Answer:
Natural selection is a mechanism, or cause, of evolution.
Explanation:
Adaptations are physical or behavioral traits that make an organism better suited to its environment. Heritable variation comes from random mutations.
Answer:
How fast and efficient the energy is released.
Explanation:
Before burning the marshmallow energy is stored in it in the form of chemical bond energy or chemical potential energy. So upon burning this energy is released. So there will be a difference in energy release from a burned marshmallow and the one we eat straight from package.
The answer is 100 my nilla.
Answer:
T’= 4/3 T
The new tension is 4/3 = 1.33 of the previous tension the answer e
Explanation:
For this problem let's use Newton's second law applied to each body
Body A
X axis
T = m_A a
Axis y
N- W_A = 0
Body B
Vertical axis
W_B - T = m_B a
In the reference system we have selected the direction to the right as positive, therefore the downward movement is also positive. The acceleration of the two bodies must be the same so that the rope cannot tension
We write the equations
T = m_A a
W_B –T = M_B a
We solve this system of equations
m_B g = (m_A + m_B) a
a = m_B / (m_A + m_B) g
In this initial case
m_A = M
m_B = M
a = M / (1 + 1) M g
a = ½ g
Let's find the tension
T = m_A a
T = M ½ g
T = ½ M g
Now we change the mass of the second block
m_B = 2M
a = 2M / (1 + 2) M g
a = 2/3 g
We seek tension for this case
T’= m_A a
T’= M 2/3 g
Let's look for the relationship between the tensions of the two cases
T’/ T = 2/3 M g / (½ M g)
T’/ T = 4/3
T’= 4/3 T
The new tension is 4/3 = 1.33 of the previous tension the answer e
Answer:
Explanation:
This is a recoil problem, which is just another application of the Law of Momentum Conservation. The equation for us is:
![[m_av_a+m_ev_e]_b=[m_av_a+m_ev_e]_a](https://tex.z-dn.net/?f=%5Bm_av_a%2Bm_ev_e%5D_b%3D%5Bm_av_a%2Bm_ev_e%5D_a)
which, in words, is
The momentum of the astronaut plus the momentum of the piece of equipment before the equipment is thrown has to be equal to the momentum of all that same stuff after the equipment is thrown. Filling in:
![[(90.0)(0)+(.50)(0)]_b=[(90.0)(v)+(.50)(-4.0)]_a](https://tex.z-dn.net/?f=%5B%2890.0%29%280%29%2B%28.50%29%280%29%5D_b%3D%5B%2890.0%29%28v%29%2B%28.50%29%28-4.0%29%5D_a)
Obviously, on the left side of the equation, nothing is moving so the whole left side equals 0. Doing the math on the right and paying specific attention to the sig fig's here (notice, I added a 0 after the 4 in the velocity value so our sig fig's are 2 instead of just 1. 1 is useless in most applications).
0 = 90.0v - 2.0 and
2.0 = 90.0v so
v = .022 m/s This is the rate at which he is moving TOWARDS the ship (negative was moving away from the ship, as indicated by the - in the problem). Now we can use the d = rt equation to find out how long this process will take him if he wants to reach his ship before he dies.
12 = .022t and
t = 550 seconds, which is the same thing as 9.2 minutes
