If you end up breathing asbestos fibers, it can increase the risk of serous diseases. Some examples are lung cancer, mesothelioma and asbestosis. It may also increase your risk for cancers of the digestive system, which includes colon cancer. Asbestos is dangerous when fibers become airborne. That is because once a person ingests or inhales asbestos, the fiber can’t be removable from the body. The fibers can cause irritation and scarring over time if it’s embed into the tissue.
Answer: 0.518
Explanation: Total Area of lake = 225 sq miles = 6.273e+9 sq feet
Change in volume per second = 43800-6200 = 37600 cfs
Change in volume per day = 37600 * 3600 * 24 = 3.249e+9 cfd
Total rise of water level per day = Change in volume per day/Total lake area
= 3.249e+9/6.273E+9 = 0.518 feet/day
Answer:
685.38 MJ
Explanation:
Given that:
mass = 10 tons = 1.0 × 10 ⁴ kg
diameter D = 10 m
radius R = 5 m
speed N = 1000 rpm
Using the formula for K.E = 
to calculate the energy stored
where;
= 104.719 rad/s
Hence, the energy stored is;
= 685379310.1
= 685.38 MJ
Answer: The work of breathing is done by the diaphragm, the muscles between the ribs (intercostal muscles), the muscles in the neck, and the abdominal muscles.
Answer:
The system is marginally stable.
Explanation:
Transfer function, M(s) = [10(s+2)]/(s³ + 3s² + 5s)
In control the stability properties of a system can be obtained from just the characteristic equation of its closed loop transfer function.
- The condition for stability is that all the roots of the characteristic equation be negative and real.
- The condition for asymptotic stability is that all the real parts of the roots must all be negative, since there'll be complex roots.
- The condition for marginal stability is that the real part of all the complex roots are negative, the roots without real parts must have distinct imaginary parts.
- The condition for instability is for at least one of the roots to be positive. Or if there are complex roots, the real part of the roots being positive indicates instability.
The characteristic equation for this transfer function is (s³ + 3s² + 5s)
Solving this polynomial
s = 0
s = [-3 - √(11i)]/2
s = [-3 + √(11i)]/2
These roots have all their real parts to be negative, and the zero root has a distinct imaginary part, hence the system is marginally stable
