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11111nata11111 [884]
3 years ago
13

Generalized global air circulation and precipitation patterns are caused by

Physics
2 answers:
kotegsom [21]3 years ago
5 0
<span> Rising, warm, moist air masses cool and release precipitation as they rise and then at high altitude, cool
and sink back to the surface as dry air masses after moving north or south of the tropics.

</span>
belka [17]3 years ago
5 0
Rising, warm, moist air masses cool and release precipitation as they rise and then at high altitude, cool and sink back to the surface as dry air masses after moving  north or south of the tropics.
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Consider this acid-base reaction:
sergeinik [125]

Answer:

r

Explanation:

8 0
2 years ago
_________statistics are measures that summarize the characteristics of a sample​
Tamiku [17]

Answer:

Explanation:

Descriptive statistics statistics are measures that summarize the characteristics of a sample.

Hope it helps

4 0
3 years ago
Two objects collide. Object A has a momentum of 49 kg.m/s, and Object B has a momentum of
Natalka [10]

Answer:

O 47 kg. m/s if they were initially headed opposite direction

O 145 kg. m/s if they were initially headed the same direction

The other two are possibilities as well if velocities are initially at an angle

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3 years ago
A charge Q is distributed uniformly around the perimeter of a ring of radius R. Determine the electric potential difference betw
il63 [147K]

Answer:

the electric potential difference between the point at the center of the ring and a point on its axis ΔV is ( 0.8356 )\frac{kQ}{R}

Explanation:

Given the data in the question;

electric potential at the center of the ring V₀ = kQ / R

electric potential on the axis point Vr = kQ / √( R² + x² )

at a distance 6R from the center,

point at x = 6R

so distance circumference r = √( R² + (6R)² )

so

electric potential on the axis point Vr = kQ / √( R² + (6R)² )

Vr = kQ / R√37

Now

ΔV = V₀ - Vr

we substitute

ΔV = ( kQ / R) - ( kQ / R√37 )

ΔV =  kQ/R( 1 - 1/√37 )

ΔV =  kQ/R( 1 - 0.164398987 )

ΔV =  kQ/R( 0.8356 )

ΔV = ( 0.8356 )\frac{kQ}{R}       { where k = \frac{1}{4\pi e_0} }

Therefore, the electric potential difference between the point at the center of the ring and a point on its axis ΔV is ( 0.8356 )\frac{kQ}{R}

5 0
3 years ago
When a pendulum is swinging, the velocity is highest at which point?
balu736 [363]

Answer:

The velocity is highest at the lowest point of the trajectory

Explanation:

We can answer the question by applying the law of conservation of energy. In fact, neglecting friction and air resistance, the mechanical energy of the pendulum is constant during the motion:

E=K+U=const.

where:

K=\frac{1}{2}mv^2 is the kinetic energy, with m being the mass of the pendulum and v the velocity

U=mgh is the potential energy, with g being the acceleration due to gravity and h the heigth of the pendulum with respect to the lowest position

Since E must remain constant, we see that when K is maximum, U is minimum, and viceversa. This also means that when the velocity (v) is maximum, then the height (h) is minimum, so the pendulum must be at its lowest point.

4 0
3 years ago
Read 2 more answers
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