Brain fission. Budding. Fragmentation
The answer is orange light reflected from the t-shirt enters human eyes.
Answer:
This question is incomplete, the complete part of the question is as follows:
This best demonstrates which type of an interaction between the plants?
A. cooperation
B. parasitism
C. commensalism
D. competition
The answer is D
Explanation:
Organisms in their natural environment interact with one another in so many ways. The ways by which this interaction occurs are; competition, predation, commensalism etc.
Competition is the interaction between two organisms in which one or both organisms are harmed due to limited resources. Competition occurs when the organisms involved occupy the same niche or utilize the same limited resources.
In this question involving corn plants and milkweed plants. They are said to grow in the same area. Over several years, the milkweed plants have taken over the field and the corn plants no longer have space to grow. In this case, there is a limited space for growth, hence, the corn plant and milkweed COMPETE.
Answer: Atmospheric pressure is caused by the weight of the column of air molecules in the atmosphere above an object, such as the tanker car. In general, pressure is defined as the force exerted on a given area
Explanation:
Answer: hundred cubic meters of carbon dioxide initially at 150◦C and 50 bar is to beisothermally compressed in a frictionless piston-and-cylinder device to a final pressure of 300 bar.Calculatei The volume of the compressed gas.ii The work done to compress the gas.iii The heat flow on compression.assuming carbon dioxide(a) Is an ideal gas.(b) Obeys the principle of corresponding states of Sec. 6.6(c) Obeys the Peng-Robinson equation of state.SolutionWe haveT1= 150◦C,P1= 50 bar,T2= 150◦C,P2= 300 bar. (1 and 2 denote the initialand final conditions in this ’snapshot’ problem, respectively - we have sometimes called themt1andt2)(a) If CO2is an idea gas, we havePV=NRT.The number of moles can be calculated from theinitial conditions:N1=P 1 V 1 RT 1 = (6 × 10 6 Pa)(100 m 3 ) (8 . 314 J/(mol K))(150 + 273 K) = 142127 mol = 142 . 1 kmol i. Since we know N 1 = N 2 , T 2 , P 2 V 2 = N 2 RT 2 P 2 = (142127 mol)(8 . 314 J/(mol K))(150 + 273 K ) 30 × 10 6 Pa = 16 . 66 m 3 ii. Since there is no shaft work, and since the gas is isothermally compressed , we only have pressure-volume work: W = - Z V 2 V 1 PdV = - Z V 2 V 1 NRT V dV = - NRT ln V 2 V 1 W = - (142127 mol)(8 . 314 J/(mol K))(150 + 273 K ) ln 16 . 66 100 = 8 . 958 × 10 8 J iii. Energy balance (integral form) for the closed system is: U 2 - U 1 = Q + W Back from homework 2, for an ideal gas, stating from equation 6.2-21, dU = C V dT + " T ∂P ∂T V - P # dV reduces to: dU = C V dT However, the process is isothermal, so dT = 0 Which gives: dU = Δ U = 1 N Δ U Therefore 0 = Q + W → Q = - W Q = - 8 . 958 × 10 8 J 3
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Explanation:As revealed above, the stimuli connections are clearly stated