

x_1 = -2.25\,\mathrm m \\\\ y_1 = -5.70\,\mathrm m \\\\ v_x = 2.70\dfrac{\rm m}{\rm s}\\\\ v_y=-2.50\dfrac{\rm m}{\rm s}

so if any of these are incorrect, you should make the appropriate adjustments to the work above.

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