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 
Explanation:
From the question we are told that
The central maxima is 
The distance from the window shade is 
The average wavelength of the sun is mathematically evaluated as

Generally the visible light spectrum has a wavelength range between 400 nm to 700 nm
So the initial wavelength of the sun is 
and the final wavelength is 
Substituting this into the above equation


The diameter is evaluated as

substituting values


Answer:
B) 2.7W
Explanation:
Converting Cal to Joule
1 cal = 4.186J
14 kcal = 14 x 1000 x 4.186
= 58604 J
Converting hour to seconds
6 hours = 6 x 60 x 60 seconds
= 21600 seconds
Power is the time rate of doing work.
Power = Work/Time
P = (58604) / (21600)
P = 2.7W
Double
Explanation:
Since the period T of a pendulum is given by

By increasing the length of the pendulum by 4, the period becomes

You can see that the period doubles when we increase the length by a factor of 4.
Answer:
Up first are Mercury and Venus. Neither of them has a moon. Because Mercury is so close to the Sun and its gravity, it wouldn't be able to hold on to its own moon. Any moon would most likely crash into Mercury or maybe go into orbit around the Sun and eventually get pulled into it.