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2 years ago

Two factors effecting the magnitude of the force of gravity between 2 objects are...

Physics

1 answer:

slavikrds [6]2 years ago

5 0

Answer:

Two factors effecting the magnitude of the force of gravity between 2 objects are the product of their masses and square of distance between them.

Explanation:

According to Newton's law of universal gravitation

$F = G\frac{m_{1}m_{2} }{r^{2} } }$

where F is the gravitational force, G is the universal gravitational constant and its value is 6.6743 × 10⁻¹¹ Nm²/kg₂ , m₁ and m₂ are masses of bodies and r is the distance between them.

It can be seen from the above equation that F is directly proportional to the product of the masses and inversely proportional to the square of distance between them.

F ∝ m₁m₂

F ∝ 1/r²

As far as the masses of the bodies increase, magnitude of the Gravitational force increases and if distance between them increase then Gravitational force between them decreases.

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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

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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 $T_e = 4.3^{o}C$

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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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