When an atom becomes electrically charged the number of electrons or protons stops and they are not equal again. The "extra" electron or proton is not balanced by something inside the atom any longer and it starts attracting itself to othet protons or electrons in other atoms.
<h3>What is atomic structure?</h3>
An atomic structure comprises of positively charged nucleus which is surrounded by negatively charged particles called electron and neutron which is neutral charged.
Unlike charges attract each other while like charges repel each other.
Therefore, When an electron is fully charged, the number of electrons will stop to be unequal again.
Learn more about Atomic charge here.
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A dissonance is an unstable tone combination ; it tension demands an onward motion to a stable chord, thus dissonant chords are active.
First year profit: $1000*12 = $12,000
Second year profit: $12,000 * 1.06
Third year profit: $12,000 * 1.06^2
Fourth year profit: $12,000 * 1.06^3
Fith year profit: $12,000 * 1.06^4 =$15,149.72
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Complete question is;
a. Two equal sized and shaped spheres are dropped from a tall building. Sphere 1 is hollow and has a mass of 1.0 kg. Sphere 2 is filled with lead and has a mass of 9.0 kg. If the terminal speed of Sphere 1 is 6.0 m/s, the terminal speed of Sphere 2 will be?
b. The cross sectional area of Sphere 2 is increased to 3 times the cross sectional area of Sphere 1. The masses remain 1.0 kg and 9.0 kg, The terminal speed (in m/s) of Sphere 2 will now be
Answer:
A) V_t = 18 m/s
B) V_t = 10.39 m/s
Explanation:
Formula for terminal speed is given by;
V_t = √(2mg/(DρA))
Where;
m is mass
g is acceleration due to gravity
D is drag coefficient
ρ is density
A is Area of object
A) Now, for sphere 1,we have;
m = 1 kg
V_t = 6 m/s
g = 9.81 m/s²
Now, making D the subject, we have;
D = 2mg/((V_t)²ρA))
D = (2 × 1 × 9.81)/(6² × ρA)
D = 0.545/(ρA)
For sphere 2, we have mass = 9 kg
Thus;
V_t = √[2 × 9 × 9.81/(0.545/(ρA) × ρA))]
V_t = 18 m/s
B) We are told that The cross sectional area of Sphere 2 is increased to 3 times the cross sectional area of Sphere 1.
Thus;
Area of sphere 2 = 3A
Thus;
V_t = √[2 × 9 × 9.81/(0.545/(ρA) × ρ × 3A))]
V_t = 10.39 m/s