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professor190 [17]
3 years ago
10

Which prediction of weather is the most accurate for the next two days if there is an occluded front over the area?

Physics
1 answer:
alisha [4.7K]3 years ago
6 0

The most accurate weather for the next two days would be dry warm weather or severe thunderstorms if there is an occluded front over the area.

<u>Explanation:</u>

The weather front that is created during the cyclogenesis process is an occluded front in meteorology. Cyclogenesis process is the development of extra-tropical cyclone and its intensification.

During the occurrence of this, the warm air is occluded (separated) from the center of cyclone at the surface of the earth.

The cold front rotates the storm as it intensifies and holds up the warm front. This develops an occluded front, that is the boundary which separates the new cold air mass and the older cool air mass that is already in warm front's north.

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Answer: The phosphoric acid in cola helps to break down the meat.

Explanation: The acid inside of the Cola breaks down the meat, making it softer. On top of this, the sweetness adds flavor. The phosphates helps in maintaining the pH and buffer properties of the meat, causes emulsification of fats, thereby making it softer.

8 0
3 years ago
Read 2 more answers
2. Two projectiles thrown from the same point at angles 0,=60° and 02=30° with the horizontal,
Liono4ka [1.6K]

Answer:

The answer is below

Explanation:

The maximum height (h) of a projectile with an initial velocity of u, acceleration due to gravity g and at an angle θ with the horizontal is given as:

h=\frac{u^2sin^2\theta}{2g}

Given that the two projectile has the same height.

For\ the\ first\ projectile\ with\ an\ angle\ \60^0 \ with\ the\ horizontal\ and initial\ velocity\ u_1:\\\\h=\frac{u_1^2sin^260}{2g} =\frac{0.75u_1^2}{2g} \\\\For\ the\ second\ projectile\ with\ an\ angle\ \30 \ with\ the\ horizontal\ and initial\ velocity\ u_2:\\\\h=\frac{u_2^2sin^230}{2g} =\frac{0.25u_2^2}{2g}\\\\\frac{0.25u_2^2}{2g}=\frac{0.75u_1^2}{2g}\\\\0.25u_2^2=0.75u_1^2\\\\\frac{u_1^2}{u_2^2} =\frac{0.25}{0.75} \\\\\frac{u_1^2}{u_2^2}=\frac{1}{3} \\\\\frac{u_1}{u_2}=\sqrt\frac{1}{3}

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3 years ago
A jet lands on an aircraft carrier at 140 mi/h. What is its acceleration if it stops in 2.0 seconds?
sergejj [24]

Answer:

Acceleration, a=-31.29\ m/s^2

Explanation:

It is given that,

Initial speed of the aircraft, u = 140 mi/h = 62.58 m/s

Finally, it stops, v = 0

Time taken, t = 2 s

Let a is the acceleration of the aircraft. We know that the rate of change of velocity is called acceleration of the object. It is given by :

a=\dfrac{v-u}{t}

a=\dfrac{0-62.58}{2}

a=-31.29\ m/s^2

So, the acceleration of the aircraft is 31.29\ m/s^2 and the car is decelerating. Hence, this is the required solution.

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