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sukhopar [10]
3 years ago
10

In​ physics, it is established that the acceleration due to​ gravity, g (in meters divided by sec squared) in meters/sec²​, at a

height h meters above sea level is given by the formula below. ​g(h) = \frac{3.99 x 10^{14}}{(6.374 x 10^6 + h)^2} where 6.374 x 10⁶⁶ is the radius of Earth in meters.
A certain building is 458 meters tall. What is the acceleration due to gravity at the top at the top of the building​? ​g(h) = _______ m/sec².
Physics
1 answer:
kompoz [17]3 years ago
5 0

Answer:

g(h)=9.81943\ m/s^2

Explanation:

The given function is

g(h)=\dfrac{3.99\times 10^{14}}{(6.374\times 10^6+h)^2}

Now h = the height from the surface of the Earth

Here the building is 458 m tall

g(458)=\dfrac{3.99\times 10^{14}}{(6.374\times 10^6+458)^2}

\Rightarrow g(458)=9.81943\ m/s^2

So,

g(h)=9.81943\ m/s^2

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Answer:

3 sigma lower control limit = 0.0429

Explanation:

Given.

n = 100

days = 100

Number of defective bulbs = 600 defective bulbs

Let p = Process Average

p = 600/(100*40)

P = 600/4000

P = 0.15

q = 1 - p

q = 1 - 0.15

q = 0.85

3 sigma lower limit = p - 3*√(pq/n)

Using the above formula

Substitute in the values

3 sigma lower control limit = 0.15 - 3 * √(0.15 * 0.85/100)

3 sigma lower control limit= 0.15 - 3√0.001275

3 sigma lower control limit = 0.15 - 3* 0.035707142142714

3 sigma lower control limit = 0.15 - 0.107121426428142

3 sigma lower control limit = 0.04287857357185

3 sigma lower control limit = 0.0429 ---- approximated

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