Question

1. the heaviest piece of equipment ever carried by an airplane was a 12,400.05 kg generator built in germany in 1993. how far above the ground was the generator when the gpe was 91,700,000.00 j?

Explanation:

Step1: Recall the formula for gravitational potential energy (GPE)

The formula for gravitational potential energy is $GPE = mgh$, where $m$ is the mass, $g$ is the acceleration due to gravity (we'll use $g = 9.8\ m/s^2$), and $h$ is the height (distance above the ground) we want to find. We need to solve this formula for $h$, so we get $h=\frac{GPE}{mg}$.

Step2: Identify the given values

We are given that the mass $m = 12400.05\ kg$, the GPE $= 91700000.00\ J$, and $g = 9.8\ m/s^2$.

Step3: Substitute the values into the formula for $h$

Substitute $m = 12400.05\ kg$, $GPE = 91700000.00\ J$, and $g = 9.8\ m/s^2$ into $h=\frac{GPE}{mg}$:

\[
h=\frac{91700000.00}{12400.05\times9.8}
\]

First, calculate the denominator: $12400.05\times9.8 = 12400.05\times(10 - 0.2)=12400.05\times10-12400.05\times0.2 = 124000.5 - 2480.01 = 121520.49$

Then, calculate the numerator divided by the denominator: $h=\frac{91700000.00}{121520.49}\approx754.6$ (rounded to one decimal place)

Answer:

The generator was approximately $\boldsymbol{755\ m}$ (or $754.6\ m$) above the ground. (The slight difference may be due to rounding during calculation steps; if we do the division more precisely: $\frac{91700000}{12400.05\times9.8}=\frac{91700000}{121520.49}\approx754.6\approx755$ when rounded to a reasonable number of significant figures or decimal places for height in this context)