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3 years ago

Determine the moment of inertia of a 14.8 kg sphere of radius 0.752 m when the axis of rotation is through its center.

Physics

1 answer:

vfiekz [6]3 years ago

6 0

Answer:

$I=3.348Kgm^2$

Explanation:

From the question we are told that:

Mass $m=14.8kg$

Radius $r=0.752m$

Generally the equation for inertia is mathematically given by

$I=\frac{2}{5}mR^2$

$I=\frac{2}{5}14.8*(0.752)^2$

$I=3.348Kgm^2$

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Naya [18.7K]

Answer:

Width = 680 [m]

Explanation:

To solve this problem we must know the speed of sound in the air, conducting an internet search, we find that this speed has a value of v = 340 [m/s].

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3 years ago

Question 1 of 4 Attempt 4 The acceleration due to gravity, ???? , is constant at sea level on the Earth's surface. However, the

Evgen [1.6K]

Answer:

g(h) = g ( 1 - 2(h/R) )

*At first order on h/R*

Explanation:

Hi!

We can derive this expression for distances h small compared to the earth's radius R.

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Remember that this law is:

$F = G \frac{m_1m_2}{r^2}$

In the present case m1 will be the mass of the earth.

Additionally, if we remember Newton's second law for the mass m2 (with m2 constant):

$F = m_2a$

Therefore, we can see that

$a(r) = G \frac{m_1}{r^2}$

With a the acceleration due to the earth's mass.

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$a(R+h) \approx a(R) + h \frac{da(r)}{dr}_{r=R}$

a(R) is actually the constant acceleration at sea level

and

$a(R) =G \frac{m_1}{R^2} \\ \frac{da(r)}{dr}_{r=R} = -2 G\frac{m_1}{R^3}$

Therefore:

$a(R+h) \approx G\frac{m_1}{R^2} -2G\frac{m_1}{R^2} \frac{h}{R} = g(1-2\frac{h}{R})$

Consider that the error in this expresion is quadratic in (h/R), and to consider quadratic correctiosn you must expand the taylor series to the next power:

$a(R+h) \approx a(R) + h \frac{da(r)}{dr}_{r=R} + \frac{h^2}{2!} \frac{d^2a(r)}{dr^2}_{r=R}$