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lidiya [134]
2 years ago
9

What happens when a cold front meets a warm front?

Physics
2 answers:
Solnce55 [7]2 years ago
8 0
These cases are rare, but what happens is something called an occluded front. There are so many things that could be talked about and results aren't typically constant, so I recommend you look up what an occluded front is. Its basically when a cold front catches up to a warm front and forces even more warm air up, causing high amounts of precipitation.
hram777 [196]2 years ago
4 0
What happens is called an occluded front.
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A bullet with a mass of 5 gramsand speed of 560 m/sis fired horizontally atablock of wood with a mass of 2 kg. The block rests o
Anni [7]

Momentum is conserved if and only if sum of all forces which are exserted on system equals zero. In our situation there are only internal forces, so by Newton's third law their vector sum is 0.

So mv=(m+M)v' \Leftrightarrow v'=\frac{mv}{m+M}.

Kinetic energy of system at first: \frac{mv^2}{2}=784\;\textbf{J}. After: \frac{(m+M)\frac{m^2v^2}{(m+M)^2} }{2}=\frac{m^2v^2}{2(m+M)}\approx 1,96\; \textbf{J}. The secret is that other energy is in work of deformation forces (they in turn heat a bullet and a block).

Answer is A)

5 0
3 years ago
A metal wire is in thermal contact with two heat reservoirs at both of its ends. Reservoir 1 is at a temperature of 563 K, and r
Schach [20]

Answer:

1.85 J/K

Explanation:

The computation of total change in entropy is shown below:-

Change in Entropy = Sum Q ÷ T

= \frac{-heat\ entering\ the\ reservoir}{Reservoir\ 1\ Temperature} + \frac{Conduction\ of\ heat}{Reservoir\ 2\ Temperature}

= \frac{-1760}{563} + \frac{1760}{354}

= -3.12 + 4.97

= 1.85 J/K

Therefore for computing the total change in entropy we simply applied the above formula.

As we can see that there is heat entering the reservoir so it will be negative while cold reservoir will be positive else the process would be impossible.

8 0
3 years ago
What morbid structure traditionally has thirteen steps?
tatiyna

Answer:

A gallow

Explanation:

6 0
3 years ago
Two cars cover the same distance in a straight line. Car a covers the distance at a constant velocity. Car b starts from rest an
Artyom0805 [142]

a) For the motion of car with uniform velocity we have , s = ut+\frac{1}{2}at^2, where s is the displacement, u is the initial velocity, t is the time taken a is the acceleration.

In this case s = 520 m, t = 223 seconds, a =0 m/s^2

Substituting

       520 = u*223\\ \\u = 2.33 m/s

 The constant velocity of car a = 2.33 m/s

b) We have s = ut+\frac{1}{2} at^2

s = 520 m, t = 223 seconds, u =0 m/s

Substituting

      520 = 0*223+\frac{1}{2} *a*223^2\\ \\ a = 0.0209 m/s^2

Now we have v = u+at, where v is the final velocity

Substituting

        v = 0+0.0209*223 = 4.66 m/s

So final velocity of car b = 4.66 m/s

c) Acceleration = 0.0209 m/s^2

7 0
3 years ago
The diagram shows movement of thermal energy. At bottom a fire has red curved lines labeled Y with arrowheads pointing upward to
Pepsi [2]

Answer:

X and Z

Explanation:

Conduction occurs through direct physical contact.  Heat transferred from the pot to the handle, and from the handle to the hand, are both examples of conduction.

6 0
2 years ago
Read 2 more answers
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