Answer:
h’ = 1/9 h
Explanation:
This exercise must be solved in parts:
* Let's start by finding the speed of sphere B at the lowest point, let's use the concepts of conservation of energy
starting point. Higher
Em₀ = U = m g h
final point. Lower, just before the crash
Em_f = K = ½ m
energy is conserved
Em₀ = Em_f
m g h = ½ m v²
v_b =
* Now let's analyze the collision of the two spheres. We form a system formed by the two spheres, therefore the forces during the collision are internal and the moment is conserved
initial instant. Just before the crash
p₀ = 2m 0 + m v_b
final instant. Right after the crash
p_f = (2m + m) v
the moment is preserved
p₀ = p_f
m v_b = 3m v
v = v_b / 3
v = ⅓ 
* finally we analyze the movement after the crash. Let's use the conservation of energy to the system formed by the two spheres stuck together
Starting point. Lower
Em₀ = K = ½ 3m v²
Final point. Higher
Em_f = U = (3m) g h'
Em₀ = Em_f
½ 3m v² = 3m g h’
we substitute
h’=
h’ =
h’ = 1/9 h
Option (b) is correct.The index of refraction for the glass is 1.52
Explanation:
velocity of light in vacuum= C= 2.99 x 10⁸m/s
Velocity of light in glass = V= 1.97 x 10⁸m/s
The refractive index is given by n=
n= 2.99 x 10⁸/1.97 x 10⁸m/s
n= 1.52
Thus the refractive index of glass is 1.52
<em>Answer:</em>
<em>When </em><em>a </em><em>body </em><em>is </em><em>moving </em><em>on </em><em>a </em><em>circle </em><em>it </em><em>is </em><em>accelerating </em><em>because </em><em>centripetal </em><em>acceleration</em><em> </em><em>is </em><em>always </em><em>acting </em><em>on </em><em>it </em><em>towards </em><em>the </em><em>center.</em>
<em>Please </em><em>see</em><em> the</em><em> attached</em><em> picture</em><em>.</em><em>.</em><em>.</em>
<em>From </em><em>the </em><em>above </em><em>diagram,</em><em>we </em><em>can </em><em>say </em><em>the </em><em>acceleration</em><em> </em><em>is </em><em>always </em><em>acting </em><em>on </em><em>the </em><em>body </em><em>when </em><em>it </em><em>moves </em><em>in </em><em>a </em><em>circle.</em>
<em>Hope </em><em>this </em><em>helps.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em>
<em>Good </em><em>luck</em><em> on</em><em> your</em><em> assignment</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em>
Answer:
Going from earth to the sun a probe would encounter the next layers in order:
- Corona
- Transition Region
- Chromosphere
- Photosphere
- Convection Zone
- Radiative Zone
- Core
A brief description of them:
Corona is the outermost layer and it cannot be seen with the naked eye, is starts at about 2100 km from the surface of the sun and it has no limit defined.
Transition Region is between the corona and the chromosphere, it has an extension of about 100km
The chromosphere is between 400 km from the surface of the sun to 2100 km. In this layer the further you get away from the sun it gets hotter.
The photosphere is the surface of the sun, the part that we can see, and extends from the surface to 400km.
The convection zone is where convection happens, hot gas rises, cools and rises again.
Radiative Zone is where the photons try to rise to move to higher layers.
The core of the Sun is where nuclear fusion occurs due to the very high temperatures.