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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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1. John has to hit a bottle with a ball to win a prize. He throws a 0.4 kg ball with a velocity of 18 m/s. It hits a 0.2 kg bott
Kobotan [32]
 0.4 x 18 = 7.2 kg m/s

The momentum of the bottle after being hit is 0.2 x 25 = 5 kg m/s

7.2 - 5 = 2.2 kg m/s is the motmentum of the ball now 

 the velocity  is 2.2/0.4 = 5.5 m/s
7 0
3 years ago
La velocidad de un tren se reduce uniformemente de 12m/s a 5m/s. Sabiendo que durante ese tiempo recorre una distancia de 100 m.
MA_775_DIABLO [31]

Responder:

Explicación:

Dados los siguientes datos

Valor inicial u = 12 m / s

velocidad final v = 5 m / s

Distancia S = 100 m

Necesario

aceleración del tren.

Usando la ecuación de movimiento

v² = u² + 2as

2as = v²-u²

a = v²-u² / 2s

Sustituyendo los valores dados para obtener la aceleración que tenemos;

a = 5²-12² / 2 (100)

a = 25-144 / 200

a = -119/200

a = -0,595 m / s²

Por tanto, la aceleración del tren durante este período es de -0,595 m / s²

b) Si el tren viaja a una parada desde 5 m / s, su velocidad final será cero y su velocidad inicial u será 5 m / s

Para obtener la distancia durante este período, sustituiremos u = 5 m / s, v = 0 m / sy a = -1,19 m / s² en la ecuación de movimiento anterior;

v² = u² + 2as

0² = 5²-2 (0.595) s

0 = 25-1,19 s

1,19 s = 25

s = 25 / 1,19

Responder:

Explicación:

Dados los siguientes datos

Valor inicial u = 12 m / s

velocidad final v = 5 m / s

Distancia S = 100 m

Necesario

aceleración del tren.

Usando la ecuación de movimiento

v² = u² + 2as

2as = v²-u²

a = v²-u² / 2s

Sustituyendo los valores dados para obtener la aceleración que tenemos;

a = 5²-12² / 2 (100)

a = 25-144 / 200

a = -119/200

a = -0,595 m / s²

Por tanto, la aceleración del tren durante este período es de -0,595 m / s²

b) Si el tren viaja a una parada desde 5 m / s, su velocidad final será cero y su velocidad inicial u será 5 m / s

Para obtener la distancia durante este período, sustituiremos u = 5 m / s, v = 0 m / sy a = -1,19 m / s² en la ecuación de movimiento anterior;

v² = u² + 2as

0² = 5²-2 (0.595) s

0 = 25-1,19 s

1,19 s = 25

s = 25 / 1,19

s = 21,0 m

Por lo tanto, la distancia que recorre hasta detenerse asumiendo la misma aceleración es 21.0 m

6 0
3 years ago
If a roller coaster train has a potential energy of 1,500 J and a kinetic energy of 500 J as it starts to travel downhill, what
DENIUS [597]

Its total mechanical energy is <em>2,000 J</em>.

We don't have enough information to say anything about its heat energy, its chemical energy, or the energy due to any electrical charge it may be carrying or any magnetic field it may have.

6 0
2 years ago
Select the correct answer.
Stels [109]

Answer:

An electric bell is placed inside a transparent glass jar. The bell can be turned on and off using a switch on the outside of the jar. A vacuum is created inside the jar by sucking out the air. Then the bell is rung using the switch. What will we see and hear?

A.

We’ll see the bell move, but we won’t hear it ring.

B.

We won’t see the bell move, but we’ll hear it ring.

C.

We’ll see the bell move and hear it ring.

D.

We won’t see the bell move or hear it ring.

E.

We’ll see the sound waves exit the vacuum pump.

Explanation:

so, the answer to the question is

A.

We'll see the bell move, but we won’t hear it ring.

5 0
3 years ago
Read 2 more answers
What frequency fapproach is heard by a passenger on a train moving at a speed of 18.0 m/s relative to the ground in a direction
Sergio039 [100]

Answer:

The frequency is 302.05 Hz.

Explanation:

Given that,

Speed = 18.0 m/s

Suppose a train is traveling at 30.0 m/s relative to the ground in still air. The frequency of the note emitted by the train whistle is 262 Hz .

We need to calculate the frequency

Using formula of frequency

f'=f(\dfrac{v+v_{p}}{v-v_{s}})

Where, f = frequency

v = speed of sound

v_{p} = speed of passenger

v_{s} = speed of source

Put the value into the formula

f'=262\times(\dfrac{344+18}{344-30})

f'=302.05\ Hz

Hence, The frequency is 302.05 Hz.

7 0
3 years ago
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