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Lena [83]
2 years ago
13

A spherical shell has an outside radius of 2.60 cm and an inside radius of a. The shell wall has uniform thickness and is made o

f a material with density 4.70 g/cm3. The space inside the shell is filled with a liquid having a density of 1.23 g/cm3. (a) Find the mass m of the sphere, including its contents, as a function of a. (b) In the answer to part (a), if a is regarded as a variable, for what value of a does m have its maximum possible value
Physics
1 answer:
lbvjy [14]2 years ago
4 0

Answer:

The answer is below

Explanation:

a) The volume of a sphere is:

Volume = (4/3)πr³; where r is the radius of the shell.

Given the outside radius of 2.60 cm and inner radius of a cm, the volume of the spherical shell is:

Volume of spherical shell = \frac{4}{3} \pi (2.6^3-a^3) cm³

But Density = mass / volume;   Mass = density * volume.

Therefore, mass of spherical shell = density * volume

mass of spherical shell = 4.70\ g/cm^3 * \frac{4}{3} \pi (2.6^3-a^3) cm³

Mass of liquid = volume of inner shell * density of liquid

Mass of liquid = \frac{4}{3} \pi a^3\ cm^3*1.23\ g/cm^3

Total mass of sphere including contents = mass of spherical shell + mass of liquid

Total mass of sphere including contents (M) = 4.70\ g/cm^3 * \frac{4}{3} \pi (2.6^3-a^3)\ cm^3  +  \frac{4}{3} \pi a^3\ cm^3*1.23\ g/cm^3 =

Total mass of sphere including contents (M) = (346 - 14.5a³) grams

b) The mass is maximum when the value of a = 0

M = 346 - 14.5a³

M' = 43.5a² = 0

43.5a² = 0

a = 0

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The complete question is;

Find the moment of inertia about each of the following axes for a rod that is 0.36 cm in diameter and 1.70m long, with a mass of 5.00 × 10 ^(−2) kg.

A) About an axis perpendicular to the rod and passing through its center in kg.m²

B) About an axis perpendicular to the rod and passing through one end in kg.m²

C) About an axis along the length of the rod in kg.m²

Answer:

A) I = 0.012 kg.m²

B) I = 0.048 kg.m²

C) I = 8.1 × 10^(-8) kg.m²

Explanation:

We are given;

Diameter = 0.36 cm = 0.36 × 10^(−2) m

Length; L = 1.7m

Mass;m = 5 × 10^(−2) kg

A) For an axis perpendicular to the rod and passing through its center, the formula for the moment of inertia is;

I = mL²/12

I = (5 × 10^(−2) × 1.7²)/12

I = 0.012 kg.m²

B) For an axis perpendicular to the rod and passing through one end, the formula for the moment of inertia is;

I = mL²/3

So,

I = (5 × 10^(−2) × 1.7²)/3

I = 0.048 kg.m²

C) For an axis along the length of the rod, the formula for the moment of inertia is; I = mr²/2

We have diameter = 0.36 × 10^(−2) m, thus radius;r = (0.36 × 10^(−2))/2 = 0.18 × 10^(−2) m

I = (5 × 10^(−2) × (0.18 × 10^(−2))^2)/2

I = 8.1 × 10^(-8) kg.m²

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Answer:

The value is   A =  2.80 *10^{-4} \  m^2

Explanation:

From the question we are told that

The  operating temperature is  T  =  2450 \  K

The emissivity is  e =  0.350

 The  power rating is  P  =  200 \  W

Generally the area is mathematically represented as

      A = \frac{P}{ e *  \sigma  *  T^2}

Where  \sigma is the Stefan Boltzmann constant  with value  

      \sigma  =  5.67 *10^{-8} \  W/m^2\cdot K^4

So

     A =  \frac{200}{0.350 *  5.67*10^{-8} *  2450^{4}}

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