Answers
Answer:
D
Explanation:
Related Questions
What are the biotic factors in this image?
Someone help thank you!!
Answers
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
Answers
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
A 10- kg ball starting from rest rolls down a 5 m tall smooth hill from one person to another person who is standing at the bottom of the hill with a big spring whose constant is 100 N/m. How far does the spring compress in order to stop the ball
Answers
Answer: 3.13 m
Explanation:
Given
mas of the ball is m=10 kg
The ball rolls down a vertical distance of 5 m
Spring constant of spring is [tex]k=100\ N/m[/tex]
Here, the potential energy of the ball converted into kinetic energy which in turn converts into elastic potential energy
[tex]\Rightarrow mgh=\frac{1}{2}kx^2\quad [\text{x=compression in the spring}]\\\\\Rightarrow 10\times 9.8\times 5=\frac{1}{2}\cdot 100\cdot x^2\\\Rightarrow x=\sqrt{9.8}\\\Rightarrow x=3.13\ m[/tex]
Thus, the spring compresses by 3.13 m.
What is Hooke's law? what is meant by elastic limit?
please answer me
Answers
Answer:
Explanation:
The law is about how object will be deformed when certain amount force is applied. Elastic limit is a point where object will not return its previous shape after being deformed
Please help ASAP with questions
Answers
Which element would have properties most like helium (He)?
A. Ar
B. Hg
C. H
D. O
Answers
Answer: A. Ar
Explanation: not anything else besides Ar
Suppose that an electron and a positron collide head-on. Both have kinetic energy of 1.20 MeV and rest energy of 0.511 MeV. They produce two photons, which by conservation of momentum must have equal energy and move in opposite directions. What is the energy Ephoton of one of these photons
Answers
Answer:
E = 1.711 MeV
Explanation:
From the law of the conservation of energy:
[tex]K.E_{e}+K.E_p + E_{e}+E_{p} = 2 E[/tex]
where,
[tex]K.E_e=K.Ep=[/tex] the kinetic energy of positron and electron = 1.2 MeV
[tex]E_e=E_p =[/tex] Rest energy of the electron and the positron = 0.511 MeV
E = Energy of Photon = ?
Therefore,
[tex]1.2\ MeV + 1.2\ MeV + 0.511\ MeV + 0.511\ MeV = 2E\\\\E = \frac{3.422\ MeV}{2}\\\\[/tex]
E = 1.711 MeV
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
Answers
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