The initial speed of the bolt is not 58.86 m/s.
Let a be the acceleration of the rocket.
During the 4 sec lift off, the rocket has reached a height of
h = (1/2)*a*t^2
with t=4,
h = (1/2)*a^16
h = 8*a
Its velocity at 4 sec is
v = t*a
v = 4*a
The initial velocity of the bolt is thus 4*a.
During the 6 sec fall, the bolt has the initial velocity V0=-4*a and it drops a total height of h=8*a. From the equation of motion,
h = (1/2)*g*t^2 + V0*t
Substituting h0=8*a, t=6 and V0=-4*a into it,
8*a = (1/2)*g*36 - 4*a*6
Solving for a
a = 5.52 m/s^2
Answer:
Myocardium. That is the type. (srry i was in a rush hope this helps)
Kirchhoff's circuit laws are two equalities that deal with the current and potential difference (commonly known as voltage) in the lumped element model of electrical circuits. They were first described in 1845 by German physicist Gustav Kirchhoff. This generalized the work of Georg Ohm and preceded the work of Maxwell.
Answer:
KE = 2.535 x 10⁷ Joules
Explanation:
given,
angular speed of the fly wheel = 940 rad/s
mass of the cylinder = 630 Kg
radius = 1.35 m
KE of flywheel = ?
moment of inertia of the cylinder
=
= 574 kg m²
kinetic energy of the fly wheel
KE = 2.535 x 10⁷ Joules
the kinetic energy of the flywheel is equal to KE = 2.535 x 10⁷ Joules
