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1 month ago

In your own words, describe the three steps in ray tracing used to identify where an image forms.

Physics

2 answers:

Yuliya22 [3.3K]1 month ago

6 0

Essential aspects of image tracing include:

1) Rays emanating from a distant object converge at the focus after interacting with the lens or mirror.
2) Rays directed through the focus will run parallel to the principal axis.
3) Ensure the object faces the front of the mirror.

kicyunya [3.2K]1 month ago

6 0

Answer:

When a ray from an object approaches parallel to the principal axis, its reflection from the mirror will direct through the focus, or if it refracts through the lens, it will converge or seem to diverge from that point.
An incoming ray that intersects the focus will reflect or refract to run parallel with the principal axis.
A ray that travels through the lens's pole will continue straight through.

Any ray hitting at an angle relative to the mirror or lens will reflect or refract according to the respective laws of reflection and Snell's law for refraction.

You might be interested in

State two advantages of a lead-acid accumulator over a leclanche cell

kicyunya [3294]

Respuesta:

Se pueden recargar.

Poseen una vida útil mayor.

Son reutilizables.

Explicación:

Las baterías de plomo-ácido y las pilas Leclanché son tipos de células electroquímicas.

Las pilas Leclanché son células primarias. Estas células producen reacciones químicas que generan corriente eléctrica de manera irreversible.

Por su parte, las baterías de plomo-ácido son células secundarias que permiten la reversibilidad de la corriente eléctrica generada.

Esto hace que las baterías de plomo-ácido sean reutilizables, con mayor durabilidad y un tiempo de vida más prolongado.

5 0

2 months ago

A 4.5-m-long wooden board with a 24-kg mass is supported in two places. One support is directly under the center of the board, a

Ostrovityanka [3204]

All the weight of the wooden board rests solely on the support situated at the center of the rod, while the other support positioned at one end experiences no reaction force, resulting in a 0 reaction force.

Thus, the reaction force at the central support corresponds to the weight of the board, whereas the end support has 0 reaction force.

8 0

1 month ago

Read 2 more answers

The deflection plates in an oscilloscope are 10 cm by 2 cm with a gap distance of 1 mm. A 100 volt potential difference is sudde

Maru [3345]

Explanation:

Data provided:

Area A = 10 cm×2 cm = 20×10⁻⁴ m²

Separation distance d between the plates = 1 mm = 1×10⁻³ m

Battery voltage, or emf = 100 V

Resistance = 1025 ohm

Solution:

In an RC circuit, the voltage across the plates varies with time t. At the outset, the voltage matches that of the battery, V₀ = emf = 100V. However, after a certain time t, both the resistance and capacitance alter this, leading to a final voltage V expressed as

$V = V_{0}(1-e^{\frac{-t}{RC} } )\\\frac{V}{V_{0} } = 1-e(^{\frac{-t}{RC} }) \\e^{\frac{-t}{RC} } = 1- \frac{V}{V_{0} }$

Applying the natural logarithm to both sides,

$e^{\frac{-t}{RC} } = 1- \frac{V}{V_{0} } \\\frac{-t}{RC} = ln(1-\frac{V}{V_{0} } )\\t = -RCln(1 - \frac{V}{V_{0} })$

$t = -RC ln (1-\frac{V}{V_{0} })$ (1)

Next, we can determine the capacitance using the plates' area.

C = ε₀A/d

= $\frac{(8.85*10^{-12))} (20*10^{-4}) }{1*10^{-3} }$

= 18×10⁻¹²F

We can now find the time it takes for the voltage to drop from 100 to 55 V by substituting C, V₀, V, and R values into equation (1)

$t = -RC ln (1-\frac{V}{V_{0} })$

= -(1025Ω)(18×10⁻¹² F) ln( 1 - 55/100)

= 15×10⁻⁹s

= 15 ns

5 0

14 days ago

1. Susie wondered if the height of a hole punched in the side of a quart-size milk carton would affect how far from the containe

Sav [3153]

Hypothesis: The liquid will project far.
Independent Variable: Height of the hole.
Dependent Variable: Distance of the squirt.
Constant: All other factors aside from the independent variable, such as the liquid volume.
Control: None that I recognize.
Number of groups: 4
Trials per group: 4

7 0

1 month ago

Read 2 more answers

Suppose that A’, B’ and C’ are at rest in frame S’, which moves with respect to S at speed v in the +x direction. Let B’ be loca

Keith_Richards [3271]

Response:

1) An observer in B 'perceives the two events occurring at the same time

2) Observer B recognizes that the events happen at different times

3) Δt = Δt₀ /√ (1 + v²/c²)

Clarification:

This scenario illustrates the concept of simultaneity in special relativity. It is important to keep in mind that light's speed remains constant across all inertial frames

1) Since the events are stationary within the frame S ', they propagate at the constant speed of light, resulting in them reaching observation point B'—located equidistantly between both events—simultaneously

Thus, an observer in B 'observes the two events occurring at the same time

2) For an observer B situated within frame S attached to the Earth, both events at A and B appear to take place at the same moment. However, the event at A covers a shorter distance, while the event at B travels a longer distance, since frame S 'is in motion at velocity + v. Hence, with a constant speed, the event covering the lesser distance is perceived first.

Consequently, observer B perceives that the events do not occur simultaneously

3) Let's determine the timing for each event

Δt = Δt₀ /√ (1 + v²/c²)

where t₀ represents the time in the S' frame, which remains at rest for the events

8 0

1 month ago