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Crank
3 years ago
8

What happens in a cold front? A.A cold air mass catches up to a moving warm air mass, sliding over it. B.A warm air mass becomes

a cold air mass as it moves. C.A warm air mass collides with a stationary cold air mass, and slides under it. D.A cold air mass comes in under a warm air mass.
Physics
2 answers:
svlad2 [7]3 years ago
6 0

Answer:

A cold air mass comes in under a warm air  mass

Explanation:

Sav [38]3 years ago
4 0
In meteorology (the study of weather), a "front" is the leading edge
of an air mass, that's moving in and replacing a mass of a different
kind of air.  Like warm moving in and replacing cold, or cold moving
in and replacing warm.

Remember that cold air is heavier than warm air.  So whenever a
warm or cold front moves in, the cold air always stays low and the 
warm air always lifts up over it.

With that little bit of an explanation, you can see that choice-D is
the only choice that makes sense.
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Answer:

A

Explanation:

There are three basic forces in aerodynamics: acceleration, which moves an airplane forward; drag, which holds it back; and height, which keeps it airborne. Lift is generally explained by three theories: Bernoulli's principle, the Coanda effect, and Newton's third law of motion.

8 0
3 years ago
Which of the following is an important step in the process of conflict resolution?
4vir4ik [10]
A would be the correct answer. Its the only one to make sense since you are trying to solve the conflict!
7 0
3 years ago
Read 2 more answers
Two point charges, a +45nC charge X and a +12nC charge Y are separated by a distance of 0.5m.
Gnoma [55]

A) Calculate the resultant electric field strength at the midpoint between the charges.

Qx is the charge at X and Qy is the charge at Y.

E at midpoint = k×Qx/0.25² - k×Qy/0.25²

k = 9×10⁹Nm²C⁻², Qx = 45nC, Qy = 12nC

E = 4752N/C

Well done.

B) Calculate the distance from X at which the electric field strength is zero.

Let D be some point between X and Y for which the net E field is 0.

Let d be the distance from X to D.

Set up the following equation:

E at D = k×Qx/d² - k×Qy/(0.5-d)² = 0

Do some algebra to solve for d:

k×Qx/d² = k×Qy/(0.5-d)²

Qx/d² = Qy/(0.5-d)²

Qx(0.5-d)² = Qyd²

(0.5-d)√Qx = d√Qy

0.5√Qx-d√Qx = d√Qy

d(√Qx+√Qy) = 0.5√Qx

d = (0.5√Qx)/(√Qx+√Qy)

Plug in Qx = 45nC, Qy = 12nC

d ≈ 330mm

C) Calculate the magnitude of the electric field strength at the point P on the diagram below.

First determine the angles of the triangle. The sides of the triangle are 0.3m, 0.4m, and 0.5m, so this is a right triangle where the angle between the 0.3m and 0.4m sides is 90°

∠Y = tan⁻¹(0.4/0.3) = 53.13°

∠X = 90-∠Y = 36.87°

Determine the horizontal component of E at P:

Ex = E from Qx × cos(∠X) - E from Qy × cos(∠Y)

Ex = k×Qx/0.4²×cos(36.87°) - k×Qy/0.3²×cos(53.13°)

Ex = 1305N/C

Determine the vertical component of E at P:

Ey = E from Qx × sin(∠X) - E from Qy × sin(∠Y)

Ey = k×Qx/0.4²×sin(36.87°) - k×Qy/0.3²×sin(53.13°)

Ey = 2479N/C

Use the Pythagorean theorem to determine the magnitude of E at P:

E = √(Ex²+Ey²)

E ≈ 2802N/C

4 0
3 years ago
2. Determine the current in a 120-watt bulb plugged into a 120-volt outlet.
Sergio039 [100]

Answer:

1 ampere

Explanation:

Power = VI

V = voltage

I = current

Given

Power = 120 watts

V = 120 volts

Therefore

120 = 120 x I

Divide both sides by 120

120/120 = 120/120 x I

1 = I

I = 1 ampere

8 0
3 years ago
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kirill [66]

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Newton’s second law is a quantitative description of the changes that a force can produce on the motion of a body. It states that the time rate of change of the momentum of a body is equal in both magnitude and direction to the force imposed on it.

Newton’s third law states that when two bodies interact, they apply forces to one another that are equal in magnitude and opposite in direction. The third law is also known as the law of action and reaction. This law is important in analyzing problems of static equilibrium, where all forces are balanced, but it also applies to bodies in uniform or accelerated motion.

-------------------------------------------------------------------------------------------------------------

The first law represented in the game would be the candy. If you blow it, it would move but then stop due to friction.

The second law would be represented by blowing the candy. Since the candy was light, it would be easier to blow but if it was heavier, it would be a lot harder.

The final law represented in the game would be if you decided to blow the candy with a ballon instead, the candy would move the opposite direction the ballon is moving.

8 0
1 year ago
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