Answer:
incurriculum design process the answer is designs
Complete Question
A flywheel in a motor is spinning at 510 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and diameter 75.0 cm . The power is off for 40.0 s , and during this time the flywheel slows down uniformly due to friction in its axle bearings. During the time the power is off, the flywheel makes 210 complete revolutions. At what rate is the flywheel spinning when the power comes back on(in rpm)? How long after the beginning of the power failure would it have taken the flywheel to stop if the power had not come back on, and how many revolutions would the wheel have made during this time?
Answer:
Explanation:
From the question we are told that:
Angular velocity
Mass
Diameter d
Off Time
Oscillation at Power off
Generally the equation for Angular displacement is mathematically given by
Generally the equation for Time to come to rest is mathematically given by
Therefore Angular displacement is
Answer:
a true
Explanation:
Water has the high specific heat capacity.
That means it resists to change its temperature when a considerable amount of heat lost or gained
4200J is needed to raise the temperature of 1kg substance by 1 degree Celsius
(a) 147,500 J
Newton's second law, applied to the helicopter, states that
where
F is the lifting force
mg is the weight of the helicopter, with m=500 kg being the mass of the helicopter and g=9.8 m/s^2 the acceleration due to gravity
a=2.00 m/s^2 is the acceleration of the helicopter
From the equation, we can calculate the magnitude of the lifting force:
The vertical distance covered by the helicopter is given by
So, the work done by the lifting force F is
2) -122,500 J
The magnitude of the gravitational force acting on the helicopter is
And the work done by this force on the helicopter is
And the negative sign is due to the fact that the direction of the gravitational force is opposite to the displacement of the helicopter.
3) 25,500 J
The net work done on the helicopter is given by the sum of the work done by the two forces:
Answer:
Explanation:
Given ; P1 = 100 kPa, T1 = 30°C, T2 = 90°C, P2 = 200 kPa, V1 = 350 m/s
From energy conservation at inlet and outlet
h1 + (1/2)v1² = h2 + (1/2)v2²
CpT1 + (1/2)v1² = CpT2 + (1/2)v2²
.718(90-30) x 103= 0.5x (350²-v2²)
velocity at the exit of a diffuser ; V2=190.63m/s