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Cerrena [4.2K]
3 years ago
8

How to tell cold fronts from warm fronts

Physics
1 answer:
Vika [28.1K]3 years ago
3 0
If cold air is advancing into warm air, a cold front is present. On the other hand, if a cold air mass is retreating and warm air is advancing, a warm front exists. Otherwise, a stationary front is present if the cold air is neither advancing nor retreating from the warm air mass.

This it’s from a textbook so it might be copyright so re-word (please mark me has brainlyest)
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What makes a hypothesis testable
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<span>A hypothesis is testable when you can create an experiment to study the proposition contained within the hypothesis. For example, the hypothesis ‘Santa travels slower than a unicorn’ is testable in theory by measuring the speeds of both, but it is not truly testable because neither exists in reality.</span>
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3 years ago
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Water at 20oC flows through a long elliptical duct 30 cm wide and 22 cm high. What average velocity, in m/s, would cause the wei
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Explanation:

The given data is as follows.

   Fluid is water so, density \rho = 1000 kg/m^{3}

  Weight flow rate = 500 lbf/s = 2224.11 N/sec

  Cross-sectional area (A) = \pi \times \frac{30}{2} \times \frac{22}{2}

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Hence, weight flow rate will be given as follows.

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            = 4.373 m/s

Thus, we can conclude that average velocity in the given case is 4.373 m/s.

8 0
3 years ago
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According to the third law of motion when we put an object, the object pushes back on us with an equal and opposite force. If th
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Hi pupil Here's Your answer :::


➡➡➡➡➡➡➡➡➡➡➡➡➡


Student's justification is not correct. Two equal and opposite force cancel each other if the act on the same body. According to the third law of motion action and reaction forces are equal and opposite but they both act on different bodies. Hence, they cannot cancel each other.

When we push a message track, then the applied force on the truck is not sufficient to overcome the force of friction between the tyres of truck and ground, hence, truck does not move.


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Hope this helps .
5 0
3 years ago
At a baseball game, the batter hit a fly ball at time t = 0 s. The outfielder caught the ball at t = 5.8 s. When was the ball at
Agata [3.3K]
We have the following equation for height:
 h (t) = (1/2) * (a) * t ^ 2 + vo * t + h0
 Where,
 a: acceleration
 vo: initial speed
 h0: initial height.
 The value of the acceleration is:
 a = -g = -9.8 m / s ^ 2
 For t = 0 we have:
 h (0) = (1/2) * (a) * 0 ^ 2 + vo * 0 + h0
 h (0) = h0
 h0 = 0 (reference system equal to zero when the ball is hit).
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 h (5.8) = (1/2) * (- 9.8) * (5.8) ^ 2 + vo * (5.8) + 0
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 h (t) = (1/2) * (a) * t ^ 2 + vo * t + h0
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 Rewriting:
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 The maximum height occurs when:
 h '(t) = -9.8 * t + 28.42
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 t = 28.42 / 9.8
 t = 2.9 seconds.
 Answer:
 
The ball was at maximum elevation when:
 
t = 2.9 seconds.
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