Answer:
The recoil speed of Astronaut A is 0.26 m/s.
Explanation:
Given that,
Mass of astronaut A, 
Mass of astronaut B, 
Astronaut A pushes B away, with B attaining a final speed of 0.4, 
We need to find the recoil speed of astronaut A. The momentum remains conserved here. Using the law of conservation of linear momentum as :

So, the recoil speed of Astronaut A is 0.26 m/s.
The pitch of a sound come from the frequency of the soundwave, soundwaves that have bigger frequencies sound with a high pitch and soundwaves with less frequencie have low pitch.
Answer:
the angle is about 67.79 degrees
Explanation:
We know that at its maximum height, the vertical component of the projectile's launching (initial) velocity (Vyi) is zero, so at that point it total velocity equals the horizontal component of the initial velocity (Vxi = 0.5 m/s)
We also know that the maximum height of the projectile is given by the square of its initial vertical component of the velocity (Vyi) divided by 2g, therefore half of such distance is :

we can use this information to find the y component of the velocity at that height via the formula:

Now we use the information that tells us the speed of the projectile at this height to be 1 m/s. That should be the result of the vector addition of the vertical and horizontal components:

Now we can use the arc-tangent to calculate the launching angle, since we know the two initial component of the velocity vector:

The other structures near the hypothalamus are thalamus and the pituitary gland.
Hypothalamus is located below the thalamus and is part of the limbic system. It is an integral part of the brain, it is a small cone-shaped structure that projects down ward from the brain, ending in the pituitary stalk, a tubular connection to the pituitary gland.
Answer:
Explanation:
Let the vertical height by which it descends be h . Let it acquire velocity of v .
1/2 mv² = mgh
v² = 2gh
As it leaves the surface of sphere , reaction force of surface R = 0 , so
centripetal force = mg cosθ where θ is the angular displacement from the vertex .
mv² / r = mg cosθ
(m/r )x 2gh = mg cosθ
2h / r = cosθ
cosθ = (r-h) / r
2h / r = r-h / r
2h = r-h
3h = r
h = r / 3