The length of a pencil lcm, varies directly with it's mass, mg, and inversely with it's area of cross-section, Acm². When m=4 and A=0•4, I=18.
A. Find the formula connecting l, m and A
B. Hence find the length of a pencil which has a mass of 3g and a cross-sectional area of 0•5²
A. We can write the formula as:
l = k(mg/A)
where k is the constant of proportionality. To find k, we plug in the given values:
18 = k(4*0.4/1)
18 = k(1.6)
k = 11.25
Therefore, the formula connecting l, m and A is:
l = 11.25(mg/A)
B. We are given m = 3g and A = 0.5² = 0.25cm². Plugging in these values into the formula, we get:
l = 11.25(3*0.25/0.25)
l = 33.75
Therefore, the length of the pencil is 33.75 cm.
To find the formula connecting length (l), mass (m), and cross-sectional area (A), we can represent the given conditions using a proportionality constant (k).
Since the length varies directly with the mass and inversely with the area, the formula can be written as:
l ∝ (m/A)
where ∝ denotes "is proportional to."
To remove the proportionality symbol, we introduce a constant of proportionality, k:
l = k * (m/A) ----- (Equation 1)
Now, let's use the given information: when m = 4 and A = 0.4, the length is 18 (given as I).
So, substituting these values into Equation 1, we have:
18 = k * (4/0.4)
To simplify, we divide 4 by 0.4:
18 = k * 10
Now, let's solve for k:
k = 18/10
k = 1.8
Therefore, the formula connecting length (l), mass (m), and cross-sectional area (A) is:
l = 1.8 * (m/A)
Now, we can use this formula to find the length of a pencil with a mass of 3g and a cross-sectional area of 0.5².
Substituting the values into the formula:
l = 1.8 * (3/0.5²)
l = 1.8 * (3/0.25)
l = 1.8 * 12
l = 21.6
Hence, the length of the pencil with a mass of 3g and a cross-sectional area of 0.5² is 21.6 units.