Question

convert 60.0 kg into m³. use 3 significant figures and units. a) using the graph, not the equation. starting at 60.0 kg on the y - axis, then following across to the line, then straight down to the x - axis, gives 1 b) use the equation to solve for x when y = 60.0 g, show your work and use units let y = 60.0 kg. then, solving the equation for the line for x 2 - 3 x = ____________ = 4 5__

Explanation:

Step1: Identify the relationship

Since no equation is given for part - b, assume the relationship between \(y\) (mass in kg) and \(x\) (volume in \(m^{3}\)) is of the form \(y = mx + c\). For a linear relationship passing through the origin (\(c = 0\)), \(y=mx\) where \(m\) is the slope. From the graph, we can estimate values. But for part - b, we need to assume a general linear form. However, since the problem is incomplete without the equation, we'll focus on part - a. Reading from the graph, when \(y = 60.0\) kg on the \(y\) - axis, moving horizontally to the line and then vertically down to the \(x\) - axis.

Step2: Read the value

By visually inspecting the graph, we estimate the value on the \(x\) - axis. Let's assume the scale of the \(x\) - axis is such that we get a value. But since the graph's scale is not clear, assume we get \(x = 0.0600m^{3}\) (this is an example value based on a general assumption of a linear relationship between mass and volume where density is constant).

Answer:

a) \(0.0600m^{3}\) (Note: The actual value depends on the accurate reading of the graph which is not fully specified in the problem. This is a sample value for illustration purposes)
b) Since the equation is not given, we cannot solve part - b.