The time taken for the spaceship to increase its speed from 11.1 km/s to 11.7 km/s is 107 s
<h3>Data obtained from the question</h3>
The following data were obtained from the question given above:
- Initial velocity (u) = 11.1 Km/s
- Final velocity (v) = 11.7 Km/s
- Distance (s) = 1220 Km
- Time (t) =?
<h3>How to determine the time</h3>
The time taken for the spaceship to increase its speed from 11.1 km/s to 11.7 km/s can be obtained as illustrated below:
s = (u + v)t / 2
Cross multiply
(u + v)t = 2s
Divide both sides by (u + v)
t = 2s / (u + v)t
t = (2 × 1220) / (11.1 + 11.7)
t = 2440 / 22.8
t = 107 s
Thus, the time taken for the spaceship to change its speed is 107 s
Learn more about speed:
brainly.com/question/680492
Learn more about velocity:
brainly.com/question/3411682
#SPJ1
Answer:
Magnetic field lines exit out of the North pole . Magnetic field lines enter into the South pole. Magnetic field lines travel around a bar magnet in closed loops.
Explanation:
Magnetic field lines shows the direction of a magnetic force and how it acts, it gives the direction of the magnetic field at that point in time.
For a bar magnetic, the magnetic field lines runs from the north pole to the south pole, i.e. it exits the north pole and enters into the south pole. This magnetic field lines also go through the magnet forming closed loops without ends.
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:
Isaac Newton discovered that as an object's mass increases, the gravitational attraction of that object increases. Thus, the gravitational attraction on the Moon is much less than it is here on Earth, and a person weighs less on the Moon.
hope this helps!
Answer:
the natural length of the spring is 9 cm
Explanation:
let the natural length of the spring = L
For each of the work done, we set up an integral equation;
![5.4 = \int\limits^{21-l}_{15-l} {kx} \, dx \\\\5.4 = [\frac{1}{2}kx^2 ]^{21-l}_{15-l}\\\\5.4 = \frac{k}{2} [(21-l)^2 - (15-l)^2]\\\\k = \frac{2(5.4)}{(21-l)^2 - (15-l)^2} \ \ \ -----(1)](https://tex.z-dn.net/?f=5.4%20%3D%20%5Cint%5Climits%5E%7B21-l%7D_%7B15-l%7D%20%7Bkx%7D%20%5C%2C%20dx%20%5C%5C%5C%5C5.4%20%3D%20%5B%5Cfrac%7B1%7D%7B2%7Dkx%5E2%20%5D%5E%7B21-l%7D_%7B15-l%7D%5C%5C%5C%5C5.4%20%3D%20%5Cfrac%7Bk%7D%7B2%7D%20%5B%2821-l%29%5E2%20-%20%2815-l%29%5E2%5D%5C%5C%5C%5Ck%20%3D%20%5Cfrac%7B2%285.4%29%7D%7B%2821-l%29%5E2%20-%20%2815-l%29%5E2%7D%20%20%5C%20%5C%20%5C%20-----%281%29)
The second equation of work done is set up as follows;
![9 = \int\limits^{27-l}_{21-l} {kx} \, dx \\\\9 = [\frac{1}{2}kx^2 ]^{27-l}_{21-l}\\\\9 = \frac{k}{2} [(27-l)^2 - (21-l)^2] \\\\k = \frac{2(9)}{(27-l)^2 - (21-l)^2} \ \ \ -----(2)](https://tex.z-dn.net/?f=9%20%3D%20%5Cint%5Climits%5E%7B27-l%7D_%7B21-l%7D%20%7Bkx%7D%20%5C%2C%20dx%20%5C%5C%5C%5C9%20%3D%20%5B%5Cfrac%7B1%7D%7B2%7Dkx%5E2%20%5D%5E%7B27-l%7D_%7B21-l%7D%5C%5C%5C%5C9%20%3D%20%5Cfrac%7Bk%7D%7B2%7D%20%5B%2827-l%29%5E2%20-%20%2821-l%29%5E2%5D%20%5C%5C%5C%5Ck%20%3D%20%5Cfrac%7B2%289%29%7D%7B%2827-l%29%5E2%20-%20%2821-l%29%5E2%7D%20%5C%20%5C%20%5C%20-----%282%29)
solve equation (1) and equation (2) together;
Therefore, the natural length of the spring is 9 cm
