Answer:
KE = 1/2mv squared (im not sure how to write the squared on top of the V but just add it there)
m=8 kg
Explanation:
Solve for m, plug in v = m/s and KE = 100J. The answer is m=8kg
Answer:
The first step in the Scientific Method is to make objective observations. These observations are based on specific events that have already happened and can be verified by others as true or false. Step 2. Form a hypothesis.
Explanation:
7. Succession, 6. Extinction, 5. Niche, 3. Ecosystem, 2. Population, 1. Habitat, 4. Environment, 8. Community
The given question is incomplete, the complete question is:
During germination, peas sprout and grow. The data table shows the carbon dioxide produced during the germination period of peas under different conditions. Condition Rate of carbon dioxide produced (mL/min) Germinating peas, 10ºC 0.01 Germinating peas, 20ºC 0.02 What is the best conclusion? The rate of cellular respiration in germinating peas is exactly one thousand times the rate of ATP production. Germinating peas at 10ºC create carbon dioxide at a rate of 0.01 mL/min during ATP production. Germinating peas at 20ºC have a higher rate of cellular respiration than germinating peas at 10ºC. The rate of cellular respiration cannot be measured without knowing the rate of ATP production.
Answer:
The correct statement is that at 20 degree C, the germinating peas exhibits a higher rate of cellular respiration in comparison to the germinating peas at 10 degree C.
Explanation:
The process of respiration results in the production of carbon dioxide, respiration refers to a chemical reaction that generates water, carbon dioxide, and energy by undergoing oxidation of the glucose molecules. This phenomenon plays an essential role in the life of the organisms for obtaining energy from the food they consume to perform daily activities.
From the question, it is evident that the peas, which were germinating at 20 degrees C exhibit a higher rate of cellular respiration as they are generating 0.02 milliliters of carbon dioxide in a minute, while on the other hand, the germinating peas at 10 degrees C are giving rise to 0.01 milliliters of carbon dioxide in a minute.
Answer:
(a) Frequency of M = 0.64
Frequency of N = 0.04
Frequency of MN= 0.32
(b) Expected frequencies of M = 0.648
Expected frequencies of MN = 0.304
Expected frequencies of N = 0.048
Explanation:
(a) If random mating takes place in the population, then the expected frequencies are
f(L(M)) = p = 0.8
F(L(N)) = q
q= 1 - p
= 1 - 0.8
= 0.2
Frequency of M = p^2 = ( 0.8)^2 = 0.64
Frequency of N = q^2 = (1-p)^2 = (1 - 0.8)^2 = (0.2)^2 = 0.04
Frequency of MN = 2pq = 2 * 0.8 * 0.2 = 0.32
(b)
F = inbreeding coefficient = 0.05
f(L(M)L(M)) = p^2 + Fpq = (0.8)^2 + 0.05 * 0.8 * 0.2 = 0.648
f(L(M)L(N)) = 2 pq - 2Fpq = 2 * 0.8 * 0.2 - ( 2 * 0.05 * 0.8 * 0.2) = 0.304
f(L(N)L(N)) = q^2 + Fpq = (0.2)^2 + ( 0.05 * 0.8 * 0.2) = 0.048
