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3 years ago

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The Earth can be approximated as a sphere of uniform density, rotating on its axis once a day. The mass of the Earth is 5.97×102

4 kg , the radius of the Earth is 6.38×106 m , and the period of rotation for the Earth is 24.0 hrs . Part A What is the moment of inertia of the Earth? Use the uniform-sphere approximation described in the introduction. Express your answer in kilogram meters squared to three significant figures. View Available Hint(s)

1 answer:

svetlana [45]3 years ago

3 0

Answer:

I = 97.2 10³⁶ kg m²

Explanation:

The moment of inertia of a body the expression of inertia in the rotational movement and is described by the expression

I = ∫ r² dm

In this problem we are told to use the moment of inertia of a uniform sphere, the expression of this moment of inertia is

I = 2/5 M r²

where m is the mass of the earth and r is the radius of the earth.

Let's calculate

I = 2/5 5.97 10²⁴ (6.38 10⁶)²

I = 97.2 10³⁶ kg m²

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A horizontal 745 N merry-go-round of radius

Arturiano [62]

Answer:

The kinetic energy of the merry-goround after 3.62 s is 544J

Explanation:

Given :

Weight w = 745 N

Radius r = 1.45 m

Force = 56.3 N

To Find:

The kinetic energy of the merry-go round after 3.62 = ?

Solution:

Step 1: Finding the Mass of merry-go-round

$m = \frac{ weight}{g}$

$m = \frac{745}{9.81 }$

m = 76.02 kg

Step 2: Finding the Moment of Inertia of solid cylinder

Moment of Inertia of solid cylinder I = $0.5 \times m \times r^2$

Substituting the values

Moment of Inertia of solid cylinder I

=> $0.5 \times 76.02 \times (1.45)^2$

=> $0.5 \times 76.02\times 2.1025$

=> $79.91 kg.m^2$

Step 3: Finding the Torque applied T

Torque applied T = $F \times r$

Substituting the values

T = $56.3 \times 1.45$

T = 81.635 N.m

Step 4: Finding the Angular acceleration

Angular acceleration , $\alpha = \frac{Torque}{Inertia}$

Substituting the values,

$\alpha = \frac{81.635}{79.91}$

$\alpha = 1.021 rad/s^2$

Step 4: Finding the Final angular velocity

Final angular velocity , $\omega = \alpha \times t$

Substituting the values,

$\omega = 1.021 \times 3.62$

$\omega = 3.69 rad/s$

Now KE (100% rotational) after 3.62s is:

KE = $0.5 \times I \times \omega^2$

KE = $0.5 \times 79.91 \times 3.69^2$

KE = 544J

6 0

3 years ago

Problem 8: Consider an experimental setup where charged particles (electrons or protons) are first accelerated by an electric fi

yanalaym [24]

Answer:

10581.59 V

Explanation:

We are given that

Magnetic field=B=0.65 T

Speed of electron= $v=6.1\times 10^7m/s$

Charge on electron, $q=e=1.6\times 10^{-19} C$

Mass of electron, $m_e=9.1\times 10^{-31} kg$

We have to find the potential difference in volts required in the first part of the experiment to accelerate electrons.

$V=\frac{v^2m_e}{2e}$

Where V=Potential difference

$m_e=$ Mass of electron

v=Velocity of electron

Using the formula

$V=\frac{(6.1\times 10^7)^2\times 9.1\times 10^{-31}}{2\times 1.6\times 10^{-19}}$

$V=10581.59 V$

Hence, the potential difference=10581.59 V

8 0

3 years ago