Answer:
I think a loan is not a asset.
Explanation:
Answer:
Total time= 2.56 s
Explanation:
Given that
So distance = 386242560 m
Velocity of light
We know that
Distance = velocity x time
Now by putting then values
t=1.28 s
So the total time = 2 x 1.28
total time= 2.56 s
Answer:
the height of the image ÷ by the height of the object.
Explanation:
Question
Rutherford tracked the motion of tiny, positively charged particles shot through a thin sheet of gold foil. Some particles travelled in a straight line and some were deflected at different angles.
Which statement best describes what Rutherford concluded from the motion of the particles?
A) Some particles travelled through empty spaces between atoms and some particles were deflected by electrons.
B) Some particles travelled through empty parts of the atom and some particles were deflected by electrons.
C) Some particles travelled through empty spaces between atoms and some particles were deflected by small areas of high-density positive charge in atoms.
D) Some particles travelled through empty parts of the atom and some particles were deflected by small areas of high-density positive charge in atoms.
Answer:
The right answer is C)
Explanation:
In the experiment described above, a piece of gold foil was hit with alpha particles, which have a positive charge. Alpha particles <em>α</em> were used because, if the nucleus was positive, then it would deflect the positive particles. The principles of physics posit that electric charges of the same orientation repel.
So most as expected some of the alpha particles went right through meaning that the gold atoms comprised mostly empty space except the areas that were with a dense population of positive charges. This area became known as the "nucleus".
Due to the presence of the positive charges in the nucleus, some particles had their paths bent at large angles others were deflected backwards.
Cheers!
Answer:
the longest time needed to read an arbitrary sector located anywhere on the disk is 2971.24 ms
Explanation:
Given the data in the question;
first we determine the rotational latency
Rotational latency = 60/(3600×2) = 0.008333 s = 8.33 ms
To get the longest time, lets assume the sector will be found at the last track.
hence we will access all the track, meaning that 127 transitions will be done;
so the track changing time = 127 × 15 = 1905 ms
also, we will look for the sectors, for every track rotations that will be done;
128 × 8.33 = 1066.24 ms
∴The Total Time = 1066.24 ms + 1905 ms
Total Time = 2971.24 ms
Therefore, the longest time needed to read an arbitrary sector located anywhere on the disk is 2971.24 ms
