Atomic emission spectra are like fingerprints for the elements, because it can show the number of orbits in that elements as well as the energy levels of that element. As each emission of atomic spectra is unique, it is the fingerprint of element.
<u>Explanation:
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Each element has unique arrangement of electrons in different energy levels or orbits. So depending upon the difference in energy of the orbital, the emission spectra will be varying for each element. As the binding energy and excitation energy is not common for any two elements, so the spectra obtained when those excited electrons will release energy to ground state will also be unique.
As in atomic emission spectra, the incident light will be absorbed by the electrons of those elements making the electron to excite, then the excited electron will return to ground state on emission of radiation of energy. Thus, this energy of emission is equal to the difference between the energy of initial and final orbital. So the spectra will act like fingerprints for elements.
Answer:
11.87 ms⁻¹
Explanation:
You can use the kinematic equation
v² = u² + 2as
Where v = final velocity
u = initial velocity
a = acceleration
s = displacement
v² = 9² + 2×1.5×20
So you get v = 11.87 ms⁻¹
Answer:
Mary will have to wait for 63.2 seconds
Explanation:
Time required for the apple to drop from a height of 17.0 m above the ground to 1.65 m above the ground is given by the formula below:
t = √2h/g where h is height through which the object falls, g is acceleration due to gravity
h = 17.0 - 1.65 = 15.35 m
g = 9.8 m/s²
t = √(2 * 15.35/9.8)
t = 1.77 s or approximately 1.8 s
Time taken for bill to get to the point below Mary's window is given below;
time taken = distance/velocity
distance = 130 m; velocity = 2.0 m/s
time taken by Bill = 130/2.0 = 65 s
Therefore, Mary will have to wait for (65 - 1.8) s = 63.2 seconds
Answer:
1752.14 tonnes per year.
Explanation:
To solve this exercise it is necessary to apply the concepts related to power consumption and power production.
By conservation of energy we know that:
Where,
Production of Power
Consumption of power
Where the production of power would be,
Where,
m = Total mass required
Energy per Kilogram
Efficiency
The problem gives us the aforementioned values under a production efficiency of 45%, that is,
Replacing the values we have,
Solving for m,
We have the mass in kilograms and the time in seconds, we need to transform this to tons per year, then,
tonnes per year.
