Difference between G and g
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Answer
The answer and procedures of the exercise are attached in the following archives.
Step-by-step explanation:
You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.
The force of attraction between the opposite charges of the ions in an ionic compound is an ionic bond.
<u>Explanation:</u>
The transfer process of valence electron between atoms referred as ionic bond. This is a kind of chemical bonds which can create two oppositely charged ions. In the presence of ionic bonds, the metal loses electrons and becomes a positive charge cation, while non-metal accepts these electrons and becomes a negative charge anion.
Here, more than 1 electron can be emitted or received to meet the octet principle and the net charge of the compound should be zero. For example: Table salt. In this compound, sodium loses the electron to become , while the chlorine loses the electron to become
.
Answer:
The magnitude of the electric field at the center of curvature of the arc is 3.87 N/C
Explanation:
Please see the attachments below
Refer to the diagram shown below.
M = 5.97 x 10²⁴ kg, mass of the earth
h = 5.69 x 10⁵ m, height of HST above the earth's surface
R = 6.38 x 10⁶ m, radius of the earth
Note that
G = 6.67 x 10⁻¹¹ (N-m²)/kg², gravitational acceleration constant.
R + h = 6.38 x 10⁶ + 5.69 x 10⁵ = 6.949 x 10⁶ m
The force between the earth and HST is
F = (GMm)/(R+h)²
Let v = tangential velocity of the HST.
The centripetal force acting on HST is equal to F.
Therefore
m*[v²/(R+h)] = (GMm)(R+h)²
v² = (GM)/(R+h)
= [(6.67 x 10⁻¹¹ (N-m²)/kg²)*(5.97 x 10²⁴ kg)]/(6.949 x 10⁶ m)
= 5.7303 x 10⁷ (m/s)²
v = 7.5699 x 103 m/s
Answer
The tangential speed of HST is about 7,570 m/s
M = 5.97 x 10²⁴ kg, mass of the earth
h = 5.69 x 10⁵ m, height of HST above the earth's surface
R = 6.38 x 10⁶ m, radius of the earth
Note that
G = 6.67 x 10⁻¹¹ (N-m²)/kg², gravitational acceleration constant.
R + h = 6.38 x 10⁶ + 5.69 x 10⁵ = 6.949 x 10⁶ m
The force between the earth and HST is
F = (GMm)/(R+h)²
Let v = tangential velocity of the HST.
The centripetal force acting on HST is equal to F.
Therefore
m*[v²/(R+h)] = (GMm)(R+h)²
v² = (GM)/(R+h)
= [(6.67 x 10⁻¹¹ (N-m²)/kg²)*(5.97 x 10²⁴ kg)]/(6.949 x 10⁶ m)
= 5.7303 x 10⁷ (m/s)²
v = 7.5699 x 103 m/s
Answer
The tangential speed of HST is about 7,570 m/s