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Natasha2012 [34]
3 years ago
11

What is the difference between virtual images produced by concave, plane, and convex mirrors?

Physics
1 answer:
julsineya [31]3 years ago
7 0

Answer:

Explanation:

A concave mirror is a curved mirror that is coated outwards. The outer part of the mirror is always coated. The nature of image formed by an object placed in front of a concave mirror can be real or virtual depending on the distance of the object on the axis from the mirror. The only time the object produces a virtual image is when it is placed between the focus and the pole of the mirror. The virtual image formed is a "MAGNIFIED and upright image"

For a convex mirror, the inner part is always coated and the nature of the image formed by the object doesn't depend on the distance between the image and the mirror. No matter where the object is placed, the image formed will always be virtual, upright and DIMINISHED. This means that magnification is always less than 1.

For a plane mirror, the nature of the image produced by a plane mirror also virtual because it is always formed behind the mirror. The size of the image formed is always THE SAME as that of the object. This means that the magnification is always equal to 1.

a) In summary, the difference between virtual images produced by concave, plane, and convex mirrors is that virtual images produced by concave mirror are MAGNIFIED, virtual images produced by plane mirror are THE SAME SIZE as that of the object and virtual images produced by convex mirrors are always DIMINISHED.

b) Magnification is defined as the ratio of the image distance to the object distance. Mag = v/u

Note that object distances are always positive, hence it is only the image distance that can either be positive or negative which in turn affects the magnification causing it to be positive or negative.

Negative image distance shows that the image is virtual while positive image distance shows that the image is real.

A negative magnification therefore shows that the nature of the image is a virtual image.

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Answer:

The maximum height is determined by the initial vertical velocity. Since steeper launch angles have a larger vertical velocity component, increasing the launch angle increases the maximum height.

Explanation:

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The axis of rotation of the Earth is tilted at an angle of 23.5 degrees away from vertical, perpendicular to the plane of our planet's orbit around the sun. The tilt of the Earth's axis is important, in that it governs the warming strength of the sun's energy.

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How much heat is required to change 200 mL of ice at -22°C (at typical Freezer temperature) into steam
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Answer:

611.064 kJ

Explanation:

Given :

m = 200 mL = 200 g

Specific heat of ice = 2.06 J/g°C

Q = mcΔt

Δt = 0 - (-22) = 22

Q1 = 200 * 2.06 * 22 = 9064 J

Q2 = Melt 0 °C solid ice into 0 °C liquid water:

Q2 = m · ΔHf ; ΔHf = heat of fusion of water = 334j/g

Q2 = 200 * 334 = 66800 J

Q3 : Heat to convert from 0°C to 100°C

Q3 = mcΔt ; c = 4.19 J/g°C ; Δt = (100 - 0) = 100

Q3 = 200 * 4.19 * 100 = 83800 J

Q4: heat required to boil water to steam

Q = m · ΔHv

Hv = heat of vaporization of water = 2257 J/g

Q4 = 200 * 2257 = 451400 J

Total Q = Q1 + Q2 + Q3 + Q4

Q = 9064 + 66800 + 83800 + 451400

Q = 611,064 Joules

Q = 611.064 kJ

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Determine the angular velocity ω of the telescope as it orbits around the Sun.
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A 7.00-kg object accelerates from rest to a final velocity in 55 seconds. If the magnitude of the
Len [333]

Answer:

The final velocity of the object is 330 m/s.

Explanation:

To solve this problem, we first must find the acceleration of the object.  We can do this using Newton's Second Law, given by the following equation:

F = ma

If we plug in the values that we are given in the problem, we get:

42 = 7 (a)

To solve for a, we simply divide both sides of the equation by 7.

42/7 = 7a/7

a = 6 m/s^2

Next, we should write out all of the information we have and what we are looking for.

a = 6 m/s^2

v1 = 0 m/s

t  = 55 s

v2 = ?

We can use a kinematic equation to solve this problem.  We should use:

v2 = v1 + at

If we plug in the values listed above, we should get:

v2 = 0 + (6)(55)

Next, we should solve the problem by performing the multiplication on the right side of the equation.

v2 = 330 m/s

Therefore, the final velocity reached by the object is 330 m/s.

Hope this helps!

7 0
3 years ago
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