Physics question Water flowing through a garden hose of diameter 2.78 cm fills a 22.0-L bucket in 1.20 min.?
Flow rate = velocity * cross-section area Flow rate = volume/time = 22,000 cm^3 / 1.2 min = 18,333.33 cm^3 / min Cross-Section = Pi * r^2 = 3.14... * (2.78/2)^2 = 6.07 cm^2 18,333.33 cm^3/min = 6.07 cm^2 * V V = 3020 cm/min Assuming the nozzle does not reduce the flow rate, work b the same way.
Physics Question: Water flowing through a garden hose of diameter 2.73 cm fills a 23.0-L bucket in 1.30 min.?
(a) What is the speed of the water leaving the end of the hose? _____m/s (b) A nozzle is now attached to the end of the hose. If the nozzle diameter is one-third the diameter of the hose, what is the speed of the water leaving the nozzle? ________m/s
Why does end of hose with flowing water stick to flat surface?
So I just now tried this with a garden hose, and got no "sticking" effect at all. The flat surface (I used a plastic kitchen plate) was just pushed away. If the effect does occur, it may relate to the OD/ID ratio of the pipe at the open end which, for a garden hose, may be like 0.8/0.7 (each term in inches). Perhaps if this ratio is much larger, so that the water flow velocity >decelerates< dramatically as the water flows out radially, there may be a net suction effect?Oops, I just reread your question, and I do see that you referred to some kind of flange. I'll repeat the experiment, but it requires some prep. (Actually, I now rather like the experiment!)
Water in the hose has a speed of 0.95 m/s, at what speed does it leave the sprinkler holes?
The volume of water passing through the hose must equal the volume of water leaving the sprinkler. Use the volume of a cylinder. s = the speed from the sprinkler Vh = 0.95 * pi * (1.5 / 2) ^ 2 = 0.534375 * pi Vs = s * 21 * pi * (0.19 / 2) ^ 2 = 0.189525 * s * pi Vh = Vs 0.534375 * pi = 0.189525 * s * pi 0.534375 = 0.189525 * s 0.534375 / 0.189525 = s 2.82 = s
Physics Problem about a water hose?
I used two of the kinematic equations to solve this problem: d=vt+½at² and d=rt First, I will use d=rt and we are going to work with the distance it travels horizontally. 4.1=rt since it is at an angle, it will be: 4.1=cos21(rt) Now, for the vertical motion: Since the hose is 1.4m above the ground, it will end up -1.4 m from the starting point. -1.4=rt - 4.905t² In this case, r= sin21r -1.4=sin21(rt) - 4.905t² You know have 2 equations and 2 variables. Solving this step by step is going to be challenging, due to the lack of characters on this keyboard. You can start by finding the value of t. T= 4.1/[cos21(v)] substitute this value into the second equation we came out with. If you had algebra 1, you should be able to solve it. the answer should come out to be about 5.64 m/s
Physics Question- Filling a Pool?
First of all, let's find out how much water you need to fill a 5.62 m diameter pool to a depth of 1.44 m You end up with a cylinder of water of height (h) = 1.44 m and radius r = (5.62 / 2) = 2.81 m {radius is half the diameter} Volume (V) of a cylinder is given by: V = πr²h Substituting in your values: V = π * 2.81² * 1.44 V = 35.72 cubic metres Now we know how much water we need, let's find out how quickly it comes out of the hose. It's a bit mean giving you the hose diameter in inches when the rest of the problem is in metric! We need to convert 3/8 inch to metric. 1 inch = 25.4 mm, so 3/8 inch = 3 * 25.4 / 8 = 9.525 mm {That's a pretty small diameter hose. I hope it really is the diameter and not the radius} So the radius of the hose = (9.525 / 2) = 4.7625 mm The water is travelling along the pipe at a speed of 0.218 m/s, so in one second a cylinder of water of radius (r = 4.7625 mm) and height (h = 0.218 m) comes out of the end. What is the volume of that? We need to be careful with units and work in metres, not mm. 4.7625 mm = 0.0047625 m V = πr²h V = π * 0.0047625² * 0.218 V = 1.55 * 10^-5 cubic metres each second {0.00000155 cubic metres/s} So how long will it take to supply the 35.72 m³ needed to fill the pool if your hose supplies 1.55 * 10^-5 m³ per second? 35.72 / (1.55 * 10^-5) = 2299509 seconds There are 3600 seconds in an hour, so thet's (2299509 /3600) = 638.75 hours There are 24 hours in a day, so that's: (638.75 / 24) = 26.6 days
Can metal pipes be kinked up like a water hose/garden hose?
As far as copper line is concerned, there are to types; soft and hard. Hard pipe will kink immediatly if u bend it. Soft copper can bbe bent with a slide-spring tool. If you find a kink in any line,it needs repaired
How do I make water to directly flow up in a water hose (syphoning doesn't work)?
Try connecting a 3/4” direct discharge centrifugal pump such as ‘little giant’ brand.
A water hose 2 cm in diameter is used to fill a 20-L bucket. If it takes 1 min to fill the bucket, what is the?
That means the volume flow is 20 L/min. 20 L = 20000 cm³ dividing by the cross-sectional area will give us the length. Area is πr² = π1² = 3.14 cm² 20000 cm³ / 3.14 cm² = 6370 cm so the speed is 6370 cm/min or 106 cm/s
Why does a water hose swing around when you let go of it?
There are two obvious forces acting on the hose, and they are usually in an inverse ratio.One is the pressure inside the hose, trying to make a long, floppy cylinder into a rigid one. Depending in the type of hose material, it’s tensile strength and rigidly it will basically stiffen under the pressure and weight of thevwater inside it.The other is the release of the pressure in one direction. When you open the valve on a hose, it acts like a rocket nozzle to some degree. A garden hose will push back on your palm, and kinda have a tendency to wander of you don’t hold it steady. A fire hose will require a qualified person, or two or three, to wrangle and direct an open hose because of the amount of pressure being released.So, back to your question. When you snap shut the valve, the hose immediately tries to reinforce its previously more rigid state and in places where less internal pressure allowed a tighter bend will be try to return the the least bent position in can achieve, barring friction with the ground and collapsing under its own weight.