Hi!
The answers would be <u>chunking</u> & <u>short-term</u>
1. <u>Chunking </u>involves organizing and breaking down information into easier groups to expand capacity.
<h3>Explanation:</h3>
Chunking is a mental process that is observed to increase short-term memory by taking the information and categorizing it into small groups. For instance, a longer number taken as a single unit is harder to recall then when it is divided into smaller units. 235469350 is harder to instantly recall as compared to when it is chunked into 3 groups: 235 469 350.
This allows more information to be stored in, thereby increasing the capacity of the mind to store information.
2. Rehearsal is the verbal repetition of information. These techniques are especially important for the improvement of <u>short-term</u> memory.
<h3>Explanation: </h3>
Short-term memory is lost after a couple of seconds or minutes, for instance even if you chunk the information, you might not recall it after 30 seconds. Rehearsing or repetition of information, either loudly or mentally, extends the time a particular information is retained.
So you depending on the number of times you repeat the number 235 469 350, the more your short term memory improves .
Hope this helps!
A and D are definitely wrong because the two rocks have different masses, so it leaves us with B and C. The most logical answer to the question is
C <span>The one with greater mass takes more force to stop.
</span>
A. 9 J
In a force-distance graph, the work done is equal to the area under the curve in the graph.
In this case, we need to extrapolate the value of the force when the distance is x=30 cm. We can easily do that by noticing that there is a direct proportionality between the force and the distance:
where k is the slope of the line. We can find k, for instance chosing the point at x=5 cm and F=10 N:
And now we can calculate the work by calculating the area under the curve until x=30 cm, F=60 N:
B. 24.5 m/s
The mass of the arrow is m=30 g=0.03 kg. The kinetic energy of the arrow when it is released is equal to the work done by pulling back the bow for 30 cm:
where m is the mass of the arrow and v is its speed. By re-arranging the formula and using W=9 J, we find the speed:
