Answer:
The distance of the object placed on the principal axis from the concave mirror.
Explanation:
In a concave mirror, the nature of the image formed formed by the object placed in front of the mirror depends on the position of the object placed in from of the mirror. It all depends on the distance between the mirror and the object placed on the principal axis.
The closer the object is to the lens, the more larger or magnified the image formed will be. For example an object placed between the focal point and the pole of a concave produces a much larger image than an object placed beyond the centre of curvature of such mirror.
Answer: They create calcuim chloride, CaCl2
Answer: a) 11.76 m/s b) 7.056 m
Explanation:
The described situation is as follows:
An object is dropped from the top of a tower and when measuring the time it takes to reach the ground that turns out to be 0.02 minutes.
This situation is related to free fall, this also means we have constant acceleration, hence the equations we will use are:
(1)
(2)
Where:
Is the final velocity of the object
Is the initial velocity of the object (it was dropped)
is the acceleration due gravity
is the height of the tower
is the time it takes to the object to reach the ground
b) Begining with (1):
(3)
(4)
(5) This is the final velocity of the object
a) Substituting (5) in (2):
(6)
Clearing
:
(7)
(8) This is the height of the tower
Your question kind of petered out there towards the end and you didn't specify
the terms, so I'll pick my own.
The "Hubble Constant" hasn't yet been pinned down precisely, so let's pick a
round number that's in the neighborhood of the last 20 years of measurements:
<em>70 km per second per megaparsec</em>.
We'll also need to know that 1 parsec = about 3.262 light years.
So the speed of your receding galaxy is
(Distance in LY) x (1 megaparsec / 3,262,000 LY) x (70 km/sec-mpsc) =
(150 million) x (1 / 3,262,000) x (70 km/sec) =
<em>3,219 km/sec </em>in the direction away from us (rounded)