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NemiM [27]
3 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]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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CHAPTER 6: KINETICS OF A PARTICLE
vekshin1

Explanation:

Work done by winch = kinetic energy of car

∫ T ds = ½ mv²

∫ 225s ds = ½ mv²

225/2 s² = ½ mv²

225 s² = mv²

v = 15s / √m

Given s = 10 m and m = 2500 kg:

v = 15 (10) / √2500

v = 3 m/s

8 0
3 years ago
In which phase of mitosis are chromosome first seen as a result of chromatin coiling?
Kruka [31]
<span>Mitosis is a a means for cells to split and produce exact copies of themselves. The process produces two identical copies of the original cell and occurs throughout the human body. Mitosis is divided up into four main phases known as prophase, metaphase, anaphase and telophase. The chromosomes first become visible in early prophase.</span>
7 0
3 years ago
A 70 kg student stands on top of a 5.0 m platform diving board . how much gravitational potential energy does the student have?
Marianna [84]

Answer:

a. P.E = 3430Joules.

b. Workdone = 3430Nm

Explanation:

<u>Given the following data;</u>

Mass = 70kg

Distance = 5m

We know that acceleration due to gravity is equal to 9.8m/s²

To find the potential energy;

Potential energy = mgh

P.E = 70*9.8*5

<em>P.E = 3430J</em>

b. To find the workdone;

Workdone = force * distance

But force = mass * acceleration

Force = 70*9.8

Force = 686 Newton.

Workdone = 686 * 5

<em>Workdone = 3430Nm</em>

6 0
2 years ago
PLEASE HELP, LOOK AT PHOTO PLZ!!!
PIT_PIT [208]

Answer:

a) 17 km

b) 9 km

Explanation:

The distance is the length of the path.

A to C: 5 km

B to C: 4 km

C to B: 4 km

B to C: 4 km

Total distance = 5 km + 4 km + 4 km + 4 km = 17 km

Displacement is the difference between the starting point and ending point.

Displacement = 9 km − 0 km = 9 km

3 0
3 years ago
A 50-kg person stands 1.5 m away from one end of a uniform 6.0-m-long scaffold of mass 70.0 kg.
babymother [125]

Answer

given,

mass of the person, m = 50 Kg

length of scaffold = 6 m

mass of scaffold, M= 70 Kg

distance of person standing from one end = 1.5 m

Tension in the vertical rope = ?

now equating all the vertical forces acting in the system.

T₁ + T₂ = m g + M g

T₁ + T₂ = 50 x 9.8  + 70 x 9.8

T₁ + T₂ = 1176...........(1)

system is equilibrium so, the moment along the system will also be zero.

taking moment about rope with tension T₂.

now,

T₁ x 6 - mg x (6-1.5) - M g x 3 = 0

'3 m' is used because the weight of the scaffold pass through center of gravity.

6 T₁ = 50 x 9.8 x 4.5 + 70 x 9.8 x 3

6 T₁ = 4263

    T₁ = 710.5 N

from equation (1)

T₂ = 1176 - 710.5

 T₂ = 465.5 N

hence, T₁ = 710.5 N and T₂ = 465.5 N

4 0
3 years ago
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