I think it would be Fusing of 2 or more small atoms into a larger one
Well, I guess you can come close, but you can't tell exactly.
It must be presumed that the seagull was flying through the air
when it "let fly" so to speak, so the jettisoned load of ballast
of which the bird unburdened itself had some initial horizontal
velocity.
That impact velocity of 98.5 m/s is actually the resultant of
the horizontal component ... unchanged since the package
was dispatched ... and the vertical component, which grew
all the way down in accordance with the behavior of gravity.
98.5 m/s = √ [ (horizontal component)² + (vertical component)² ].
The vertical component is easy; that's (9.8 m/s²) x (drop time).
Since we're looking for the altitude of launch, we can use the
formula for 'free-fall distance' as a function of acceleration and
time:
Height = (1/2) (acceleration) (time²) .
If the impact velocity were comprised solely of its vertical
component, then the solution to the problem would be a
piece-o-cake.
Time = (98.5 m/s) / (9.81 m/s²) = 10.04 seconds
whence
Height = (1/2) (9.81) (10.04)²
= (4.905 m/s²) x (100.8 sec²) = 494.43 meters.
As noted, this solution applies only if the gull were hovering with
no horizontal velocity, taking careful aim, and with malice in its
primitive brain, launching a remote attack on the rich American.
If the gull was flying at the time ... a reasonable assumption ... then
some part of the impact velocity was a horizontal component. That
implies that the vertical component is something less than 98.5 m/s,
and that the attack was launched from an altitude less than 494 m.
Answer:
W = 30 N
Explanation:
Applying the summation of torques about the wedge for equilibrium, taking the clockwise direction as negative. Since the ruler is balanced horizontally about the wedge. Therefore, the summation of all torques acting about the wedge must be equal to zero.
<u>W = 30 N</u>
1). D. Combined Gas Law
3.) D. Pressure and temperature
4.) C. Boyles law
