Total energy associated with a spring:
When x = 0.5a:
The ratio:
The vessel contains a total air mass of 235.34 kilograms. Assuming the air behaves like an ideal gas, the air's density (in kilograms per cubic meter) can be expressed using the following formula:
(1)
Where: - Pressure is in kilopascals. - Molar mass is expressed in kilomoles per kilogram. - The ideal gas constant is measured in kilopascal-cubic meters per kilomole-Kelvin. - Temperature is indicated in Kelvin. Given the values of pressure, molar mass, the gas constant, and temperature, we can find the air density. The mass of the air inside the vessel is calculated from the density definition as follows:
(2)
Where m represents the mass, given in kilograms. With the known values of density and volume in mind, we derive the mass of air stored in the vessel as 235.34 kilograms.
Since the roundabout operates at a constant angular velocity, the input power equals the frictional power. Given that the frictional power is 2.5 kW, we can express this as frictional torque multiplied by angular velocity:
frictional torque x 0.47 = 2.5 kW. Therefore, solving for frictional torque gives us 2.5 / 0.47 kN.m, which amounts to approximately 5.32 kN.m, leading to a rounded value of 5 kN.m.
When the power supply is interrupted, the roundabout experiences deceleration due to the influence of the frictional torque.
<span>a. To determine the velocity at which the camera strikes the ground:
v^2 = (v0)^2 + 2ay = 0 + 2ay
v = sqrt{ 2ay }
v = sqrt{ (2)(3.7 m/s^2)(239 m) }
v = 42 m/s
The camera impacts the ground with a speed of 42 m/s.
b. To calculate the duration it takes for the camera to reach the bottom:
y = (1/2) a t^2
t^2 = 2y / a
t = sqrt{ 2y / a }
t = sqrt{ (2)(239 m) / 3.7 m/s^2 }
t = 11.4 seconds
The camera descends for 11.4 seconds before hitting the ground.</span>
