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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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You could move something across the Earth with a little push. It would make fuel really efficient on those pathways. You could make a floor that is impossible to walk on. Everybody would just fall without traction.

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3 years ago
A 26 foot ladder is lowered down a vertical wall at a rate of 3 feet per minute. The base of the ladder is sliding away from the
lakkis [162]

Answer:

(i) 7.2 feet per minute.

(ii) No, the rate would be different.

(iii) The rate would be always positive.

(iv) the resultant change would be constant.

(v) 0 feet per min

Explanation:

Let the length of ladder is l, x be the height of the top of the ladder from the ground and y be the length of the bottom of the ladder from the wall,

By making the diagram of this situation,

Applying Pythagoras theorem,

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Differentiating with respect to t ( time ),

0=2x\frac{dx}{dt} + 2y\frac{dy}{dt}  ( l = 26 feet = constant )

\implies 2y\frac{dy}{dt} = -2x\frac{dx}{dt}

\implies \frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}

We have,

y = 10, \frac{dx}{dt}= -3\text{ feet per min}

\frac{dy}{dt}=\frac{3x}{10}-----(X)

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676=x^2 + 100

576 = x^2

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From equation (X),

\frac{dy}{dt}=\frac{3\times 24}{10}=7.2\text{ feet per min}

(ii) From equation (X),

\frac{dy}{dt}\propto x

Thus, for different value of x the value of \frac{dy}{dt} would be different.

(iii) Since, distance = Positive number,

So, the value of y will always a positive number.

Thus, from equation (X),

The rate would always be a positive.

(iv) The length of the ladder is constant, so, the resultant change would be constant.

i.e. x = increases ⇒ y = decreases

y = decreases ⇒ y = increases

(v) if ladder hit the ground x = 0,

So, from equation (X),

\frac{dy}{dt}=0\text{ feet per min}

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3 years ago
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2 years ago
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ivanzaharov [21]

Answer:

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