The centripetal acceleration of an object is given by the relation,

where Ac = centripetal acceleration =
R = radius of rotation = 15 m
V = speed of astronaut
Hence, 
solving this we get, V = 38.34 m/s
Answer:
Distance: -30.0 cm; image is virtual, upright, enlarged
Explanation:
We can find the distance of the image using the lens equation:

where:
f = 15.0 cm is the focal length of the lens (positive for a converging lens)
p = 10.0 cm is the distance of the object from the lens
q is the distance of the image from the lens
Solving for q,

The negative sign tells us that the image is virtual (on the same side of the object, and it cannot be projected on a screen).
The magnification can be found as

The magnification gives us the ratio of the size of the image to that of the object: since here |M| = 3, this means that the image is 3 times larger than the object.
Also, the fact that the magnification is positive tells us that the image is upright.
To solve this problem it is necessary to apply the concepts related to linear momentum, velocity and relative distance.
By definition we know that the relative velocity of an object with reference to the Light, is defined by

Where,
V = Speed from relative point
c = Speed of light
On the other hand we have that the linear momentum is defined as
P = mv
Replacing the relative velocity equation here we have to







Therefore the height with respect the observer is



Therefore the height which the observerd measure for her is 0.56m
Answer:
the ratio of the bubble’s volume at the top to its volume at the bottom is 1.019
Explanation:
given information
h = 0.2 m
= 1.01 x
Pa


=
+ ρgh, ρ = 1000 kg/
= 1.01 x
Pa + (1000 x 9.8 x 0.2) = 1,0296 x
Pa
=
=
Pa
thus,
/
= 1.019