That force is what most people would call the object's "weight".
Wherever the object is, its weight is
(mass) x (acceleration due to local gravity) .
On Earth, the acceleration due to gravity is 9.8 m/s² . (rounded)
The object's weight is (20 kg) x (9.8 m/s²) = 196 newtons .
(about 44.1 pounds)
If each lightbulb is identical they should have approximately the same resistance therefore the voltage would be split equally so the Voltage over each light bulb would be 3V.
No. it is not good for people to live on Mars.
A). Earth completes one rotation on its rotation axis
every (3 minutes and 56 seconds less than 24 hours).
This causes night and day. It's rotation, not precession.
B). Earth completes one orbital revolution around the sun
every (365 plus a little bit) days. It's revolution, not precession.
C). I'm not sure what happens on a cycle of 512 years.
It may have something to do with the moon. But whatever
it is, it's not precession.
D). When you spin a top and it starts to run down, the first
thing that happens is that the handle of the top, or the point
on top of the top, starts moving in a circle. That's precession.
That's what the Earth's axis is doing. The north pole points
toward the north star (Polaris) now. But the north pole is
actually drawing out a big circle in the sky, and other stars
will become the "north star" far in the future.
The north pole draws out a full circle in the sky, roughly
once every 26,000 years. That's precession.
Answer:
515 nm
Explanation:
The equation that gives the position of the maxima (bright fringes) in the interference pattern produced on a distant screen by the diffraction of light through two slits is
where
y is the distance of the maximum (bright fringe) from the central maximum
m is the order of the maximum
is the wavelength of the light used
D is the distance of the screen
d is the separation between the slits
In this problem we have:
is the separation between the slits
D = 3.30 m is the distance of the screen
m = 1 (we are analyzing the 1st bright fringe)
is the distance of the 1st bright fringe from the central maximum
Solving for , we find the wavelength of the light used:
