A block with mass M = 3 kg is moving on a flat surface with constant speed v1 =

Each part of the mixture has the same appearance.

Luden [163]

What is that supposed to mean?

5 0

3 years ago

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

Olegator [25]

Answer:

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

Explanation:

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

8 0

2 years ago

Select the correct answer.

motikmotik

I'm not sure. But it can be A or C.

4 0

3 years ago

Which phase of matter makes up stars?

nydimaria [60]

I believe it’s gas because that’s all that stars are really made of

4 0

3 years ago

Read 2 more answers

A brass rod with a length of 1.22 m and a cross-sectional area of 2.19 cm2 is fastened end to end to a nickel rod with length L

GaryK [48]

Answer:

a) L₂ = 0.676 m

b) σ₁ = 2.28*10⁸ N/m²

σ₂ = 9.62*10⁸ N/m²

c) ε₁ = 0.00253678

ε₂ = 0.00457875

Explanation:

Given info

L₁ = 1.22 m

A₁ = 2.19 cm² = 2.19*10⁻⁴ m²

L₂ = ?

A₂ = 0.52 cm² = 0.52*10⁻⁴ m²

P = 5.00*10⁴ N

E₁ = 9*10¹⁰ N/m²

E₂ = 2.1*10¹¹ N/m²

In order to get the length L of the nickel rod if the elongations of the two rods are equal, we can say that

ΔL₁ = ΔL₂ ⇒ P*L₁/(A₁*E₁) = P*L₂/(A₂*E₂)

⇒ L₂ = A₂*E₂*L₁ / (A₁*E₁)

⇒ L₂ = (0.52*10⁻⁴ m²)*(2.1*10¹¹ N/m²)*(1.22 m) / (2.19*10⁻⁴ m²*9*10¹⁰ N/m²)

⇒ L₂ = 0.676 m

The stress in the brass rod is obtained as follows

σ₁ = P/A₁ ⇒ σ = 5.00*10⁴ N / 2.19*10⁻⁴ m² = 2.28*10⁸ N/m²

The stress in the niquel rod is obtained as follows

σ₂ = P/A₂ ⇒ σ = 5.00*10⁴ N / 0.52*10⁻⁴ m² = 9.62*10⁸ N/m²

The strain in the brass rod is obtained as follows

σ₁ = E₁*ε₁ ⇒ ε₁ = σ₁ / E₁

⇒ ε₁ = 2.28*10⁸ N/m² / 9*10¹⁰ N/m² = 0.00253678

The strain in the niquel rod is obtained as follows

σ₂ = E₂*ε₂ ⇒ ε₂ = σ₂ / E₂

⇒ ε₂ = 9.62*10⁸ N/m² / 2.1*10¹¹ N/m² = 0.00457875

3 0

2 years ago