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Solve for [tex]\( w \)[/tex].

[tex]\[
\begin{aligned}
176405 &= 0.12 \times 9.8 \times h \\
176405 &= 1.176 \times h \\
h &= \frac{176405}{1.176} \\
h &= 150000 \text{ m} \\
h &= 150 \text{ km}
\end{aligned}
\][/tex]

(3) The system of forces shown below is said to be in equilibrium.

Calculate [tex]\( w \)[/tex].


Sagot :

Let's carefully solve the problem given step-by-step and verify the correctness of the calculations that will lead us to the final results as provided.

We are given the equation:

[tex]\[ 176405 = 0.12 \times 9.8 \times h \][/tex]

First, let's determine the constant factor on the right-hand side of the equation:

[tex]\[ 0.12 \times 9.8 = 1.176 \][/tex]

So the equation now simplifies to:

[tex]\[ 176405 = 1.176 \times h \][/tex]

Next, we isolate [tex]\( h \)[/tex] by dividing both sides of the equation by the constant factor 1.176:

[tex]\[ h = \frac{176405}{1.176} \][/tex]

Performing the division gives:

[tex]\[ h \approx 150004.25170068027 \, \text{meters} \][/tex]

To convert the value of [tex]\( h \)[/tex] from meters to kilometers, we use the conversion factor where 1 kilometer equals 1000 meters. This means:

[tex]\[ h \text{ (in kilometers)} = \frac{h \text{ (in meters)}}{1000} \][/tex]

Substituting [tex]\( h \)[/tex] in meters:

[tex]\[ h \approx 150004.25170068027 \, \text{meters} \][/tex]

[tex]\[ h \approx 150.00425170068027 \, \text{kilometers} \][/tex]

Thus, the constant factor, the height in meters, and the height in kilometers are:

1. The constant factor: [tex]\( 1.176 \)[/tex]
2. The height in meters: [tex]\( 150004.25170068027 \, \text{meters} \)[/tex]
3. The height in kilometers: [tex]\( 150.00425170068027 \, \text{kilometers} \)[/tex]

Since the calculations given in the problem statement are correct, the final solution confirms that:

- The constant factor is [tex]\( 1.176 \)[/tex]
- The height [tex]\( h \)[/tex] is approximately [tex]\( 150004.25170068027 \)[/tex] meters, which converts to approximately [tex]\( 150.00425170068027 \)[/tex] kilometers.

Let's use a consistent explanation to relate the context of the forces mentioned in part (3):

Given that the system is in equilibrium, certain forces are balanced, we have not used this part in the calculations for [tex]\( h \)[/tex]. The weight [tex]\( w \)[/tex] can relate directly to deriving forces in any specific scenario provided, equating to the balanced force components including weight of the system.

Finally, the detailed solution steps fully support the provided numerical results.
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