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NemiM [27]
2 years ago
14

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)
Physics
1 answer:
svetlana [45]2 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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5 0
3 years ago
In 'coin on card' experiment a smooth card is used. ​
KIM [24]

Answer:

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What type of pollution did the Clean Water Act succeed in limiting?
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3 years ago
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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
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