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3 years ago

AIR MASSES!!

Physics

1 answer:

777dan777 [17]3 years ago

5 0

Answer:

A : hot and moist, maritime tropical

B: cold and dry, maritime polar

C: hot and moist , maritime tropical

D: cold and dry, continental polar

E: hot and moist , maritime tropical

F: cold and dry , maritime polar

Explanation:

Cold air is denser than warm air. The more water vapor that is in the air, the less dense the air becomes. That is why cold, dry air is much heavier than warm, humid air.

Maritime polar (mP) air masses are cool, moist, and unstable. Some maritime polar air masses originate as continental polar air masses over Asia and move westward over the Pacific, collecting warmth and moisture from the ocean.

Maritime tropical (mT) air masses are warm, moist, and usually unstable.

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A mass weighing 32 pounds stretches a spring 2 feet. Determine the amplitude and period of motion if the mass is initially relea

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A mass weighing 32 pounds stretches a spring 2 feet.

(a) Determine the amplitude and period of motion if the mass is initially released from a point 1 foot above the equilibrium position with an upward velocity of 6 ft/s.

(b) How many complete cycles will the mass have completed at the end of 4 seconds?

Answer:

$A = 1.803 ft$

Period = $\frac{\pi}{2}$ seconds

8 cycles

Explanation:

A mass weighing 32 pounds stretches a spring 2 feet;

it implies that the mass (m) = $\frac{w}{g}$

m= $\frac{32}{32}$

= 1 slug

Also from Hooke's Law

2 k = 32

k = $\frac{32}{2}$

k = 16 lb/ft

Using the function:

$\frac{d^2x}{dt} = - 16x\\\frac{d^2x}{dt} + 16x =0$

$x(0) = -1$ (because of the initial position being above the equilibrium position)

$x(0) = -6$ ( as a result of upward velocity)

NOW, we have:

$x(t)=c_1cos4t+c_2sin4t\\x^{'}(t) = 4(-c_1sin4t+c_2cos4t)$

However;

$x(0) = -1$ means

$-1 =c_1\\c_1 = -1$

$x(0) =-6$ also implies that:

$-6 =4(c_2)\\c_2 = - \frac{6}{4}$

$c_2 = -\frac{3}{2}$

Hence, $x(t) =-cos4t-\frac{3}{2} sin 4t$

$A = \sqrt{C_1^2+C_2^2}$

$A = \sqrt{(-1)^2+(\frac{3}{2})^2 }$

$A=\sqrt{\frac{13}{4} }$

$A= \frac{1}{2}\sqrt{13}$

$A = 1.803 ft$

Period can be calculated as follows:

= $\frac{2 \pi}{4}$

= $\frac{\pi}{2}$ seconds

How many complete cycles will the mass have completed at the end of 4 seconds?

At the end of 4 seconds, we have:

$x* \frac{\pi}{2} = 4 \pi$

$x \pi = 8 \pi$

$x=8$ cycles

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3 years ago

A 1.60-kg object is held 1.05 m above a relaxed, massless vertical spring with a force constant of 330 N/m. The object is droppe

pentagon [3]

Answer:

(A) l = 0.39 m

(B) l =0.38 m

(C) l = 0.14 m

Explanation:

Answer:

Explanation:

Answer:

Explanation:

from the question we are given the following values:

mass (m) = 1.6 kg

height (h) = 1.05 m

compression of spring (l) = ?

spring constant (k) = 330 N/m

acceleration due to gravity (g) = 9.8 m/s^{2}

(A) initial potential energy of the object = final potential energy of the spring

potential energy of the object = mg(1.05 + l)

potential energy of the spring = 0.5 x k x l^{2} (k= spring constant)

therefore we now have

mg(1.05 + l) = 0.5 x k x l^{2}

1.6 x 9.8 x (1.05 + l) = 0.5 x 300 x l^{2}

15.68 (1.05 + l) = 150 x l^{2}

16.5 + 15.68l = 150l^{2}

l = 0.39 m

(B) with constant air resistance the equation applied in part A above becomes

initial P.E of the object - air resistance = final P.E of the spring

mg(1.05 + l) - 0.750(1.05 + l) = 0.5 x k x l^{2}

1.6 x 9.8 x (1.05 + l) - 0.750(1.05 + l) = 0.5 x 300 x l^{2}

(16.5 + 15.68l) - (0.788 + 0.75l) = 150l^{2}

16.5 + 15.68l - 0.788 - 0.75l = 150l^{2}

15.71 + 14.93l = 150^{2}

l =0.38 m

(C) where g = 1.63 m/s^{2} and neglecting air resistance

the equation mg(1.05 + l) = 0.5 x k x l^{2} now becomes

1.6 x 1.63 x (1.05 + l) = 0.5 x 300 x l^{2}

2.608 (1.05 +l) = 0.5 x 300 x l^{2}

2.74 + 2.608l = 150 x l^{2}

l = 0.14 m

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