Answer:
The time required for sucrose transportation through the tube is 8.4319 sec.
Explanation:
Given:
L = 0.025 m
A = 6.5×10^-4 m^2
D = 5×10^-10 m^2/s
ΔC = 5.2 x 10^-3 kg/m^3
m = 5.7×10^-13 kg
Solution:
t = m×L / D×A×ΔC
t = (5.7×10^-13) × (0.025) / (5×10^-10)×(6.5×10^-4)×(5.2 x 10^-3)
t = 8.4319 sec.
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
Answer:
Explanation:
The maximum efficient power plant will be the plant based on carnot cycle whose efficiency is given by the following formula
Efficiency = (T₁ - T₂) / T₁
T₁ is temperature of hot reservoir and T₂ is temperature of cold reservoir.
Putting the given values
efficiency of power plant = (35 - 5) / (273 + 35 )
= 30 / 308
= .097
= 9.7 %
Okay, I don't know if this question is supposed to be a trick question or not. The weight of the apple does not change as the plane travels up the atmosphere, but the MASS changes. Weight doesn't change no matter what environment you're in, but the mass changes in different environments. In this case, the weight is constant but the mass is decreasing as you go higher up.
Answer:
I believe it is a transfer of energy from the rubber band to the car
Explanation:
However it could be a transfer of potential energy from the stretched rubber band that is released into kinetic energy into the car. Depends as there is little context to the question.
