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Masteriza [31]
3 years ago
8

Which is true about a concave mirror? Incident rays that are parallel to the central axis are dispersed but will be perceived as

originating from a point on the near side of the mirror.
Physics
1 answer:
Reil [10]3 years ago
5 0

Answer:

'Incident rays that are parallel to the central axis are sent through a point on the near side of the mirror'.

Explanation:

The question is incomplete, find the complete question in the comment section.

Concave mirrors is an example of a curved mirror. The outer surface of a concave mirror is always coated. On the concave mirror, we have what is called the central axis or principal axis which is a line cutting through the center of the mirror. The points located on this axis are the Pole, the principal focus and the centre of curvature. <em>The focus point is close to the curved  mirror than the centre of curvature.</em>

<em></em>

During the formation of images, one of the incident rays (rays striking the plane surface) coming from the object and parallel to the principal axis, converges at the focus point after reflection because all incident rays striking the surface are meant to reflect out. <em>All incident light striking the surface all converges at a point on the central axis known as the focus.</em>

Based on the explanation above, it can be concluded that 'Incident rays that are parallel to the central axis are sent through a point on the near side of the mirror'.

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Introduction: The specific heat capacity of a substance is the amount of energy needed to change the temperature of that substan
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Answer:

A) 8,368 J

B) ) 0.893 J/gºC

Explanation:

A)

  • The heat gained by the water can be obtained solving the following equation:

       q_{g} = c_{w} * m *  \Delta T (1)

  • where cw = specific heat of water = 4.184 J/gºC
  • m= mass of water = 1,000 g
  • ΔT = 2ºC
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       q_{g} = c_{w} * m *  \Delta T = 4.184 J/gºC*1,000 g* 2ºC = 8,368 J (2)

B)

  • Assuming that the heat energy gained by the water is equal to the one lost by the aluminum, we can use the same equation, taking into account that the energy is lost by the aluminum, so the sign is negative:  -8,368 J.
  • Replacing by the mass of aluminum (125 g), and the change in temperature (-74.95ºC), in (1), we can solve for the specific heat of aluminum, as follows:

       q_{l} = c_{Al} * m_{Al} *  \Delta T  (3)

⇒    -8,368 J = c_{Al}* 125 g * (-74.95ºC) (4)

       c_{Al} = \frac{-8,368J}{125g*(-74.95ºC} = 0.893 J/gºC (5)

  • which is pretty close to the Aluminum's accepted specific heat value of 0.900 J/gºC.

8 0
3 years ago
A diver initially moving horizontally with speed v dives dives offthe edge of a vertical cliff and lands in the water a distance
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Answer:

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Explanation:

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d = ??

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Since the times are same, equate the results.

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v*d1 = 2v*d The v's cancel because they are related.

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