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ch4aika [34]
2 years ago
14

What is the approximate size of the Earth's magnetic field? (dont ask me to specify thats what the question is and im as confuse

d as heck too)
Physics
1 answer:
Olegator [25]2 years ago
8 0

Answer:

The Earth's magnetic field intensity is roughly between 25,000 - 65,000 nT (.25 -.65 gauss).

Explanation:

<em>To measure the Earth's magnetism in any place, we must measure the direction and intensity of the field. The Earth's magnetic field is described by seven parameters. These are declination (D), inclination (I), horizontal intensity (H), the north (X), and east (Y) components of the horizontal intensity, vertical intensity (Z), and total intensity (F). The parameters describing the direction of the magnetic field are declination (D) and inclination (I). D and I are measured in units of degrees, positive east for D and positive down for me. The intensity of the total field (F) is described by the horizontal component (H), vertical component (Z), and the north (X) and east (Y) components of the horizontal intensity. These components may be measured in units of gauss but are generally reported in nanoTesla (1nT * 100,000 = 1 gauss). </em><em>The Earth's magnetic field intensity is roughly between 25,000 - 65,000 nT (.25 - .65 gauss). </em><em>Magnetic declination is the angle between magnetic north and true north. D is considered positive when the angle measured is east of true north and negative when west.  The magnetic inclination is the angle between the horizontal plane and the total field vector, measured positive into Earth. In older literature, the term “magnetic elements” is often referred to as D, I, and H.</em>

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marissa [1.9K]

Answer:

35.20 m

Explanation:

By the law of conservation of energy we have,

mg(H-h)=\frac{1}{2}mv^2

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H= maximum height attained

Also, the horizontal distance covered by the skier is

D=vt

=v\frac{2g}{h}

\Rightarrow v^2=\frac{gD^2}{2h}

thus, height H in terms of D  is given by

H=\frac{D^2}{2h}+h

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H=35.20 m

4 0
3 years ago
Arocket launches at an angle of 33.6 degrees from the horizontal at a
babymother [125]

Answer:

Y component = 32.37

Explanation:

Given:

Angle of projection of the rocket is, \theta=33.6

Initial velocity of the rocket is, u=58.5

A vector at an angle \theta with the horizontal can be resolved into mutually perpendicular components; one along the horizontal direction and the other along the vertical direction.

If a vector 'A' makes angle \theta with the horizontal, then the horizontal and vertical components are given as:

A_x=A\cos \theta(\textrm{Horizontal or X component})\\A_y=A\sin \theta(\textrm{Vertical or Y component})

Here, as the velocity is a vector quantity and makes an angle of 33.6 with the horizontal, its Y component is given as:

u_y=u\sin \theta

Plug in the given values and solve for u_y. This gives,

u_y=(58.5)(\sin 33.6)\\u_y=58.5\times 0.55339\\u_y=32.373\approx32.37(\textrm{Rounded to two decimal places})

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5 0
2 years ago
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Energy that is passed on from one trophic level to the next is called what?
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Answer:

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Explanation:

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