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Svet_ta [14]
3 years ago
8

Pressure is about 1000 hPa at sea level and about 500 hPa at an altitude of 5.5 km. Why doesn’t this vertical pressure gradient

cause permanent upward acceleration and motion?A. The upward pressure gradient force is balanced by gravity.B. The upward pressure gradient force is balanced by friction.C. The upward pressure gradient force is balanced by the Coriolis force.D. The upward pressure gradient force is too small to have any noticeable effect.E. Upward motion is prevented by the temperature inversion.
Physics
1 answer:
saul85 [17]3 years ago
6 0

Answer:

A. The upward pressure gradient force is balanced by gravity.

Explanation:

A. is correct because the pressure difference is actually generated by gravity. As in the following formula for the pressure at different points:

p = p_0 + \rho g h

where p, p_0 are the pressure at 2 points, ρ is the density of the fluid, g is the gravitational constant, and h is the height difference.

B is incorrect because friction in air is too small to make an effect.

C is incorrect because the Coriolis force is horizontal, not vertical.

D is incorrect because a difference of 500 hPa = 50000 Pa, this is half of the atmospheric pressure.

E is incorrect because temperature cannot generate force.

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A car drives past a pole at 40km/hr. Describe the motion from the point of view of a) the car, and b) the pole. Thanks in advanc
ki77a [65]
I was going to beg off until tomorrow, but this one is nothing like those others.
Why, at only 40km/hr, we can ignore any relativistic correction, and just go with Newton.

To put a finer point on it, let's give the car a direction.  Say it's driving North.

a).  From the point of view of the car, its driver, and passengers if any,
the pole moves past them, heading south, at 40 km/hour .

b).  From the point of view of the pole, and any bugs or birds that may be
sitting on it at the moment, the car and its contents whiz past them, heading
north, at 40 km/hour.

c).  A train, steaming North at 80 km/hour on a track that exactly parallels
the road, overtakes and passes the car at just about the same time as
the drama in (a) and (b) above is unfolding.

The rail motorman, fireman, and conductor all agree on what they have
seen. From their point of view, they see the car moving south at 40 km/hr,
and the pole moving south at 80 km/hr.

Now follow me here . . .

The car and the pole are both seen to be moving south.  BUT ... Since the
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That's what everybody on the train sees.

==============================================

Finally ... since you posed this question as having something to do with your
fixation on Relativity, there's one more question that needs to be considered
before we can put this whole thing away:

You glibly stated in the question that the car is driving along at 40 km/hour ...
AS IF we didn't need to know with respect to what, or in whose reference frame.
Now I ask you ... was that sloppy or what ? ! ? 

Of course, I came along later and did the same thing with the train, but I am
not here to make fun of myself !  Only of others.

The point is . . . the whole purpose of this question, obviously, is to get the student accustomed to the concept that speed has no meaning in and of itself, only relative to something else.  And if the given speed of the car ...40 km/hour ... was measured relative to anything else but the ground on which it drove, as we assumed it was, then all of the answers in (a) and (b) could have been different.

And now I believe that I have adequately milked this one for 50 points worth.


7 0
3 years ago
A 1000 kg car moving at 108 km/h jams on its brakes and comes to a stop. How much work was done by friction?
Nostrana [21]

Answer:

The work done by friction was -4.5\times10^{5}\ J

Explanation:

Given that,

Mass of car = 1000 kg

Initial speed of car =108 km/h =30 m/s

When the car is stop by brakes.

Then, final speed of car will be zero.

We need to calculate the work done by friction

Using formula of work done

W=\Delta KE

W=K.E_{f}-K.E_{i}

W=\dfrac{1}{2}mv_{f}^2-\dfrac{1}{2}mv_{f}^2

Put the value of m and v

W=0-\dfrac{1}{2}\times1000\times(30)^2

W=-450000
\ J

W=-4.5\times10^{5}\ J

Hence, The work done by friction was -4.5\times10^{5}\ J

3 0
3 years ago
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Allisa [31]
<span> B. A person moving a ball through a stream of water</span>
5 0
3 years ago
In a game of egg-toss, you and a partner are throwing an egg back and forth trying not to break it. Given your knowledge of mome
lutik1710 [3]
F=dP/dt.  So you want the momentum to change as slowly as possible in time to minimize the force.  So as you catch the egg, let your hand move backward with it for awhile, slowly bringing it to a stop.  If you hold your hand steady when you catch it the force due to the impact could break it.
3 0
3 years ago
* CRIMINOLOGY*
Setler [38]

I think the answer is B.

Hope this helps.

5 0
3 years ago
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