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3 years ago

Which air mass would produce cold dry weather in the winter?

Physics

2 answers:

quester [9]3 years ago

8 0

The correct answer is - Continental polar air mass.

The continental polar air masses are very big air masses that are very cold, and dry. They are much heavier than the air masses that surround them, especially the ones that are further south of them, so they move with ease and push them away, thus spreading over a large territory very quickly and bringing in very cold and dry weather during the winter. This types of air masses are forming over Siberia , and the central part of Canada.

Solnce55 [7]3 years ago

6 0

It is Continental polar ( only if this is for apex)

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In what order does light pass through structures of the eye? lens, cornea, retina cornea, pupil, lens pupil, iris, lens pupil, c

Mrrafil [7]

Answer: Light passes through the front of the eye (cornea) to the lens. The cornea and the lens help to focus the light rays onto the back of the eye (retina). The cells in the retina absorb and convert the light to electrochemical impulses which are transferred along the optic nerve and then to the brain.

Explanation:

When light rays reflect off an object and enter the eyes through the cornea (the transparent outer covering of the eye), you can then see that object. The cornea bends, or refracts, the rays that pass through the round hole of the pupil.

4 0

3 years ago

Planets are not uniform inside. Normally, they are densest at the center and have decreasing density outward toward the surface.

SSSSS [86.1K]

Answer:

a = 9.94 m/s²

Explanation:

given,

density at center= 1.6 x 10⁴ kg/m³

density at the surface = 2100 Kg/m³

volume mass density as function of distance

$\rho(r) = ar^2 - br^3$

r is the radius of the spherical shell

dr is the thickness

volume of shell

$dV = 4 \pi r^2 dr$

mass of shell

$dM = \rho(r)dV$

$\rho = \rho_0 - br$

now,

$dM = (\rho_0 - br)(4 \pi r^2)dr$

integrating both side

$M = \int_0^{R} (\rho_0 - br)(4 \pi r^2)dr$

$M = \dfrac{4\pi}{3}R^3\rho_0 - \pi R^4(\dfrac{\rho_0-\rho}{R})$

$M = \pi R^3(\dfrac{\rho_0}{3}+\rho)$

we know,

$a = \dfrac{GM}{R^2}$

$a = \dfrac{G( \pi R^3(\dfrac{\rho_0}{3}+\rho))}{R^2}$

$a =\pi RG(\dfrac{\rho_0}{3}+\rho)$

$a =\pi (6.674\times 10^{-11}\times 6.38 \times 10^6)(\dfrac{1.60\times 10^4}{3}+2.1\times 10^3)$

a = 9.94 m/s²

7 0

2 years ago

A 20 newton force moves an object southeast. What two properties of force have been described?

Galina-37 [17]

Answer:

The answer is C

Explanation:

Force is a vector quantity so it has both magnitude and direction. The question shows 20N act as magnitude and Southeast act as direction.

7 0

3 years ago

8. A rectangle is measured to be 6.4 +0.2 cm by 8.3 $0.2 cm.

mamaluj [8]

Answer:

a) The perimeter of the rectangle is 29.4 centimeters.

b) The uncertainty in its perimeter is 0.8 centimeters.

Explanation:

a) From Geometry we remember that the perimeter of the rectangle ( $p$ ), measured in centimeters, is represented by the following formula:

$p = 2\cdot (w+l)$ (1)

Where:

$w$ - Width, measured in centimeters.

$l$ - Length, measured in centimeters.

If we know that $w = 6.4\,cm$ and $l = 8.3\,cm$ , then the perimeter of the rectangle is:

$p = 2\cdot (6.4\,cm+8.3\,cm)$

$p = 29.4\,cm$

The perimeter of the rectangle is 29.4 centimeters.

b) The uncertainty of the perimeter ( $\Delta p$ ), measured in centimeters, is estimated by differences. That is:

$\Delta p = 2\cdot (\Delta w + \Delta l)$ (2)

Where:

$\Delta w$ - Uncertainty in width, measured in centimeters.

$\Delta l$ - Uncertainty in length, measured in centimeters.

If we know that $\Delta w = 0.2\,cm$ and $\Delta l = 0.2\,cm$ , then the uncertainty in perimeter is:

$\Delta p = 2\cdot (0.2\,cm+0.2\,cm)$

$\Delta p = 0.8\,cm$

The uncertainty in its perimeter is 0.8 centimeters.

5 0

2 years ago

A kayakeris paddling 2.50 m/s at an angle of 45° (northeast) and the current is moving 1.25 m/s at an angle of 315° (southeast).

PIT_PIT [208]

The kayaker has velocity vector

v = (2.50 m/s) (cos(45º) i + sin(45º) j )

v ≈ (1.77 m/s) (i + j )

and the current has velocity vector

w = (1.25 m/s) (cos(315º) i + sin(315º) j )

w ≈ (0.884 m/s) (i - j )

The kayaker's total velocity is the sum of these:

v + w ≈ (2.65 m/s) i + (0.884 m/s) j

That is, the kayaker has a velocity of about ||v + w|| ≈ 2.80 m/s in a direction θ such that

tan(θ) = (0.884 m/s) / (2.65 m/s) → θ ≈ 18.4º

or about 18.4º north of east.

5 0

2 years ago