Answer:
D. Calculate the area under the graph.
Explanation:
The distance made during a particular period of time is calculated as (distance in m) = (velocity in m/s) * (time in s)
You can think of such a calculation as determining the area of a rectangle whose sides are velocity and time period. If you make the time period very very small, the rectangle will become a narrow "bar" - a bar with height determined by the average velocity during that corresponding short period of time. The area is, again, the distance made during that time. Now, you can cover the entire area under the curve using such narrow bars. Their areas adds up, approximately, to the total distance made over the entire span of motion. From this you can already see why the answer D is the correct one.
Going even further, one can make the rectangular bars arbitrarily narrow and cover the area under the curve with more and more of these. In fact, in the limit, this is something called a Riemann sum and leads to the definition of the Riemann integral. Using calculus, the area under a curve (hence the distance in this case) can be calculated precisely, under certain existence criteria.
Answer:
D. n=6 to n=2
Explanation:
Given;
energy of emitted photon, E = 3.02 electron volts
The energy levels of a Hydrogen atom is given as; E = -E₀ /n²
where;
E₀ is the energy level of an electron in ground state = -13.6 eV
n is the energy level
From the equation above make n, the subject of the formula;
n² = -E₀ / E
n² = 13.6 eV / 3.02 eV
n² = 4.5
n = √4.5
n = 2
When electron moves from higher energy level to a lower energy level it emits photons;
For a photon to be emitted, electron must move from higher energy level to a lower energy level. The higher energy level is 6 while the lower energy level is 2
Therefore, The electron energy-level transition is from n = 6 to n = 2
<span>C. Reduce the friction between its moving parts.</span>
This cab be solve using Newtons second law of motion :
F = ma
Where F is the force applied
A is the accelaration = v/s
V is the velosity
S is the time
M is the mass
F = (2100 kg)*( 96km/hr)( 1000m/ 1km)( 1 hr/3600 s) / 3.5 s
F = 16000 N is the force applied
