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
Answer:
Fc=5253
N
Explanation:
Answer:
Fc=5253
N
Explanation:
sequel to the question given, this question would have taken precedence:
"The 86.0 kg pilot does not want the centripetal acceleration to exceed 6.23 times free-fall acceleration. a) Find the minimum radius of the plane’s path. Answer in units of m."
so we derive centripetal acceleration first
ac (centripetal acceleration) = v^2/r
make r the subject of the equation
r= v^2/ac
ac is 6.23*g which is 9.81
v is 101m/s
substituing the parameters into the equation, to get the radius
(101^2)/(6.23*9.81) = 167m
Now for part
( b) there are two forces namely, the centripetal and the weight of the pilot, but the seat is exerting the same force back due to newtons third law.
he net force that maintains circular motion exerted on the pilot by the seat belts, the friction against the seat, and so forth is the centripetal force.
Fc (Centripetal Force) = m*v^2/r
So (86kg* 101^2)/(167) =
Fc=5253
N
Answer:
<h3>0.445</h3>
Explanation:
In friction, the coefficient of friction formula is expressed as;

Ff is the frictional force = Wsinθ
R is the reaction = Wcosθ
Substitute inti the equation;

Given
θ = 24°

Hence the coefficient of kinetic friction between the box and the ramp is 0.445
Answer:
i put this in the calculator and my answer is 600. hope this helps
Explanation:
Answer:
The answer is A good luck :P