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A platinum sphere with radius 1.65 cm is totally immersed in mercury. find the weight of the sphere, the buoyant force acting on the sphere, and the sphere\'s apparent weight. the densities of platinum and mercury are 2.14 × 104 kg/m3 and 1.36 × 104 kg/m3, respectively.

Respuesta :

W0lf93
Gvn that, Radius r = 1.65cm then the Diameter d = 3.3 cm Volume V = 4/3pi*(D/3)^3 = 4/3*3.14*(3.3)^3/8 = 4/3*3.14*35.937/8 = 4/3*3.14*4.492125 = 4/3*14.1052725 = 56.421.9/3 = 18.80703 dm^3 We know that gravity g=9.806m/s^2 p pt = 2.14 * 104 kg/m^3 = 21.4 p hg = 1.36 * 104 kg/m^3 = 13.6 Real Weight Wr = V*p pt*g = 18.80703*21.4*9.806 = 3946.625 N Buoyant force Fb= V*p hg*g = 18.80703*13.6*9.806 = 2508.136 N Apparent weight = Wr-Fb = 3946.625-2508.136 = 1438.489 N Therefore Apparent weight is 1438.489 N
johanrusli

(a) weight of the sphere = 3.95 N

(b) buoyant force = 2.51 N

(c) sphere's apparent weight = 1.44 N

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Further explanation

The basic formula of pressure that needs to be recalled is:

Pressure = Force / Cross-sectional Area

or symbolized:

[tex]\boxed {P = F \div A}[/tex]

P = Pressure (Pa)

F = Force (N)

A = Cross-sectional Area (m²)

Let us now tackle the problem !

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Given:

Radius Sphere = R = 1.65 cm = 0.0165 m

Density of Platinum = ρ_p = 2.14 × 10⁴ kg/m³

Density of Mercury = ρ_m = 1.36 × 10⁴ kg/m³

Asked:

(a) weight of the sphere = w = ?

(b) buoyant force = F = ?

(c) sphere's apparent weight = w' = ?

Solution:

Part (a):

[tex]w = mg[/tex]

[tex]w = \rho_p V g[/tex]

[tex]w = \rho (\frac{4}{3} \pi R^3) g[/tex]

[tex]w = \frac{4}{3} \pi \rho g R^3[/tex]

[tex]w = \frac{4}{3} \pi \times 2.14 \times 10^4 \times 9.8 \times 0.0165^3[/tex]

[tex]\boxed {w = 3.95 \texttt{ N}}[/tex]

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Part (b):

[tex]F = \rho_m g V[/tex]

[tex]F = \rho_m g (\frac{4}{3} \pi R^3} )[/tex]

[tex]F = \frac{4}{3} \pi \rho_m g R^3[/tex]

[tex]F = \frac{4}{3} \pi \times 1.36 \times 10^4 \times 9.8 \times 0.0165^3[/tex]

[tex]\boxed {F = 2.51 \texttt{ N} }[/tex]

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Part (c):

[tex]w' = w - F[/tex]

[tex]w' = 3.95 - 2.51[/tex]

[tex]\boxed {w' = 1.44 \texttt{ N}}[/tex]

[tex]\texttt{ }[/tex]

Learn more

  • Minimum Coefficient of Static Friction : https://brainly.com/question/5884009
  • The Pressure In A Sealed Plastic Container : https://brainly.com/question/10209135
  • Effect of Earth’s Gravity on Objects : https://brainly.com/question/8844454

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Answer details

Grade: High School

Subject: Physics

Chapter: Pressure

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