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sdas [7]
3 years ago
5

What are the properties of a virtual image? What are the properties of a real image?

Physics
1 answer:
aleksandrvk [35]3 years ago
4 0

Answer:

Explanation:

<u>Properties of a virtual image:</u>

1. Image formed cannot be projected or focused on a screen.

2. The distance of the object to the mirror is the same as the distance from the image to the mirror.

3. The size of the image formed is the same as the size of the object.

4. The image formed is laterally inverted. That is the right becomes left and vice versa.

5. The image is upright.

<u>Properties of a real image:</u>

1. Image formed can be projected on a screen.

2, The distance from the image to the mirror is not the same as the distance from the object to the mirror.

3. The size of the image is not the same as the size of the object.

4. Image formed is upside down.

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A 567-g empty iron kettle is put in a hot stove the kettle absorbs 18,100 j of heat to raise its temperature from 15.0c to a fin
ipn [44]

Answer:

T2 = 355.92 Kelvin or 82.92°C

Explanation:

Given the following data;

Mass = 567g to kilograms = 567/1000 = 0.567 kg

Quantity of heat = 18,100J

Initial temperature = 15°C to Kelvin = 15 + 273 = 288K

Specific heat capacity of iron = 470j/kg•k

To find the final temperature;

Heat capacity is given by the formula;

Q = mcdt

Where;

  • Q represents the heat capacity or quantity of heat.
  • m represents the mass of an object.
  • c represents the specific heat capacity of water.
  • dt represents the change in temperature.

Making dt the subject of formula, we have;

dt = \frac {Q}{mc}

Substituting into the equation, we have;

dt = \frac {18100}{0.567*470}

dt = \frac {18100}{266.49}

dt = 67.92K

Now, the final temperature T2 is;

But, dt = T2 - T1

T2 = dt + T1

T2 = 67.92 + 288

T2 = 355.92 Kelvin or 82.92°C

7 0
3 years ago
The student draws an arrow on the paper to mark the incident ray. She marks the
garri49 [273]

Answer:

Explanation:

Snell's law

1.00sin66 = nsin38

<u>n = 1.48</u>

3 0
3 years ago
In a semiclassical model of the hydrogen atom, the electron orbits the proton at a distance of 0.053 nm. Part A What is the elec
Bezzdna [24]

Answer with Explanation:

We are given that

r=0.053 nm=0.053\times 10^{-9} m

1 nm=10^{-9} m

Charge on proton,q=1.6\times 10^{-19} C

a.We have to find the electric  potential of the proton at the position of the electron.

We know that the electric potential

V=\frac{kq}{r}

Where k=9\times 10^9

V=\frac{9\times 10^9\times 1.6\times 10^{-19}}{0.053\times 10^{-9}}

V=27.17 V

B.Potential energy of electron,U=\frac{kq_e q_p}{r}

Where

q_e=-1.6\times 10^{-19} c=Charge on electron

q_p=q=1.6\times 10^{-19} C=Charge on proton

Using the formula

U=\frac{9\times 10^9\times (-1.6\times 10^{-19}\times 1.6\times 10^{-19}}{0.053\times 10^{-9}}

U=-4.35\times 10^{-18} J

8 0
3 years ago
Where is the centre of mass of a system of two particles is situated?​
Sever21 [200]

Answer:

In a two particle system, the center of mass lies on the center of the line joining the two particles.

4 0
3 years ago
Violet light of wavelength 405 nm ejects electrons with a maximum kinetic energy of 0.890 eV from a certain metal. What is the b
AleksandrR [38]

Answer:

  Ф = 2.179 eV

Explanation:

This exercise has electrons ejected from a metal, which is why it is an exercise on the photoelectric effect, which is explained assuming the existence of energy quanta called photons that behave like particles.

            E = K + Ф

the energy of the photons is given by the Planck relation

            E = h f

we substitute

           h f = K + Ф

           Ф= hf - K

the speed of light is related to wavelength and frequency

            c = λ f

            f = c /λ

            Φ = \frac{hc}{\lambda } - K

let's reduce the energy to the SI system

            K = 0.890 eV (1.6 10⁻¹⁹ J / 1eV) = 1.424 10⁻¹⁹ J

calculate

           Ф = 6.63 10⁻³⁴ 3 10⁸/405 10⁻⁹  -1.424 10⁻¹⁹

           Ф = 4.911 10⁻¹⁹ - 1.424 10⁻¹⁹

           Ф = 3.4571 10⁻¹⁹ J

         

we reduce to eV

           Ф = 3.4871 10⁻¹⁹ J (1 eV / 1.6 10⁻¹⁹ J)

           Ф = 2.179 eV

4 0
3 years ago
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