Answer:
5.0-5.5 is the answer to your question
You are studying a population of flowering plants for several years. When you present your research findings you make the statement that, "Increased allocation of resources to reproduction relative to growth diminished future fecundity." Which of the following graph descriptions could accurately present your data?
a) With seeds in the current year on the y-axis and seeds in the previous year on the x-axis, you would see a line that increased from left to right
b) With survivorship on the y-axis and number of seeds produced on the x-axis, you would see a line that decreased left to right.
c) With leaf area on the y-axis and number of seeds produced on the x-axis, you would see a line that increased left to right
d) With survivorship on the y-axis and number of seeds produced on the x-axis, you would see a line that increased left to right.
e) With seeds in the current year on the y-axis and seeds in the previous year on the x-axis, you would see a line that decreased from left to right
Answer:
Option A
Explanation:
The graph for this problem must depict the following ""Increased allocation of resources to reproduction relative to growth diminished future fecundity."
Hence, the survivor ship must be on the Y axis and the resources on the X axis.
Here the resources include the number of seeds produced.
hence, the higher is the number of seeds (resource), the lower is the survivorship (future fecundity)
Hence, option A is correct
What are the biotic factors in this image?
Someone help thank you!!
A spinning disc with a mass of 2.5kg and a radius of 0.80m is rotating with an angular velocity of 1.5 rad/s. A ball of clay with unknown mass is dropped onto the disk and sticks to the very edge causing the angular velocity of the disk to slow to 1.13 rad/s. What is the mass of the ball of clay
Answer:
M = 1.90 Kg
Explanation:
Given data: mass = 2.5 Kg
radius R = 0.8 m
angular velocity ω = 1.5 rad/s
Angular momentum L =0.5×Iω^2
Where, I is the moment of inertia of the spinning disc.
I = 0.5MR^2
I = 0.5×2.5×0.8^2
I = 0.8 Kg/m^2
Then L = 0.5×0.8×1.5^2 = 0.8×2.25 = 0.9 Kg-m^2/sec
Let unknown mass be M
New mass of disc = (2.5+M) Kg, R = 0.8 m
New I = 0.5(2.5+M)(0.8)^2
Since, angular momentum is conserved
Angular momentum before = angular momentum after
0.5×0.5(2.5+M)(0.8)^2×(1.13)^2 = 0.9
Solving for M we get
0.204304(2.5+M)=0.9
M = 1.90 Kg