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4 years ago

What mirror or Lents can created an image upside down ?

Physics

1 answer:

ser-zykov [4K]4 years ago

3 0

Answer:

mirrors : convergent mirror

lens : convex lens

Explanation:

<em>mirrors : concave mirror gives upside down images everytime except when the object is located in front of F (focal point)</em>

The object is located beyond C

<em> </em>image will be located in between the F and C
inverted image.
reduced in size
real image

2. The object is located between C and F or on C

Image will be located between C and F
inverted
larger than the object dimensions.
real image
<em>(when the object is @ C , image size = object size )</em>

3. The object is located in front of F

image will be located behind the mirror
upright image
magnified
virtual image
<em>(when the object is located @ F ,no image is formed )</em>

<em>lens : convex lens gives upside down images everytime except when the object is located in between in front of F (focal point)</em>

<u>The properties of images</u>

varies depending on the position of the object.

distance > 2F (when the object is far away)

Real
Diminished
Inverted
smaller than the object
occurs in the opposite side of the lens

F < distance < 2F (object approaches the lens)

Real
Inverted
gets magnified when approaching the focal point (Image size = object size at 2F point. this becomes gradually enlarged and get infinitely large @ F point )

distance < F ( object passes the focal point )

Virtual
Erected
Magnified
occurs in the same side of the lens.(magnifying glass is one example which uses this formation, mentioned in graph 2 )

Characteristics of the images of each mirror and lens are mentioned in tables for further reference

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Capacitor 2 has half the capacitance and twice the potential difference as capacitor 1. What is the ratio (U_{\rm C})_1/\,(U_{\r

IrinaK [193]

Answer:

1/2

Explanation:

The energy stored in a capacitor is given by

$U=\frac{1}{2}CV^2$

where

C is the capacitance

V is the potential difference

Calling $C_1$ the capacitance of capacitor 1 and $V_1$ its potential difference, the energy stored in capacitor 1 is

$U=\frac{1}{2}C_1 V_1^2$

For capacitor 2, we have:

- The capacitance is half that of capacitor 1: $C_2 = \frac{C_1}{2}$

- The voltage is twice the voltage of capacitor 1: $V_2 = 2 V_1$

so the energy stored in capacitor 2 is

$U_2 = \frac{1}{2}C_2 V_2^2 = \frac{1}{2}\frac{C_1}{2}(2V_1)^2 = C_1 V_1^2$

So the ratio between the two energies is

$\frac{U_1}{U_2}=\frac{\frac{1}{2}C_1 V_1^2}{C_1 V_1^2}=\frac{1}{2}$

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3 years ago

Question: Self-test 3.12 Calculate the change in G for ice at -10°C, with density 917 kg mº, when the pressure is increased from

Akimi4 [234]

The change in the Gibb's free energy per mole (G) is 1.96 J.

The given parameters:

Density of the ice, ρ = 917 kg/m³
Initial pressure, P₁ = 1.0 bar
Final pressure, P₂ = 2.0 bar
Temperature, T = - 10 C
Mass of water = 18 g

The change in the Gibb's free energy per mole (G) is calculated as follows;

$\Delta G = V(P_2-P_1) \\\\$

where;

V is the volume of the ice

$Density = \frac{Mass}{Volume} \\\\Volume = \frac{Mass}{Density} \\\\Volume = \frac{18 \times 10^{-3} \ kg}{917 \ m^3} \\\\Volume = 1.96 \times 10^{-5} \ m^3\\\\Volume = 1.96 \times 10^{-5} \ m^3 \times \frac{1000 \ L}{m^3} \\\\Volume = 0.0196 \ L$

Change in pressure;

$P_2 - P_1 = 2.0 \ bar \ - \ 1.0 \ bar = 1.0 \ bar = 0.987 \ atm$

The change in the Gibb's free energy per mole (G);

$\Delta G= V(P_2-P_1)\\\\\Delta G = 0.0196\ L \times 0.987\ atm \\\\\Delta G = 0.0193 \ L.atm\\\\1 \ L.atm = 101.325 \ J\\\\\Delta G = 0.0193 \ L.atm \times \frac{101.325 \ J}{1 \ L.atm} \\\\\Delta G = 1.96 \ J$

Thus, the change in the Gibb's free energy per mole (G) is 1.96 J.

Learn more about Gibb's free energy here: brainly.com/question/10012881

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3 years ago

1. What is the momentum of a golf ball with a mass of 62 g moving at 73 m/s?

Anit [1.1K]

Answer:

<h3>The answer is 4.53 kgm/s</h3>

Explanation:

The momentum of an object can be found by using the formula

<h3>momentum = mass × velocity</h3>

From the question

mass = 62 g = 0.062 kg

velocity = 73 m/s

We have

momentum = 0.062 × 73 = 4.526

We have the final answer as

Hope this helps you

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3 years ago

A beaker contain 200mL of water<br> What is its volume in cm3 and m3

Bogdan [553]

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4 years ago

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Based on the information you know about friction discuss with your group how you could experimentally determine the coefficient

Alona [7]

Answer:

The experiment to determine the coefficient of static friction between a rough block and an incline plane is detailed in the explanation.

Explanation:

A simple experiment is as follows:

Take a weight in which its mass is known and hold it with a roller at the end of the inclined plane with a rope.

Place the weight on the plane inclined horizontally with the ground, forming a zero degree angle.

Gradually lift the weight so that the angle of incline begins to increase.

Observe that the weight will begin to slide. This angle will be the angle theta.

In this way, the coefficient of friction will be calculated as the tangent of the angle theta.

Take a different weight and repeat the procedure to get different values of the coefficient of friction.

By averaging the coefficients of friction, a final value will be obtained.

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3 years ago