An object that could be considered as negatively charged would be when it has an excess of an electron in its atom. However, when it loses an electron, it could go back to its stable state which is "uncharged" or when there is an excess proton, it could be a positively charged object.
Answer:
The electronic transition of an electron back to a lower energy level generates an emission spectrum.
Explanation:
The atomic emission spectrum¹ of an element has its origin when an electronic transition² occurs. An electron in an atom or ion³will absorb energy coming from a source and pass to a higher energy level, the electron, upon returning to its base state will emit a photon⁴ or a series of photons.
Hence, that leads to the formation of an emission spectrum.
Remember that an electron has energy levels in an atom or ion, at which each energy level has a specific value.
The energy values will differ from one element to another. So, it can be concluded that each element has a unique pattern of emission lines.
Key terms:
¹Spectrum: Decomposition of light in its characteristic colors.
²Electronic transition: When an electron passes from one energy level to another, either for the emission or absorption of a photon.
³Ion: An atom electrically charged due to the gain or loss of electrons.
⁴Photon: Elementary particle that constitutes light.
<h2>
a) Initial velocity = 83 ft/s</h2><h2>
b) Object's maximum speed = 99.4 ft/s</h2><h2>
c) Object's maximum displacement = 153.64 ft</h2><h2>
d) Maximum displacement occur at t = 2.59 seconds.</h2><h2>e)
The displacement is zero when t = 5.70 seconds</h2><h2>
f) Object's maximum height = 153.64 ft</h2>
Explanation:
We have velocity
v(t)= -32t + 83
Integrating
s(t) = -16t²+83t+C
At t = 0 displacement is 46 feet
46 = -16 x 0²+83 x 0+C
C = 46 feet
So displacement is
s(t) = -16t²+83t+46
a) Initial velocity is
v(0)= -32 x 0 + 83 = 83 ft/s
Initial velocity = 83 ft/s
b) Maximum velocity is when the object reaches ground, that is s(t) = 0 ft
Substituting
0 = -16t²+83t+46
t = 5.70 seconds
Substituting in velocity equation
v(t)= -32 x 5.70 + 83 = -99.4 ft/s
Object's maximum speed = 99.4 ft/s
c) Maximum displacement is when the velocity is zero
That is
-32t + 83 = 0
t = 2.59 s
Substituting in displacement equation
s(2.59) = -16 x 2.59²+83 x 2.59+46 = 153.64 ft
Object's maximum displacement = 153.64 ft
d) Maximum displacement occur at t = 2.59 seconds.
e) Refer part b
The displacement is zero when t = 5.70 seconds
f) Same as option d
Object's maximum height = 153.64 ft
