The working equation for this problem is a derived equation from the rectilinear motion at constant acceleration. The equation is written below:
2ax = v² - v₀²
where
a is the acceleration
x is the distance
v is the final velocity
v₀ is the initial velocity
Of course, the plane starts from rest, so v₀ = 0. The final velocity is the liftoff speed which is v = 120 km/h. Substituting the values:
2a(270 m) = (120 km/h)² - 0²
Solving for a,
a = 26.67 km/h²
Compute the work done on the table:
<em>W</em> = <em>Fd</em> = (320 N) (32 m) = 10,240 J
Divide this by the given time duration to get the power output:
<em>P</em> = <em>W</em>/∆<em>t</em> = (10,240 J) / (150 s) ≈ 63.3 W
In order to solve this problem, there are two equations that you need to know to solve this problem and pretty much all of kinematics. The first is that d=0.5at^2 (d=vertical distance, a=acceleration due to gravity and t=time). The second is vf-vi=at (vf=final velocity, vi=initial velocity, a=acceleration due to gravity, t=time). So to find the time that the ball traveled, isolate the t-variable from the d=0.5at^2. Isolate the t and the equation now becomes

. Solving the equation where d=8 and a=9.8 makes the time

=1.355 seconds. With the second equation, the vi=0 m/s, the vf is unknown, a=9.8 m/s^2 and t=1.355 sec. Substitute all these values into the equation vf-vi=at, this makes it vf-0=9.8(1.355). This means that the vf=13.28 m/s.
Answer:
Explanation:
Work required = q x V
where q is charge on electron and V is potential difference
= 1.6 x 10⁻¹⁹ x 12
= 19.2 x 10⁻¹⁹ J
Answer:
true
Explanation:
I just had a test on this