An object weighs 43 N. It displaces 9 N of water.  While in water, what is the object's weight? Input a whole number answer excluding the unit.

1. Break Down the Problem

We need to determine the weight of an object while it is submerged in water. The information provided indicates the object's weight in air and the weight of the water it displaces.

2. Relevant Concepts

According to Archimedes' principle, when an object is submerged in a fluid, it experiences an upward buoyant force equal to the weight of the fluid displaced. The effective weight of the object when submerged can be calculated using the formula: $W_{\text{effective}} = W_{\text{object}} - W_{\text{displaced water}}$

3. Analysis and Detail

Given:

• Weight of the object, $W_{\text{object}} = 43 \, \text{N}$
• Weight of the displaced water, $W_{\text{displaced water}} = 9 \, \text{N}$

Now we can substitute these values into the formula: $W_{\text{effective}} = 43 \, \text{N} - 9 \, \text{N}$

4. Verify and Summarize

Calculating: $W_{\text{effective}} = 34 \, \text{N}$ So, the effective weight of the object while it is in water is $34 \, \text{N}$.

Final Answer

34