the amount a spring will stretch s varies directly with the weight F attached to the spring of spring scratches 2.22 inches with 37 pounds attached how far will it stretch when 25 pounds are attached?
0.6in
16.67in
0.95in
1.5in***
2.22/37=?/25
?=2.22*25/37=1.5 inch.
no.
To solve this problem, we need to use the concept of direct variation. If the amount a spring will stretch (s) varies directly with the weight attached to the spring (F), we can write the equation as:
s = kF
where "k" is the constant of variation.
Now, we are given that the spring stretches 2.22 inches when 37 pounds are attached. We can use this information to find the value of "k".
2.22 = k * 37
To find "k", we divide both sides of the equation by 37:
k = 2.22 / 37
Now that we have the value of "k", we can use it to find the stretch of the spring when 25 pounds are attached. We will plug in the given weight value (F = 25) into the equation:
s = k * F
s = (2.22 / 37) * 25
Calculating this expression, we find:
s ≈ 1.5 inches
Therefore, the correct answer is 1.5 inches.