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Harman [31]
3 years ago
7

The universal force that is effective over the longest distance between very massive objects like planets is

Physics
1 answer:
fenix001 [56]3 years ago
8 0
The universal force that is effective over the longest distance between very massive objects like planets is Gravity
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One day, after pulling down your window shade, you notice that sunlight is passing through a pinhole in the shade and making a s
DedPeter [7]

Complete Question

One day, after pulling down your window shade, you notice that sunlight is passing through a pinhole in the shade and making a small patch of light on the far wall. Having recently studied optics in your physics class, you're not too surprised to see that the patch of light seems to be a circular diffraction pattern. It appears that the central maximum is about 2 cm across, and you estimate that the distance from the window shade to the wall is about 5 m.

Required:

Estimate the diameter of the pinhole.  

Answer:

The diameter is  d =0.000336 m

Explanation:

     From the question we are told that

            The central maxima is D= 2cm = \frac{2}{100} = 0.02m

            The distance from the window shade is L = 5m

     The  average wavelength of the  sun is mathematically evaluated as

                         \lambda_{ave } = \frac{\lambda_i  + \lambda_f}{2}

 Generally the visible light spectrum  has a wavelength  range  between  400 nm  to 700 nm  

        So  the initial wavelength of the sun is \lambda _i = 400nm

           and the final wavelength is  \lambda_f = 700nm

  Substituting this into the above equation

                 \lambda_{sun} = \frac{400nm  +700nm}{2}

                        = 550nm

The diameter is evaluated as

              d = \frac{2.44 \lambda_{sun} L}{D}

substituting values

              d = \frac{2.44 * 550*10^{-9} * 5 }{0.02}

                d =0.000336 m

5 0
3 years ago
A 10.0 cm object is 5.0 cm from a concave mirror that has a focal length of 12 cm. What is the distance between the image and th
fiasKO [112]
Let's use the mirror equation to solve the problem:
\frac{1}{f}= \frac{1}{d_o}+ \frac{1}{d_i}
where f is the focal length of the mirror, d_o the distance of the object from the mirror, and d_i the distance of the image from the mirror.
For a concave mirror, for the sign convention f is considered to be positive. So we can solve the equation for d_i by using the numbers given in the text of the problem:
\frac{1}{12 cm}= \frac{1}{5 cm}+ \frac{1}{d_i}
\frac{1}{d_i}= -\frac{7}{60 cm}
d_i = -8.6 cm
Where the negative sign means that the image is virtual, so it is located behind the mirror, at 8.6 cm from the center of the mirror.
6 0
3 years ago
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QUESTION 1<br> 67.2 kilometers = how many decameters
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Answer: the answer should be 6,720 decameters.

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3 years ago
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6. (a) Suppose the earth is revolving round the sun in a circular orbit of radius one b astronomical unit (1.5% 10 km). Find the
kkurt [141]

Answer:

tough

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Explanation:

5 0
2 years ago
ID like to say thx everyone who is on brainly u guys helped me get through my quiz so thx
VARVARA [1.3K]

Answer:

I want to know how your experience has been on brainly. I hope it is good. I really like spending time helping others here. If you have any other questions please ask me :)

Explanation:

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2 years ago
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