Answer:
Explanation:
an object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force.
1. The bag's velocity immediately before hitting the ground.
Recall this kinematics equation:
Vf = Vi + aΔt
Vf is the final velocity, Vi is the initial velocity, a is the acceleration, and Δt is the time elapsed.
Given values:
Vi = 0m/s (you assume this because the bag is dropped, so it falls starting from rest)
a is 9.81m/s² (this is the near-constant acceleration of objects near the surface of the earth)
Δt = 2s
Plug in the values and solve for Vf:
Vf = 0 + 9.81×2
Vf = 19.62m/s
2. The height of the helicopter.
Recall this other kinematics equation:
d = ViΔt + 0.5aΔt²
d is the distance traveled by the object, Vi is the initial velocity, a is the acceleration, and Δt is the time elapsed.
Given values:
Vi = 0m/s (bag is dropped starting from rest)
a = 9.81m/s² (acceleration due to gravity of the earth)
Δt = 2s
Plug in the values and solve for d:
d = 0×2 + 0.5×9.81×2²
d = 19.62m
3. Time of the bag's fall and its velocity immediately before hitting the ground... if it started falling at 2m/s
Reuse the equation from question 2:
d = ViΔt + 0.5aΔt²
Given values:
d = 19.6m (height of the helicopter obtained from question 2)
Vi = 2m/s
a = 9.81m/s² (acceleration due to earth's gravity)
Plug in the values and solve for Δt:
19.6 = 2Δt + 0.5×9.81Δt²
4.91Δt² + 2Δt - 19.6 = 0
Use the quadratic formula to get values of Δt (a quick Google search will give you the formula and how to use it to solve for unknown values):
Δt = 1.8s, Δt = −2.2s
The formula gives us 2 possible answers for Δt but within the situation of our problem, only the positive value makes sense. Reject the negative value.
Δt = 1.8s
Now we can use this new value of Δt to get the velocity before hitting the ground:
Vf = Vi + aΔt
Given values:
Vi = 2m/s
a = 9.81m/s²
Δt = 1.8s (result from previous question)
Plug in the values and solve for Vf:
Vf = 2 + 9.81×1.8
Vf = 19.66m/s
<span>A cloud of gas and dust begins to contract under the force of gravity. In regions of star birth, we find gaseous nebulae and molecular clouds. These sites of pre-birth are dark patches called globules.The protosun collapsed. As it did, its temperature rose to about 150,000 degrees and the sun appeared very red. Its radius was about 50 present solar radii.When the central temperature reaches 10 million degrees, nuclear burning of hydrogen into helium commences.The star settles into a stable existence on the Main Sequence, generating energy via hydrogen burning. This is the longest single stage in the evolutionary history of a star, typically lasting 90% of its lifetime. Thermonuclear fusion within the Sun is a stable process, controlled by its internal structure.</span><span>The hydrogen in the core is completed burned into helium nuclei. Initially, the temperature in the core is not hot enough to ignite helium burning. With no additional fuel in the core, fusion dies out. The core cannot support itself and contracts; as it shrinks, it heats up. The rising temperature in the core heats up a thin shell around the core until the temperature reaches the point where hydrogen burning ignites in this shell around the core. With the additional energy generation in the H-burning shell, the outer layers of the star expand but their temperature decreases as they get further away from the center of energy generation. This large but cool star is now a red giant, with a surface temperature of 3500 degrees and a radius of about 100 solar radii.<span>The helium core contracts until its temperature reaches about 100 million degrees. At this point, helium burning ignites, as helium is converted into carbon (C) and oxygen (O). However, the core cannot expand as much as required to compensate for the increased energy generation caused by the helium burning. Because the expanion does not compensate, the temperature stays very high, and the helium burning proceeds furiously. With no safety valve, the helium fusion is uncontrolled and a large amount of energy is suddenly produced. This<span>helium flash </span>occurs within a few hours after helium fusion begins.The core explodes, the core temperature falls and the core contracts again, thereby heating up. When the helium burns now, however, the reactions are more controlled because the explosion has lowered the density enough. Helium nuclei fuse to form carbon, oxygen, etc..</span>The star wanders around the red giant region, developing its distinct layers, eventually forming a carbon-oxygen core.When the helium in the core is entirely converted into C, O, etc., the core again contracts, and thus heats up again. In a star like the Sun, its temperature never reaches the 600 million degrees required for carbon burning. Instead, the outer layers of the star eventually become so cool that nuclei capture electrons to form neutral atoms (rather than nuclei and free electrons). When atoms are forming by capturing photons in this way, they cause photons to be emitted; these photons then are readily available for absorption by neighboring atoms and eventually this causes the outer layers of the star to heat up. When they heat up, the outer layers expand further and cool, forming more atoms, and releasing more photons, leading to more expansion. In other words, this process feeds itself.The outer envelope of the star blows off into space, exposing the hot, compressed remnant core. This is a <span>planetary nebula </span>.</span><span>The core contacts but carbon burning never ignites in a one solar mass star. Contraction is halted when the electrons become degenerate, that is when they can no longer be compressed further. The core remnant as a surface temperature of a hot 10,000 degrees and is now a <span>white dwarf </span>.With neither nuclear fusion nor further gravitational collapse possible, energy generation ceases. The star steadily radiates is energy, cools and eventually fades from view, becoming a black dwarf.</span>
