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
B you will see the objective outside the vehicle not moving will’l you are moveing inside the vehicle
Electromagnetic waves need no matter to travel - they can travel through empty space (a vacuum). In a vacuum, all electromagnetic waves travel at approximately 3 x 108 m/s - which is the fastest speed possible. ...
Light traveling value through an optical Fibre is, 2 x 108 m/s. Hope that helps.
Answer:
2.82 s
Explanation:
The ball will be subject to the acceleration of gravity which can be considered constant. Therefore we can use the equation for uniformly accelerated movement:
Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2
Y0 is the starting position, 2.3 m in this case.
Vy0 is the starting speed, 13 m/s.
a will be the acceleration of gravity, -9.81 m/s^2, negative because it points down.
Y(t) = 2.3 + 13 * t - 1/2 * 9.81 * t^2
It will reach the ground when Y(t) = 0
0 = 2.3 + 13 * t - 1/2 * 9.81 * t^2
-4.9 * t^2 + 13 * t + 2.3 = 0
Solving this equation electronically gives two results:
t1 = 2.82 s
t2 = -0.17 s
We disregard the negative solution. The ball spends 2.82 seconds in the air.
Micah knows that a car had a change in velocity of 15 m/s.To determine acceleration ,Micah also needs the TIME of the total trip in seconds.
