A ray diagram without the produced image is shown.

Physics

2 answers:

tia_tia [17]3 years ago

5 0

Answer: Real and smaller than the object.

Explanation:

It is a convex lens which converges light. The object lies beyond 2F. The light rays would converge and focus at point between F and 2F. As can be seen in the diagram given in the question. The image would be formed where the two rays are intersecting between F and 2F. The image formed would be <u>real, inverted and smaller </u>than the object. Hence, the correct option is real and smaller than the object.

Mashutka [201]3 years ago

3 0

The answer is real and smaller than the object. The image point of the top of the object is the point where the two refracted rays intersect. Tracing the entire image having the same distance from the mirror as the image of the top of the object and with the bottom on the principal axis. Hence, a real inverted image will be formed for an object outside the focal point.

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ElenaW [278]

Answer:

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5 0

3 years ago

A positively charged objectwith a mass of 0.114 kg oscillates at the end of a spring, generating ELF (extremely low frequency) r

katen-ka-za [31]

Answer:

k = 167.33 N/m

Explanation:

The radio waves have a fixed relationship between the propagation speed (the speed of light in vacuum), the frequency and the wavelength, as follows:
v = c = λ*f

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$\omega_{o}^{2} =\frac{k}{m} (1)$

Now, we know that there exists a fixed relationship between the angular frequency and the frequency, as follows:

$\omega = 2*\pi *f (2)$

We also know that f in (2) is the same that we got for the radio waves, so replacing (2) in (1), and rearranging terms, we can solve for k, as follows:
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