A wooden ball with a weight of 24.0 N hangs from a string tied to a spring scale

Question

A wooden ball with a weight of 24.0 N hangs from a string tied to a spring scale. When the ball is at rest, exactly 50% submerged in water, the spring scale reads 5.00 N. For this problem, we will use a density of water of 1000 kg/m3, and we will use g 10.0 m/s2 (a) If the ball was only 20.0% submerged in water instead, what would the spring scale read? 16.4 (b) Determine the density of the ball 750 X kg/m3 (c) Determine the volume of the ball m3 Additional Materials a section 9-3: Archimedes' Principle Section 9-5: An Example Buoyancy Problem eBook

Explanation / Answer

Buoyant force
In the first case
0.5V*dg = 24- 5 = 19 ---------------1
Vdg = 38 N

In the second case
0.2*Vdg = (24 – F)
0.2*38 = (24 – F)
F = 24-7.6 = 16.4N

Answer for a = 16.4N
==================
b)
V = 38/ (d g ),

d is density of water=1000 kg/m3 and hence density of the ball = (24/g) / ( 38/dg) =0.63d = 630 kg/m^3

===============
c )
v = 38 /(9800) = 0.0038 m^3
=============================

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.