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3 years ago

Which of the following diagrams involves a virtual image ?

Physics

1 answer:

sergiy2304 [10]3 years ago

7 0

Answer:

The third diagram

Explanation:

A virtual image is an image that can not be formed on a screen.
A convex lens can form both virtual and real image depending on the position of the object from the lens.
A virtual image in convex lens is formed when the object is placed between the focus and the optical center of the lens.
In the third diagram, a virtual image is formed because the position of the object is between the focus and the optical center of the convex lens.

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The sun will eventually become a

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The sun will eventually become a

>white dwarf 5 billion years from now then 2 billion years after that the core will crystallize, leaving a giant diamond in the center of our solar system

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3 years ago

A 1.2-kg mass is projected from ground level with a velocity of 30 m/s at some unknown angle above the horizontal. A short time

Dmitriy789 [7]

Answer:

$K_f$ = 351.84 J

Explanation:

Using the conservation of energy K:

$E_i = E_f$

so:

$\frac{1}{2}mv^2 = K_f + mgh$

where m is the mass, v the initial velocity, $K_f$ is the kinetic energy of the mass as it clears the fence, g the gravity and h the altitude.

Then, replacing values, we get:

$\frac{1}{2}(1.2kg)(30m/s)^2 = K_f + (1.2kg)(9.8m/s^2)(16m)$

solving for $K_f$ :

$K_f$ = 351.84 J

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3 years ago

Monochromatic light of wavelength λ=620nm from a distant source passes through a slit 0.450 mm wide. The diffraction pattern is

Elan Coil [88]

Answer:

The intensity of light from the 1mm from the central maximu is $I = 0.822I_o$

Explanation:

From the question we are told that

The wavelength is $\lambda = 620 nm = 620 *10^{-9}m$

The width of the slit is $w = 0.450mm = \frac{0.45}{1000} = 0.45*10^{-3} m$

The distance from the screen is $D = 3.00m$

The intensity at the central maximum is $I_o$

The distance from the central maximum is $d_1 = 1.00mm = \frac{1}{1000} = 1.0*10^{-3}m$

Let z be the the distance of a point with intensity I from central maximum

Then we can represent this intensity as

$I = I_o [\frac{sin [\frac{\pi * w * sin (\theta )}{\lambda} ]}{\frac{\pi * w * sin (\theta )}{\lambda } } ]^2$

Now the relationship between D and z can be represented using the SOHCAHTOA rule i.e

$sin \theta = \frac{z}{D}$

if the angle between the the light at z and the central maximum is small

Then $sin \theta = \theta$

Which implies that

$\theta = \frac{z}{D}$

substituting this into the equation for the intensity

$I = I_o [\frac{sin [\frac{\pi w}{\lambda} \cdot \frac{z}{D} ]}{\frac{\pi w z}{\lambda D\frac{x}{y} } } ]$

given that $z =1mm = 1*10^{-3}m$

We have that

$I = I_o [\frac{sin[\frac{3.142 * 0.45*10^{-3}}{(620 *10^{-9})} \cdot \frac{1*10^{-3}}{3} ]}{\frac{3.142 * 0.45*10^{-3}*1*10^{-3} }{620*10^{-9} *3} } ]^2$

$=I_o [\frac{sin(0.760)}{0.760}] ^2$

$I = 0.822I_o$

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3 years ago

thin films of soap sometime display an array of colours. this display is the result of: a) reflection, diffraction, and interfer

kumpel [21]

The array of colors observed when viewing films of soap are caused by the interference, and from the reflections of light from two nearby surfaces (outer surface and inner surface). Among the choices, the closest answer would be B. reflection, refraction, and interference.

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3 years ago

Read 2 more answers

A person is riding a motorized tricycle. They weigh 180kg and are moving at 3 m/s over a distance of 300 m. How much work is don

agasfer [191]

If I am to understand this question correctly this is what asks you:

If a person is riding a motorized tricycle how much work do they do?

You may ask yourself, why did I only use part of the question. Simple, the rest is not relevant to what is being asked. The weight, speed, and distance wont affect the person riding any motorized vehicle other than the time it takes to get from one place to another.

So to answer this question I would say:

Not much, all they really have to do is to steer and set the motorized tricycle to cruise control. Just like any rode certified vehicle.

If you have any questions about my answer please let me know and I will be happy to clarify any misunderstandings. Thanks and have a great day!

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3 years ago