Answer:
BC and DE
Explanation:
In the given figure, the velocity time graph is shown. We know that the area under v-t curve gives the displacement of the particle.
Area under AB, 
Area under BC, 
Area under CD, 
Area under DE, 
Area under EF, 
So, form above calculations it is clear that, during BC and DE undergo equal displacement. Hence, the correct option is (c) "BC and DE = 4 meters".
Answer:
Required energy Q = 231 J
Explanation:
Given:
Specific heat of copper C = 0.385 J/g°C
Mass m = 20 g
ΔT = (50 - 20)°C = 30 °C
Find:
Required energy
Computation:
Q = mCΔT
Q = 20(0.385)(30)
Required energy Q = 231 J
Answer: The pressure that one experiences on the Mount Everest will be different from the one, in a classroom. It is because pressure and height are inversely proportional to each other. This means that as we move up, the height keeps on increasing but the pressure will keep on decreasing. This is the case that will be observed when one stands on the Mount Everest as the pressure is comparatively much lower there.
It is because as we move up, the amount of air molecules keeps on decreasing but all of the air molecules are concentrated on the lower part of the atmosphere or on the earth's surface.
Thus a person in a low altitude inside a classroom will experience high pressure and a person standing on the Mount Everest will experience low pressure.
Answer:
Research towards increasing the understanding of progressive illnesses, such as Parkinson’s and Alzheimer’s diseases.
Explanation:
The Yerkes Primate center was established in 1930 by Robert Yerkes, in Orange Park, Florida. But was moved to its present location in Emory University.
The center has its focus on two major area of research, which are Immunology and Vaccine, and research towards increasing the understanding of progressive illnesses, such as Parkinson’s and Alzheimer’s diseases.
Answer:
See explanation below
Explanation:
The equation to use for this is the following:
dU = q + w
As the heat is being release, this value is negative, and same here happens with the work done, because it's in the surroundings.
Therefore the change in the energy would be:
dU = -2.59x10^4 - 6.46^4
dU = -9.05x10^4 kJ
