Starting from rest, the plane travels a distance
<em>x</em> = 1/2 <em>at</em>²
with acceleration <em>a</em> after time <em>t</em>. In this instance, it travels
<em>x</em> = 1/2 (3.20 m/s²) (32.8 s)²
<em>x</em> ≈ 1720 m
vf=vi+at
No need for rearranging because it is already set up for Vf (final velocity)
a= -9.8m/s² (because it is falling)
vi= 0
t= 2.7
0 + -9.8(2.7) = vf
vf = -26.5 (-26.46)
Answer:
negative work is done.
Explanation:
Here a football player pushes against another player so the force is applied by the football player so that it will move ahead Now here the force of another player is opposite to his motion so the work done on player by another player is negative.
So work done is negative because it is done on the system.
Its breathing that is moving it is digesting and
The solution would be like this for this specific problem:
Given:
Oil spill radius constant rate increase = 1 m/s
r = 34m
Let the area of the spill
be A and then let its radius be r. <span>
Then A = π r². </span>
Differentiating with
respect to t: <span>
dA / dt = 2π r dr / dt. </span>
Substituting r = 34 and dr
/ dt = 1:
dA/dt = 2 π * r * dr / dt
dA/dt = 2π * 34 * 1 = 68π
= 214 m²/s, to 3 significant figures.
<span>So, given that the radius
is 34m, then the area of the spill is increasing at 214 m²/s.</span>
