<span>Maritime tropical air masses develop over warm waters present in the tropics and Gulf of Mexico, where heat and moisture are carried to to the overlying air from the water below.
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</span><span> Tropical air masses having northward movement carry warm moist air into the United States, thus increasing the potential for condensation. Generally the southern states experience tropical air masses. But, in winter season, southerly winds ahead of migrating cyclones <span>sometimes transport tropical air mass towards north.
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</span></span><span><span>The counterclockwise winds related to northern hemisphere mid latitude cyclones play an important role in the movement air masses, carrying warm moist air towards north ahead of a low while dragging colder and drier air towards south.</span></span>
At the bottom of the rotation, the kinetic far exceeds the potential. However, at both tops, potential exceeds kinetic.
Answer:
charge on each
Q1 = 2.06 ×
C
Q2 = 7.23 × C
when force were attractive
Q1 = 1.07 × 
C
Q2 = -1.39 × C
Explanation:
given data
total charge = 93.0 μC
apart distance r = 1.14 m
force exerted F = 10.3 N
to find out
What is the charge on each and What if the force were attractive
solution
we know that force is repulsive mean both sphere have same charge
so total charge on two non conducting sphere is
Q1 + Q2 = 93.0 μC = 93 ×
C
and
According to Coulomb's law force between two sphere is
Force F = 
.........1
Q1Q2 = 
here F is force and r is apart distance and k is 9 × 
N-m²/C² put all value we get
Q1Q2 = 
Q1Q2 = 1.49 × 
C²
and
we have Q2 = 93 × C - Q1
put here value
Q1² - 93 × Q1 + 1.49 × = 0
solve we get
Q1 = 2.06 × C
and
Q1Q2 = 1.49 ×
2.06 × Q2 = 1.49 ×
Q2 = 7.23 × C
and
if force is attractive we get here
Q1Q2 = - 1.49 × C²
then
Q1² - 93 × Q1 - 1.49 × = 0
we get here
Q1 = 1.07 × C
and
Q1Q2 = - 1.49 ×
2.06 × Q2 = - 1.49 ×
Q2 = -1.39 × C
Its letter C. 5N to the left. Since Jeremy's force in Newtons are higher than Amanda's (in newtons), and since Jeremy's force directs to the left, then the direction of the force will be to the LEFT. Then subtract the higher one to the lower one so that would be: 10N-5N=5N. So it is C. 5N to the left.
Answer:
The potential energy stored in the spring is 0.018 J.
Explanation:
Given;
spring constant, k = 90 N/m
extension of the spring, x = 2 cm = 0.02 m
The potential energy stored in the spring is calculated as;
U = ¹/₂kx²
where;
U is the potential energy stored in the spring
Substitute the given values in the equation above;
U = ¹/₂ x 90 N/m x (0.02 m)²
U = 0.018 J
Therefore, the potential energy stored in the spring is 0.018 J.
