Calculate the mass of an 86kg object on the moon

For heat transfer purposes, an egg can be considered to be a 5.5-cm-diameter sphere having the properties of water. An egg that

Harrizon [31]

Answer:

The time taken is $t = 40007 sec$

Explanation:

From the question we are told that

The diameter of the egg is $d_e = 5.5cm = \frac{5.5}{100} = 5.5*10^{-2}m$

The initial temperature of egg the $T_e = 4.3^{o}C$

The temperature of the boiling water $T_b = 100^oC$

The heat transfer coefficient is $H = 800 W/m^2 \cdot K$

The final temperature is $T_e_f = 74^oC$

The thermal conductivity of water is $k = 0.607 W/m^oC$

The diffusivity of the egg $\alpha = 0.146 * 10^{-6} m^2 /s$

Using one term approximation

We have the

$\frac{T_e_f - T_b}{T_e - T_b} = Ae^{-\lambda ^2 \tau}$

The radius is $r = \frac{5.5*10^ {-2}}{2} =2.75*10^{-2}m$ Note that this radius is approximation to that of a real egg

Now we need to obtain the Biot number which help indicate the value of $A \ and \ \lambda$ to use in the above equation

The Biot number is mathematically represented as

$Bi = \frac{H r}{k}$

Substituting values

$Bi = \frac{800 * 2.75 *10^{-2}}{0.607}$

$= 36.24$

So for this value which greater than 0.1 the coefficient $\lambda_1 \ and \ A_1$ is

$\lambda = 3.06632$

$A = 1.9942$

Substituting this into equation 1 we have

$\frac{74- 100}{4.3 - 100} = 1.9942 e^{-(3.0632^2) \tau}$

$0.2717= 1.9942 e^{-(3.0632^2) \tau}$

$0.2717= 1.9942 e^{-9.383 \tau}$

$0.13624 = e^{-9.383 \tau}$

Taking natural log of both sides

$-1.993 = -9.383\ \tau$

$\tau = 0.2124$

The time required for the egg to be cooked is mathematically represented as

$t = \frac{\tau r^2}{\alpha }$

substituting value is

$= \frac{0.2124 * 2.75 *10^{-2}}{0.146 *10^{-6}}$

$t = 40007 sec$

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3 years ago

Your starship, the Aimless Wanderer,lands on the mysterious planet Mongo. As chief scientist-engineer,you make the following mea

Setler [38]

Answer:

a) M = 4,997 10²⁰ kg , b) T = 1.43 10³ s

Explanation:

a) This exercise should be solved in several parts, let's start by calculating the acceleration of gravity of this planet from kinematics

v = v₀ - a t

As it indicates that there is no atmosphere, the friction force is zero and the initial and final velocity have the same module, but the opposite direction

a = (v₀ - v) / t

a = (15 - (-15)) /9.00 = 30/9

a = 3.33 m / s²

Now we use Newton's second law where force is the force of universal attraction

F = m a

G m M / r² = m a

M = a r² / G

Let's calculate

M = 3.33 (1.00 10⁵)² / 6.67 10⁻¹¹

M = 4,997 10²⁰ kg

b) The period of the ship's orbit

In this case we have a centripetal acceleration

The radius of the orbit is the radius of the plant plus the height of the ship from the surface

R = $R_{m}$ + h

R = 1 10⁵ + 2.00 10⁴

R = 12 10⁴ m

F = m a

G m M / R² = m a

Centripetal acceleration is

a = v² / R

The orbit is circular therefore the velocity module is constant, so we can use the equation of uniform motion, where the distance is the length of the orbit, for a circle

d = 2π R

v = d / t

v = 2π R / T

Let's replace

G m M / R² = m (2π R / T)² / R

G M = R³ 4π² / T²

T² = 4π² R³ / G M

T² = (4π² (12 10⁴)³ / (6.67 10⁻¹¹ 4,997 10²⁰)

T² = 6.82 10¹⁶ / 3.33 10¹⁰

T = √ (2,048 10⁶)

T = 1.43 10³ s

3 0