Flash drivers are more robust, if you drop it, it is still most likely to work. Hard drivers provide more storage. There’s not much difference
Answer:
20 m/s^2
Explanation:
given,
final velocity (v) = 6000m/s
initial velocity (u) = 0m/s
time taken (t) = 5 minutes
= 5×60second
= 300second
acceleration(a) = ?
we know that,
a = (v-u)/t
= (6000-0)/300
= 20 m/s^2
Explanation:
Relation between entropy change and specific heat is as follows.
The given data is as follows.
mass = 500 g, 
= 24.4 J/mol K
= 500 K, 
= 250 K
Mass number of copper = 63.54 g /mol
Number of moles = 
= 
= 7.86 moles
Now, equating the entropy change for both the substances as follows.
![7.86 \times 24.4 \times [T_{f} - 250]](https://tex.z-dn.net/?f=7.86%20%5Ctimes%2024.4%20%5Ctimes%20%5BT_%7Bf%7D%20-%20250%5D)
= ![7.86 \times 24.4 \times [500 -T_{f}]](https://tex.z-dn.net/?f=7.86%20%5Ctimes%2024.4%20%5Ctimes%20%5B500%20-T_%7Bf%7D%5D)
= 750
So, 
= 
- For the metal block A, change in entropy is as follows.
= ![24.4 log [\frac{375}{500}]](https://tex.z-dn.net/?f=24.4%20log%20%5B%5Cfrac%7B375%7D%7B500%7D%5D)
= -3.04 J/ K mol
- For the block B, change in entropy is as follows.
= ![24.4 log [\frac{375}{250}]](https://tex.z-dn.net/?f=24.4%20log%20%5B%5Cfrac%7B375%7D%7B250%7D%5D)
= 4.296 J/Kmol
And, total entropy change will be as follows.
= 4.296 + (-3.04)
= 1.256 J/Kmol
Thus, we can conclude that change in entropy of block A is -3.04 J/ K mol and change in entropy of block B is 4.296 J/Kmol.
Answer:
The statement "Six turns of the cycle are required for every glucose molecule later produced in non–Calvin cycle reactions" is incorrect. It really looks not well-worded.
Explanation:
It is incorrect because Six turns of the cycle are required for every glucose molecule produced in Calvin cycle reactions, no in non-Calvin cycle reactions. This process includes the fixation of 6 molecules of carbon dioxide to produce 1 Glucose (seen as the addition of the two Phosphoglyceraldehide molecules (PGAL). Moreover, the other statements in the questions are correct:
ATP is required during carbon fixation.
The most intensive energy phase is reduction and sugar production.
Twelve NADPH are required for every six CO2 fixed.
NADPH is required for reduction and sugar production.
