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4 years ago

Low-pressure areas tend to before occluding and after occluding

Physics

1 answer:

alexgriva [62]4 years ago

4 0

Out of the following given choices;

a. decelerate; accelerate

b. accelerate; decelerate

c. stop; accelerate

d. decelerate; stop

<span>The answer is B. This is because the cold front usually rotates around the warm front as cold air mass coverage in the low-pressure system. This causes the cold air mass to accelerate to catch up with the warm front. When an occluded front ( that is the boundary that separates the older cool air mass already in place from new incoming cold air mass), that the system decelerates. </span>

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3 years ago

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Answer:

Average acceleration, $a=2.77\ m/s^2$

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4 years ago

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horsena [70]

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8 0

4 years ago

The maximum speed of a mass m on an oscillating spring is vmax . what is the speed of the mass at the instant when the kinetic a

umka21 [38]

Let
A = the amplitude of vibration
k = the spring constant
m = the mass of the object

The displacement at time, t, is of the form
x(t) = A cos(ωt)
where
ω = the circular frequency.

The velocity is
v(t) = -ωA sin(ωt)

The maximum velocity occurs when the sin function is either 1 or -1.
Therefore
$v_{max} = \omega A$
Therefore
$v(t) = -V_{max} sin(\omega t)$

The KE (kinetic energy) is given by
$KE = \frac{m}{2}v^{2} = \frac{m}{2} V_{max}^{2} sin^{2} (\omega t)$

The PE (potential energy) is given by
$PE = \frac{k}{2} x^{2} = \frac{k}{2} A^{2} cos^{2} (\omega t)$

When the KE and PE are equal, then
$v^{2} = \frac{k}{m} A^{2} cos^{2} (\omega t)$

For the oscillating spring,
$\omega ^{2} = \frac{k}{m} \\ V_{max} = \omega A = \sqrt{ \frac{k}{m} } A$
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$v^{2} = \frac{k}{m} \frac{m}{k} V_{max}^{2} cos^{2} ( \sqrt{ \frac{k}{m} t} ) \\ v = V_{max} \,cos( \sqrt{ \frac{k}{m} t} )$

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4 years ago

Low-pressure areas tend to ____ before occluding and ____ after occluding

Low-pressure areas tend to before occluding and after occluding