<h2>
Answer: (C) Radioactive dating</h2>
Explanation:
Let's begin by explaining that radioactive decay is a spontaneous process in which the nucleus of an atom disintegrates, giving way to the emission of radiation and the appearance of a new nucleus, releasing energy in the process.
<u>This process is widely used in</u><u> radioactive dating</u> (also called isotopic dating or radioisotope dating) in which radioactive impurities were selectively incorporated when the fossil materials were formed.
In this sense, when dating the age of the Earth and its components, it is useful to compare them with a naturally occurring radioisotope having a known half-life (generally uranium-238 is used and sometimes carbon-14).
Answer: Covalent Bonds or Covalent Bonding Electrons
Explanation:
A covalent bond, also called a molecular bond, is a chemical bond that involves the sharing of electron pairs between atoms. These electron pairs are known as shared pairs or bonding pairs, and the stable balance of attractive and repulsive forces between atoms, when they share electrons, is known as covalent bonding.
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
Answer:
100 A.
Explanation:
From the question given above, the following data were obtained:
Electrochemical equivalent (Z) = 0.0012 g/C
Mass (M) = 36 g
Time (t) = 5 mins
Current (I) =?
Next, we shall determine the charge. This can be obtained as follow:
Electrochemical equivalence (Z) = 0.0012 g/C
Mass (M) = 36 g
Charge (Q) =.?
M = ZQ
36 = 0.0012 × Q
Divide both side by 0.0012
Q = 36 / 0.0012
Q = 30000 C
Next, we shall convert 5 mins to s. This can be obtained as follow:
1 min = 60 s
Therefore,
5 mins = 5 × 60
5 mins = 300 s
Finally, we shall determine the current. This can be obtained as shown below:
Charge (Q) = 30000 C
Time (t) = 300 s
Current (I) =?
Q = It
30000 = I × 300
Divide both side by 300
I = 30000 / 300
I = 100 A
Therefore, the current is 100 A.