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irinina [24]
4 years ago
5

Which element in the earth’s atmosphere is essential for human life, is extremely rare or nonexistent in the atmospheres of othe

r planets in our solar system, and is primarily responsible for the development of life outside the oceans?
Physics
1 answer:
gayaneshka [121]4 years ago
3 0
That element is oxygen.  It is also highly volatile since it reacts with other elements and is flammable. 
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An upright object 2.80 cm tall is placed 16.0 cm away from the vertex of a concave mirror with a center of curvature of 24.0 cm.
horrorfan [7]

Answer:

f = 12 cm

Explanation:

<u>Center of Curvature</u>:

The center of that hollow sphere, whose part is the spherical mirror, is known as the ‘Center of Curvature’ of  mirror.

<u>The Radius of Curvature</u>:

The radius of that hollow sphere, whose part is the spherical mirror, is known as the ‘Radius of Curvature’ of  mirror. It is the distance from pole to the center of curvature.

<u>Focal Length</u>:

The distance between principal focus and pole is called ‘Focal Length’. It is denoted by ‘F’.

The focal length of the spherical (concave) mirror is approximately equal to half of the radius of curvature:

f = \frac{R}{2}

where,

f = focal length = ?

R = Radius of curvature = 24 cm

Therefore,

f = \frac{24\ cm}{2}

<u>f = 12 cm</u>

8 0
3 years ago
How does radiation differ from conduction?
Natali5045456 [20]

Answer:

b

Explanation:

3 0
3 years ago
The coefficient of static friction between a 3.00 kg crate and the 35.0o incline is 0.300. What minimum force F must be applied
Yakvenalex [24]

Answer:

So the minimum force is

32.2Newton

Explanation:

To solve for the minimum force, let us assume it to be F (N)

So

F=mgsinA

But

=>>>> coefficient of static friction x (F + mgcosA

=>3 x 9.8 x sin35 = 0.3 x (F + 3 x 9.8 x cos35)

So making F subject of formula

F + 24.0 = 56.2

F = 32.2N

3 0
3 years ago
If you could shine a very powerful flashlight beam toward the Moon, estimate the diameter of the beam when it reaches the Moon.
grin007 [14]

To develop this problem it is necessary to apply the Rayleigh Criterion (Angular resolution)criterion. This conceptos describes the ability of any image-forming device such as an optical or radio telescope, a microscope, a camera, or an eye, to distinguish small details of an object, thereby making it a major determinant of image resolution.  By definition is defined as:

\theta = 1.22\frac{\lambda}{d}

Where,

\lambda= Wavelength

d = Width of the slit

\theta= Angular resolution

Through the arc length we can find the radius, which would be given according to the length and angle previously described.

The radius of the beam on the moon is

r = l\theta

Relacing \theta

r = l(\frac{1.22\lambda}{d})

r = 1.22\frac{l\lambda}{d}

Replacing with our values we have that,

r = 1.22*(\frac{(384*10^3km)(\frac{1000m}{1km})(550*10^{-9}m)}{7*10^{{-2}}})

r = 3680.91m

Therefore the diameter of the beam on the moon is

d = 2r

d = 2 * (3690.91)

d = 7361.8285m

Hence, the diameter of the beam when it reaches the moon is 7361.82m

8 0
3 years ago
Can you label the parts of the eye?
photoshop1234 [79]

Answer:

  1. Iris
  2. Pupil
  3. Corona
  4. Anterior chamber
  5. lens
  6. Vitreous humor
  7. Blood vessels
  8. Optic nerve
  9. Hyaloid canal
  10. Retina

Hope it helps :)

7 0
3 years ago
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