Answer:
correct option is c. a testable prediction leading to design of an experiment
Explanation:
The results of raising tadpoles were estimated to be water with an atrazine level of 0.1 ppb compared to those grown in pure water. So, this is not the question. If this assumption can now be tested, an experiment can be made in which a certain number of tadpoles can be raised in pure water and the same number of tadpoles can be raised in water with a 0.1ppb atrazine level can. The difference between the two populations can be estimated or compared.
Answer:
The option that is best described as a way engineers can test and investigate how things should be under certain circumstances is;
Explanation:
Modeling is a tool an engineer can use for the physical representation of a system that will facilitate the definition, testing and analysis, communication, data generation, data verification and data validation of given concepts
Models also aid in setting specifications, supporting designs, and verification of a given system
Therefore, with modeling engineers can investigate the behavior of systems under given environmental conditions.
Answer:
Have the power company install insulated sleeves (also known as “eels”) over power lines.
Wearing PPE is the only way to prevent being electrocuted
Explanation:
To prevent electrocution at workplace, employers can ensure that the power company install insulated sleeves (also known as “eels”) over power lines. Additionally, the employees should wear PPEs which are insulators to prevent electrocution.
Answer:
The answer is below
Explanation:
Given that:
Diameter (D) = 0.03 mm = 0.00003 m, length (L) = 2.4 mm = 0.0024 m, longitudinal tensile strength
, Fracture strength
![(\sigma_f)=5100\ MPa=5100*10^6\ Pa,fiber-matrix\ stres(\sigma_m)=17.5\ MPa=17.5*10^6\ Pa,matrix\ strength=\tau_c=17\ MPa=17 *10^6\ Pa](https://tex.z-dn.net/?f=%28%5Csigma_f%29%3D5100%5C%20MPa%3D5100%2A10%5E6%5C%20Pa%2Cfiber-matrix%5C%20stres%28%5Csigma_m%29%3D17.5%5C%20MPa%3D17.5%2A10%5E6%5C%20Pa%2Cmatrix%5C%20strength%3D%5Ctau_c%3D17%5C%20MPa%3D17%20%2A10%5E6%5C%20Pa)
a) The critical length (
) is given by:
![L_c=\sigma_f*(\frac{D}{2*\tau_c} )=5100*10^6*\frac{0.00003}{2*17*10^6}=0.0045\ m=4.5\ mm](https://tex.z-dn.net/?f=L_c%3D%5Csigma_f%2A%28%5Cfrac%7BD%7D%7B2%2A%5Ctau_c%7D%20%29%3D5100%2A10%5E6%2A%5Cfrac%7B0.00003%7D%7B2%2A17%2A10%5E6%7D%3D0.0045%5C%20m%3D4.5%5C%20mm)
The critical length (4.5 mm) is greater than the given length, hence th composite can be produced.
b) The volume fraction (Vf) is gotten from the formula:
![\sigma_{cd}=\frac{L*\tau_c}{D}*V_f+\sigma_m(1-V_f)\\\\V_f=\frac{\sigma_{cd}-\sigma_{m}}{\frac{L*\tau_c}{D}-\sigma_{m}} \\\\Substituting:\\\\V_f=\frac{630*10^6-17.5*10^6}{\frac{0.0024*17*10^6}{0.00003} -17.5*10^6} \\\\V_f=0.456](https://tex.z-dn.net/?f=%5Csigma_%7Bcd%7D%3D%5Cfrac%7BL%2A%5Ctau_c%7D%7BD%7D%2AV_f%2B%5Csigma_m%281-V_f%29%5C%5C%5C%5CV_f%3D%5Cfrac%7B%5Csigma_%7Bcd%7D-%5Csigma_%7Bm%7D%7D%7B%5Cfrac%7BL%2A%5Ctau_c%7D%7BD%7D-%5Csigma_%7Bm%7D%7D%20%20%5C%5C%5C%5CSubstituting%3A%5C%5C%5C%5CV_f%3D%5Cfrac%7B630%2A10%5E6-17.5%2A10%5E6%7D%7B%5Cfrac%7B0.0024%2A17%2A10%5E6%7D%7B0.00003%7D%20-17.5%2A10%5E6%7D%20%5C%5C%5C%5CV_f%3D0.456)
Answer:
14.506°C
Explanation:
Given data :
flow rate of water been cooled = 0.011 m^3/s
inlet temp = 30°C + 273 = 303 k
cooling medium temperature = 6°C + 273 = 279 k
flow rate of cooling medium = 0.02 m^3/s
Determine the outlet temperature
we can determine the outlet temperature by applying the relation below
Heat gained by cooling medium = Heat lost by water
= ( Mcp ( To - 6 ) = Mcp ( 30 - To )
since the properties of water and the cooling medium ( water ) is the same
= 0.02 ( To - 6 ) = 0.011 ( 30 - To )
= 1.82 ( To - 6 ) = 30 - To
hence To ( outlet temperature ) = 14.506°C