<span>Maritime tropical air masses develop over warm waters present in the tropics and Gulf of Mexico, where heat and moisture are carried to to the overlying air from the water below.
</span><span>
</span><span> Tropical air masses having northward movement carry warm moist air into the United States, thus increasing the potential for condensation. Generally the southern states experience tropical air masses. But, in winter season, southerly winds ahead of migrating cyclones <span>sometimes transport tropical air mass towards north.
</span></span><span><span>
</span></span><span><span>The counterclockwise winds related to northern hemisphere mid latitude cyclones play an important role in the movement air masses, carrying warm moist air towards north ahead of a low while dragging colder and drier air towards south.</span></span>
Answer:
<u><em>The truck was moving 16.5 m/s during the time it took to stop, which was 3 seconds. </em></u>
- <u><em>Initial velocity = 33 m/s</em></u>
- <u><em>Final velocity = 0 m/s</em></u>
- <u><em>Average velocity = (33 + 0) / 2 m/s = 16.5 m/s</em></u>
Explanation:
- <u><em>First, how long does it take the truck to come to a complete stop?</em></u>
- <u><em>( 33 m/s ) / ( 11 m / s^2 ) = 3 seconds</em></u>
- <u><em>Then we can look at the average velocity between when the truck started decelerating and when it came to a complete stop. Because the deceleration is constant (always 11m/s^2) we can use this trick.</em></u>
Mollusks live
in fresh water, in marine environment, but also on land.
<span>Echinoderms
live only in water. This is the reason why they have developed different
systems for moving. Mollusks have only singular muskullus foot for walking and
Echinoderms have tube feet which they use for moving, as well for collecting
and transporting food to their mouth. </span>
<span>6.20 m/s^2
The rocket is being accelerated towards the earth by gravity which has a value of 9.8 m/s^2. Given the total mass of the rocket, the gravitational drag will be
9.8 m/s^2 * 5.00 x 10^5 kg = 4.9 x 10^6 kg m/s^2 = 4.9 x 10^6 N
Add in the atmospheric drag and you get
4.90 x 10^6 N + 4.50 x 10^6 N = 9.4 x 10^6 N
Now subtract that total drag from the thrust available.
1.250 x 10^7 - 9.4 x 10^6 = 12.50 x 10^6 - 9.4 x 10^6 = 3.10 x 10^6 N
So we have an effective thrust of 3.10 x 10^6 N working against a mass of 5.00 x 10^5 kg. We also have N which is (kg m)/s^2 and kg. The unit we wish to end up with is m/s^2 so that indicates we need to divide the thrust by the mass. So
3.10 x 10^6 (kg m)/s^2 / 5.00 x 10^5 kg = 0.62 x 10^1 m/s^2 = 6.2 m/s^2
Since we have only 3 significant figures in our data, the answer is 6.20 m/s^2</span>
The spring will come to rest 4.9 m below the natural length
Explanation:
The mass-spring system will come to rest when the restoring force on the spring (pulling upward) balances the weight of the mass (pulling downward). Mathematically, this can be written as

where
k is the spring constant
x is the elongation of the spring
m is the mass
g is the acceleration of gravity
In this problem, we have:
is the mass
is the acceleration of gravity
is the spring constant
Solving the equation for x,

Therefore, the spring will come to rest 4.9 m below the natural length.
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