bigger acceleration......................
Answer:
Explanation:
The original has hybrid 15N/14N DNA, and the second generation has both hybrid 15N/14N DNA and 14N/14N DNA. No 15N/15N DNA was observed. In this experiment:
Nitrogen is a significant component of DNA. 14N is the most bounteous isotope of nitrogen, however, DNA with the heavier yet non-radioactive and 15N isotope is likewise practical.
E. coli was developed for several generations in a medium containing NH4Cl with 15N. When DNA is extracted from these cells and centrifuged on a salt density gradient, the DNA separates at which its density equals to the salt arrangement. The DNA of the cells developed in 15N medium had a higher density than cells developed in typical 14N medium. After that, E. coli cells with just 15N in their DNA were transferred to a 14N medium.
DNA was removed and compared to pure 14N DNA and 15N DNA. Immediately after only one replication, the DNA was found to have an intermediate density. Since conservative replication would result in equal measures of DNA of the higher and lower densities yet no DNA of an intermediate density, conservative replication was eliminated. Moreso, this result was consistent with both semi-conservative and dispersive replication. Semi conservative replication would result in double-stranded DNA with one strand of 15N DNA, and one of 14N DNA, while dispersive replication would result in double-stranded DNA with the two strands having mixtures of 15N and 14N DNA, either of which would have appeared as DNA of an intermediate density.
The DNA from cells after two replications had been completed and found to comprise of equal measures of DNA with two different densities, one corresponding to the intermediate density of DNA of cells developed for just a single division in 14N medium, the other corresponding to DNA from cells developed completely in 14N medium. This was inconsistent with dispersive replication, which would have resulted in a single density, lower than the intermediate density of the one-generation cells, yet at the same time higher than cells become distinctly in 14N DNA medium, as the first 15N DNA would have been part evenly among all DNA strands. The result was steady with the semi-conservative replication hypothesis. The semi conservative hypothesis calculates that each molecule after replication will contain one old and one new strand. The dispersive model suggests that each strand of each new molecule will possess a mixture of old and new DNA.
Answer:
The maximum altitude reached with respect to the ground = 0.5 + 0.510 = 1.01 km
Explanation:
Using the equations of motion,
When the rocket is fired from the ground,
u = initial velocity = 0 m/s (since it was initially at rest)
a = 10 m/s²
The engine cuts off at y = 0.5 km = 500 m
The velocity at that point = v
v² = u² + 2ay
v² = 0² + 2(10)(500) = 10000
v = 100 m/s
The velocity at this point is the initial velocity for the next phase of the motion
u = 100 m/s
v = final velocity = 0 m/s (at maximum height, velocity = 0)
y = vertical distance travelled after the engine shuts off beyond 0.5 km = ?
g = acceleration due to gravity = - 9.8 m/s²
v² = u² + 2gy
0 = 100² + 2(-9.8)(y)
- 19.6 y = - 10000
y = 510.2 m = 0.510 km
So, the maximum altitude reached with respect to the ground = 0.5 + 0.510 = 1.01 km
Hope this helps!!!
Answer:
The earth is a vast, complex system powered by two sources of energy: an internal source (the decay of radioactive elements in the geosphere, which generates geothermal heat) and an external source (the solar radiation received from the Sun); the vast majority of the energy in the earth system comes from the Sun.
Explanation:
becuse
I found some good web pages with highly detailed answers to predicting the range of a trebuchet. A very simple model we have used in my Intro to Eng class just uses the mass of the projectile (m2), the mass of the counter weight (m1), and the height the counter weight falls (h):
Range (max) = 2 * (m1/m2) * h
Now the efficiency of the trebuchet will cause this model to be off by quite a bit. But once you have a working trebuchet, we find this model works well when we vary m1, m2, or h. We assume we have a take off angle of 45 degrees above the horizon.
This solution is based on the classic max range ballistics problem - 45 degree take off angle. It also assumes converting all the potential energy of the counter weight to kinetic energy of the projectile. That is why the efficiency issue comes up as a lot of energy is lost due to friction in the moving trebuchet. If the projectile spins a lot then it will travel a shorter distance as the potential energy is split into kinetic and rotational energy. Projectile shape and wind will also vary the results.
Good luck.
