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This is a perfect opportunity to stuff all that data into the general equation for the height of an object that has some initial height, and some initial velocity, when it is dropped into free fall.
H(t) = (H₀) + (v₀ T) + (1/2 a T²)
Height at any time 'T' after the drop =
(initial height) +
(initial velocity) x (T) +
(1/2) x (acceleration) x (T²) .
For the balloon problem ...
-- We have both directions involved here, so we have to define them:
Upward = the positive direction
Initial height = +150 m
Initial velocity = + 3 m/s
Downward = the negative direction
Acceleration (of gravity) = -9.8 m/s²
Height when the bag hits the ground = 0 .
H(t) = (H₀) + (v₀ T) + (1/2 a T²)
0 = (150m) + (3m/s T) + (1/2 x -9.8 m/s² x T²)
-4.9 T² + 3T + 150 = 0
Use the quadratic equation:
T = (-1/9.8) [ -3 plus or minus √(9 + 2940) ]
= (-1/9.8) [ -3 plus or minus 54.305 ]
= (-1/9.8) [ 51.305 or -57.305 ]
T = -5.235 seconds or 5.847 seconds .
(The first solution means that the path of the sandbag is part of
the same path that it would have had if it were launched from the
ground 5.235 seconds before it was actually dropped from balloon
while ascending.)
Concerning the maximum height ... I don't know right now any other
easy way to do that part without differentiating the big equation.
So I hope you've been introduced to a little bit of calculus.
H(t) = (H₀) + (v₀ T) + (1/2 a T²)
H'(t) = v₀ + a T
The extremes of 'H' (height) correspond to points where h'(t) = 0 .
Set v₀ + a T = 0
+3 - 9.8 T = 0
Add 9.8 to each side: 3 = 9.8 T
Divide each side by 9.8 : T = 0.306 second
That's the time after the drop when the bag reaches its max altitude.
Oh gosh ! I could have found that without differentiating.
- The bag is released while moving UP at 3 m/s .
- Gravity adds 9.8 m/s of downward speed to that every second.
So the bag reaches the top of its arc, runs out of gas, and starts
falling, after
(3 / 9.8) = 0.306 second .
At the beginning of that time, it's moving up at 3 m/s.
At the end of that time, it's moving with zero vertical speed).
Average speed during that 0.306 second = (1/2) (3 + 0) = 1.5 m/s .
Distance climbed during that time = (average speed) x (time)
= (1.5 m/s) x (0.306 sec)
= 0.459 meter (hardly any at all)
But it was already up there at 150 m when it was released.
It climbs an additional 0.459 meter, topping out at 150.459 m,
then turns and begins to plummet earthward, where it plummets
to its ultimate final 'plop' precisely 5.847 seconds after its release.
We can only hope and pray that there's nobody standing at
Ground Zero at the instant of the plop.
I would indeed be remiss if were to neglect, in conclusion,
to express my profound gratitude for the bounty of 5 points
that I shall reap from this work. The moldy crust and tepid
cloudy water have been delicious, and will not soon be forgotten.
Omitting the 1 will not change the value of the number, but will change the units at the end of the problem
Answer:
Magnetic field, B = 0.23 T
Explanation:
Given that,
Length of the copper rod, L = 0.49 m
Mass of the copper rod, m = 0.15 kg
Current in rod, I = 13 A (in +ve y direction)
When the rod is placed in magnetic field, the magnetic force is balanced by its weight such that :
So, the magnitude of the minimum magnetic field needed to levitate the rod is 0.23 T.
Answer:
<em>It matters because crystalline and amorphous materials have different properties. The arrange affects the melting point (defined in crystals and a larger range in amorphous) and shape (geometrical in crystals, no geometrical in amorphous). </em>
Explanation:
The particles that compose a solid material are held in place by strong tractive forces between them when we analyze solids we consider the position of the atoms (molecules or ions) rather than their motion (which is important in liquids and gases). This positioning can be arranged in two general ways:
- Crystalline solids have internal structures that in turn lead to distinctive flat surfaces or face, these faces intersect at angles that are characteristic of the substance, crystals tend to have sharp, well defined and high melting points because of the same distance from the same number and type of neighbors. They generally have geometric shapes, some examples are diamonds, metals, salts.
- Amorphous solids produce irregular or curved surfaces when broken and they have poorly defined patterns when exposed to x rays because of their irregular array. In contrast with crystal solids, amorphous solids soften over a wide temperature range due to the different amounts of thermal energy needed to overcome different interactions. Some examples of these solids are gels, plastics, and some polymers.
I hope you find this information useful and interesting! Good luck!
