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3 years ago

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NEED HELP ASAP 100 POINTS!!!! A woman looks out a window of a building. She is 93 feet above the ground. Her line of sight makes

an angle of theta with the building. The distance in feet of an object from the woman is modeled by the function d=93 secant theta. How far away are objects sighted at angles of 29degrees and 59degrees?

1 answer:

enyata [817]3 years ago

3 0

12 feet long hope this helped

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Answer:

a) It takes 0.0017s for the light to reach San Francisco.

b) It takes 0.033s for the light to reach London.

c) It takes 1.334s for the light to reach Mars.

d) It takes 149.7s for the light to reach Venus.

Step-by-step explanation:

Here we can solve this problem by using this following formula:

$s = \frac{d}{t}$

In which s is the speed(in km/s), d is the distance(in km) and t is the time(in s).

The light speed is 299 792 458 m / s = 299,792.458 km/s, so $s = 299,792.458$

(a) San Francisco (about 500km)

Find t when $d = 500$ . So

$s = \frac{d}{t}$

$299,792.458 = \frac{500}{t}$

$299,792.458t = 500$

$t = \frac{500}{299,792.458}$

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b) London(about 10,000km)

Find t when $d = 10,000$ . So

$s = \frac{d}{t}$

$299,792.458 = \frac{10,000}{t}$

$299,792.458t = 10,000$

$t = \frac{10,000}{299,792.458}$

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It takes 0.033s for the light to reach London.

(c) the Moon (400,000km)

Find t when $d = 400,000$ . So

$s = \frac{d}{t}$

$299,792.458 = \frac{400,000}{t}$

$299,792.458t = 400,000$

$t = \frac{400,000}{299,792.458}$

$t = 1.334s$

It takes 1.334s for the light to reach Mars.

(d) Venus (0.3 A.U. from Earth at its closest approach).

Each A.U. has 149,59,7871 km.

So 0.3 A.U. = 0.3*(149,597,871) = 44,879,361.3. This means that $d = 44,879,361.3$ . So:

$s = \frac{d}{t}$

$299,792.458 = \frac{44,879,361.3}{t}$

$299,792.458t = 44,879,361.3$

$t = \frac{44,879,361.3}{299,792.458}$

$t = 149.7s$

It takes 149.7s for the light to reach Venus.

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