DIY Solar Panels

its more common in horses, but can dogs suck air to? because my 1.2yr old American Staghound x (yes a cross bred- no abuse please) seems to suck wind or air (and caugh) when he gets to excited or runs for to long, at first i thought it was parvo (as that theres an outbrake in my area) but none of my other dogs caugh or suck. they all have been tested and are up to date with their treatments and vac’s…
im going to call my vet on monday and see what she says, he did it when i first got him from a rescue home last year he was extreamly sick so naturally i thought it was to do with that, but now that he has continued im not to sure – he is in top condition aside from this wind/air sucking/caughing.
if it is that what can i do to help him or eliminate the problem? has anyone else had this problem?
again no harsh comments im just looking for some advice here before i call the vets

That sounds like a neurological problem.
Is it possible that it is reverse sneezing?
That is a repetitive honking some dogs do, and is a genetic trait.
The next time he does it, try pressing gently on both his sides just behind the ribcage to interrupt the impulses – if that works, it’s reverse sneezing.

It could also be a collapsing trachea or even embedded Foxtails (sharp grass seeds)- so have your vet check out his upper respiratory tract. In the meantime, use a harness and no collar to prevent further possible damage.

Please help! I have a test tomorrow and this part of the review and there is is a sub teaching my class. I know the answer is 16.5 m but I don’t know how to get there

?

A hurricane wind blows across a 7.00m x 16.6m flat roof at a speed of 140 km/hr.

A) Is the air pressure above the roof higher or lower than the pressure inside the house? Explain
B) What is the pressure difference?
C) How much force is exerted on the roof? If the roof cannot withstand this much force, will it "blow in" or "blow out"?

Thanks!!! much appreciated!!!!

we use Bernoulli’s principle to analyze this problem. Since the velocity of fluid is greater above the roof, the pressure is less above the roof; therefore, if the roof fails, the force will be coming from below, and the roof blows out

bernoulli’s principle gives us

P1 + 1/2 rho v1^2 + rho g h1 = P2 + 1/2 rho v2^2 + rho g h2

subscripts 1 refer to below the roof; 2 above the roof
rho= density of fluid = 1.29 kg/m^3
P=pressure
v=velocity of fluid
h= height of fluid; we will assume the roof is sufficiently thin that we can set h1=h2, so the two terms sum to zero

we have

P1 +0 + rho g h1 = P2 + 1/2 rho v^2 + rho gh 2

P1-P2 = 1/2 rho v2^2 = 1/2 x 1.29 kg/m^3 x (38.9m/s)^2 = 975N/m^2 (convert v to m/s)

therefore, force = pressure x area = 975N/m^2 x 7mx16.6m = 1.13×10^5N

I need some advice from some experianced sailors. I own a 1976 Ericson 29 in excellent condition. It is the tall rig, 38ft mast, full modified keel, displacement is 8500lbs,Draft is 4’3" I plan on going with a partner but I’m not going to deny that I could be solo, plan to use a MR. VEE windvane for steering. IS THIS BOAT DESIGN STRONG ENOUGH TO HANDLE ROUGH SEAS? I don’t want to destroy it or damage it. What additional equipment will I need to add? I was planning to have minimal electronics and sail like a purist as much as possible, except I will have GPS, backup GPS, sexton, compass, charts and the usuals. I will also have a good battery bank with one 90 watt solar panel and probably and air x marine wind genny. I don’t have radar, can’t afford it, don’t have mechanical auto pilot

First to answer your question, absolutely. Now for the flaws. You state that the boat has a full modified keel. All the Ericson 29′s I’ve seen have fin keel and spade rudder as in the picture below. I’d choose a boat with a shoal keel rather than the fin keel and spade rudder. That rig works great for racing and day sailing, but what makes it good for those uses makes it more difficult for crossings. The rig allows it to turn abruptly. When crossing slow gradual turns are better for maintaining helm control. They’re also better because the keel protects the rudder when it’s attached to the back. A spade rudder and fin keel are easier to snap off in a collision or grounding. The other things you’ll need to do is to study weather conditions and cross when conditions are most favorable. Winter on the North Atlantic is a bear. You’ll also probably want to avoid hurricanes and other storms. Finally because you say you won’t have radar you’ll want to know the primary shipping lanes so that you can stay out of them. A fiberglass sailboat has a very small radar signature. If you plan on sleeping on the crossing you don’t want to be waken up by a freighter, tanker or destroyer bearing down on you in the fog early in the morning. This will also hinder you in the event rescue is needed, but if you have a Satellite phone which you forgot to list but should be a mandatory requirement. You’ll also probably want a second long range radio, life raft and other emergency equipment on board. You always want to be prepared for the worst.

If you put proper advanced planning in place you shouldn’t have all that much trouble with your crossing. Good luck with it..

Let East represent the positive x axis and North the positive y axis. A constant wind with a speed of 10.0 km/hr is blowing due NE. You have a model airplane that flies at 30.0 km/hr in level flight in still air. You start its engine and release it in the direction due NW. What will be the displacement of the airplane relative to your position after it flies for 10.0 minutes?

Since you want distance we can convert the speeds to distance using d = v*t

so for the plane d1 = 30km/hr*10min/60min/hr = 5km

and for the wind d2 = 10.0km/hr*10/60 = 1.67km

Now the angle between these vectors is 90o so the displacement =

sqrt(5^2 + 1.67^2) = 5.27 km

and the angle is 45o + arctan(3/1.67) = 106o N of East

Is the air pressure above the roof higher or lower than the pressure inside the house? Explain. What is the pressure difference?

bernoulli’s equation
p+ev^2=constant

p at the top of the roof is reduced to compensate for the high velocity to keep p + ev^2/2 constant
p=1.01.10^5-[50^2]/2[pascals]

p below is higher

An airplane takes 7 hours to travel a distance of 4707km against the wind. The return trip takes 6 hours with the wind. What is the rate of the plane in still air and what is the rate of the wind?

x = still air. i think x is 728 idk how to get y please help or correct me if im wrong

Let the speed of wind be y and that of plane in still air x
While going against the wind, the speed of the plane will be reduced
Net speed while against the wind = x – y
Speed = Distance/time
x – y = 4707/7 …(1)

While going with the wind, the speed will enhance.
Net speed while going with the wind = x + y
x + y = 4707/6 …(2)

I will use elimination method to solve them
Add (1) and (2),
2x = 4707/7 + 4707/6
2x = 4707 (13/42)
x = 728.4 km/hr

Put this value of x in any of the equations above to get y
Putting in (2)
728.4 + y = 4707/6
y = 4707/6 – 728.4 = 56 km/hr

drum and B hears the sound t1 time after he sees the event.A and B interchange their positions and the experiment is repeated.this time B hears the drum t2 time after he sees the event.calculate the velocity of sound in still air v and the velocity of wind u.neglect the time light takes in travelling between the friends.

The beats are a unit of time.
t = d/v

In the first experiment, the wind is assisting the speed of sound. The total speed is v + u. Therefore the time = x / (v + u)
11 = x / (v + u)
11v + 11u = x

In the second experiment, the wind is going against the travel of direction. v – u
The equation you get is
12v – 12u = x

The two x’s are an equal distance So you can begin by equating the two distances.
12v – 12u = 11v + 11u
v = 23 u

Now you can go to one of the original equations.
12v – 12u = x
12*23u – 12u = x
12u (23 – 1) = x
12u * 22 = x
u = x / 264

v = 23*u
v = 23 * x/264

The scenario is parachute deployment, I am wondering if there is such device which can basically release a weight so that when my rocket is in the air the parachute can deploy on its own. I was thinking a timed release. I.e. can I wind something up for x amount of revolutions and after it stops spinning a string connected to a mass releases?

Many hobby rocket motors have a charge at the end away from the nozzle which pops after the rocket’s burnt out. This bust of pressure can be used to pop the nose cone off, allowing a parachute to deploy

Main difficulty with a mehanical timing method is making a spring powered shaft turn at a reliable/predicatable rate, the spring will get weaker as it unwinds. If that can be solved just use a worm drive to reduce the RPM, and use a cam to pull a toggle out.

There is a 100% mechanical method http://en.wikipedia.org/wiki/Fusee_%28horology%29
A more modern solution would involve some some very clever dilatant greases which allow free rotation up to a certain RPM, then above that speed they thicken up considerably effectively braking the shaft. Would work as a very compact speed limiter for a rotating shaft.
There’s always possibility of using a propeller on the rocket’s nose to drive a shaft, and "count" that, it’s more distance flown than time, and V1 flying bombs used this method to determine when to dive.

One form of the formula meteorologists use to calculate wind chill temperature (WC) is

WC = 35.74 + 0.6215t – 35.75s^0.16 + 0.4275ts^0.16

where x is the wind speed in mph and t is the actual air temp in degrees fahrenheit. Suppose temperature is constant at 15 degrees.

I have the answer to A:
A) Express wind chill as a function of wind speed s. WC=45.0625 – 29.3375s^0.16

Now I need help with B:
B) Find the rate of change of wind chill with respect to wind speed when the wind speed is 25 mph.

Thanks in advance!

You need to take the derivative of WC

derivative of WC = (0.16)(- 29.3375) s ^ (0.16 – 1)

Then plug in s = 25 and solve.