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Millions of dollars of funding are available to our district and potentially our school from the Federal Government through this funding source. The funding may be used for the purchase of books, materials, technology, personnel, capital improvement (i.e. buildings, etc.), etc.
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subset: commoncrawl


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Aug 28, 2017
Made with ACID Music Studio 9.0, The second of three of the I64 series as voiced in his Ford Fiesta!
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At the Mardi Gras Carnival in New Orleans coloring page
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Some rights reserved. This work is licensed under a Creative Commons Attribution-NonCommercial 4.0 License.
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Author: Derek Wyatt
Open Water Judges course
Team Managers Module 1
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5 Questions to Ask When Choosing Lodging Options for Wedding Guests
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A PhD is the highest degree anyone can obtain in an academic institution. It is a more refined and in-depth study of a certain academic discipline that makes a PhD graduates an authority figure in his or her own field of expertise. However, before being an authority figure requires a graduation first to receive the PhD diploma, and academic regalia for the PhD are required to be worn during that prestigious event.
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F mg (10.1 kg)(9.80 m/s ) m
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1 Week 9 homework IMPORTANT NOTE ABOUT WEBASSIGN: In the WebAssign versions of these problems, various details have been changed, so that the answers will come out differently. The method to find the solution is the same, but you will need to repeat part of the calculation to find out what your answer should have been. WebAssign Problem : A 0.-kg uniform board is wedged into a corner and held by a spring at a 50.0 angle, as the drawing shows. The spring has a spring constant of 76 N/m and is parallel to the floor. Find the amount by which the spring is stretched from its unstrained length. REASONING AND SOLUTION In order to find the amount the spring stretches we need to calculate the force that acts on the spring. The magnitude of this force is F. Since the board is in equilibrium, the net torque acting on it is zero. Taking the axis of rotation to be at the corner and assuming the board has a length L, the net torque is Στ FL sin 50.0 mg (L/) cos Solving for F gives mg cos 50.0 mg F sin 50.0 tan 50.0 The amount x by which the spring stretches is equal to the magnitude F of the force applied to it divided by the spring constant k (see Equation 0.). Thus, F mg (0. kg)(9.80 m/s ) x k k tan 50.0 (76 N/m) tan m WebAssign Problem : Concept Simulation 0.3 at illustrates the concepts pertinent to this problem. An 0.80-kg object is attached to one end of a spring, as in Figure 0.6, and the system is set into simple harmonic motion. The displacement x of the object as a function of time is shown in the drawing. With the aid of these data, determine (a) the amplitude A of the motion, (b) the angular frequency ω (c) the spring constant k, (d) the speed of the object at t.0 s, and (e) the magnitude of the object s acceleration at t.0 s.
2 REASONING AND SOLUTION a. Since the object oscillates between ± m, the amplitude of the motion is m. 0.4, b. From the graph, the period is T 4.0 s. Therefore, according to Equation ω π T π.6 rad/s 4.0 s c. Equation 0. relates the angular frequency to the spring constant: ω k/ m. Solving for k we find k ω m (.6 rad/s) (0.80 kg).0 N/m d. At t.0 s, the graph shows that the spring has its maximum displacement. At this location, the object is momentarily at rest, so that its speed is v 0 m/s. e. The acceleration of the object at t.0 s is a maximum, and its magnitude is a max Aω (0.080 m)(.6 rad/s) 0.0 m/s WebAssign Problem 3: A 3.0-kg block is between two horizontal springs. Neither spring is strained when the block is at the position labeled x 0 m in the drawing. The block is then displaced a distance of m from the position where x 0 m and is released from rest. (a) What is the speed of the block when it passes back through the x 0 m position? (b) Determine the angular frequency ω of this system. REASONING AND SOLUTION a. When the block passes through the position x 0 m, its velocity is a maximum and can be found from Equation 0.8: v max Aω.
3 We can find the angular frequency ω from the following reasoning: When the mass is given a displacement x, one spring is stretched by an amount x, while the other is compressed by an amount x. The total restoring force on the mass is, therefore, F x k x k x (k + k )x Comparison with Equation 0. shows that the two-spring system has an effective spring constant of k eff k + k. Thus, from Equation 0. ω k k + k m m eff Combining this with Equation 0.8 we obtain v max k + k 650 N/m N/m A (0.070 m).3 m/s m 3.0 kg b. The angular frequency of the system is ω k + k 650 N/m N/m m 3.0 kg 9 rad/s WebAssign Problem 4: Concept Simulation 0. at allows you to explore the concepts to which this problem relates. A.00-kg object is hanging from the end of a vertical spring. The spring constant is 50.0 N/m. The object is pulled 0.00 m downward and released from rest. Complete the table below by calculating the translational kinetic energy, the gravitational potential energy, the elastic potential energy, and the total mechanical energy E for each of the vertical positions indicated. The vertical positions h indicate distances above the point of release, where h 0 m. h PE PE (meters) KE (gravity) (elastic) E REASONING AND SOLUTION If we neglect air resistance, only the conservative forces of the spring and gravity act on the ball. Therefore, the principle of conservation of mechanical energy applies.
4 When the.00 kg object is hung on the end of the vertical spring, it stretches the spring by an amount y, where F mg (.00 kg)(9.80 m/s ) y 0.39 m k k 50.0 N/m (0.) This position represents the equilibrium position of the system with the.00-kg object suspended from the spring. The object is then pulled down another 0.00 m and released from rest ( v 0 0 m/s). At this point the spring is stretched by an amount of 0.39 m m 0.59 m. This point represents the zero reference level ( h 0 m) for the gravitational potential energy. h 0 m: The kinetic energy, the gravitational potential energy, and the elastic potential energy at the point of release are: mv 0 m KE (0 m/s) 0 J PE mgh mg(0 m) 0 J gravity elastic ky 0 PE (50.0 N/m)(0.59 m) 8.76 J The total mechanical energy E 0 at the point of release is the sum of the three energies above: E J. h 0.00 m: When the object has risen a distance of h 0.00 m above the release point, the spring is stretched by an amount of 0.59 m 0.00 m 0.39 m. Since the total mechanical energy is conserved, its value at this point is still E 8.76 J. The gravitational and elastic potential energies are: PE mgh (.00 kg)(9.80 m/s )(0.00 m) 3.9 J gravity elastic ky PE (50.0 N/m)(0.39 m) 3.84 J Since KE + PEgravity + PEelastic E, KE E PE PE 8.76 J 3.9 J 3.84 J.00 J gravity elastic h m: When the object has risen a distance of h m above the release point, the spring is stretched by an amount of 0.59 m m 0.9 m. At
5 this point, the total mechanical energy is still E potential energies are: 8.76 J. The gravitational and elastic PE mgh (.00 kg)(9.80 m/s )(0.400 m) 7.84 J gravity The kinetic energy is elastic ky PE (50.0 N/m)(0.9 m) 0.9 J KE E PE PE 8.76 J 7.84 J 0.9 J 0 J gravity elastic The results are summarized in the table below: h KE PEgrav PEelastic E 0 m 0 J 0 J 8.76 J 8.76 J 0.00 m.00 J 3.9 J 3.84 J 8.76 J m 0.00 J 7.84 J 0.9 J 8.76 J WebAssign Problem 5: United States currency is printed using intaglio presses that generate a printing pressure of. A $0 bill is 6. in. by.6 in. Calculate the magnitude of the force that the printing press applies to one side of the bill. REASONING Pressure is the magnitude of the force applied perpendicularly to a surface divided by the area of the surface, according to Equation.3. The force magnitude, therefore, is equal to the pressure times the area. SOLUTION According to Equation.3, we have c h b gb g 4 6 F PA lb / in. 6. in.. 6 in lb WebAssign Problem 6: Two identical containers are open at the top and are connected at the bottom via a tube of negligible volume and a valve that is closed. Both containers are filled initially to the same height of.00 m, one with water, the other with mercury, as the drawing indicates. The valve is then opened. Water and mercury are immiscible. Determine the fluid level in the left container when equilibrium is reestablished.
6 REASONING AND SOLUTION The mercury, being more dense, will flow from the right container into the left container until the pressure is equalized. Then the pressure at the bottom of the left container will be P ρ w gh w + ρ m gh ml and the pressure at the bottom of the right container will be P ρ m gh mr. Equating gives ρ w gh w + ρ m g(h ml h mr ) 0 () Both liquids are incompressible and immiscible so h w.00 m and h ml + h mr.00 m Using these in () and solving for h ml gives, h ml (/)(.00 ρ w /ρ m ) 0.46 m. So the fluid level in the left container is.00 m m.46 m from the bottom. WebAssign Problem 7: Interactive Solution.33 at presents a model for solving this problem. Multiple-Concept Example 8 also presents an approach to problems of this kind. The hydraulic oil in a car lift has a density of. The weight of the input piston is negligible. The radii of the input piston and output plunger are and 0.5 m, respectively. What input force F is needed to support the N combined weight of a car and the output plunger, when (a) the bottom surfaces of the piston and plunger are at the same level, and (b) the bottom surface of the output plunger is.30 m above that of the input piston? REASONING We label the input piston as and the output plunger as. When the bottom surfaces of the input piston and output plunger are at the same level, F F A / A, applies. However, this equation is not applicable Equation.5, ( ) when the bottom surface of the output plunger is h.50 m above the input piston. In this case we must use Equation.4, P P + ρ gh, to account for the difference in heights. In either case, we will see that the input force is less than the combined weight of the output plunger and car. SOLUTION a. Using A π r for the circular areas of the piston and plunger, the input force required to support the N weight is
7 3 ( m) A π F F ( N ) 93.0 N A π ( 0.5 m) (.5) b. The pressure P at the input piston is related to the pressure P at the bottom of the output plunger by Equation.4, P P + ρgh, where h is the difference in heights. Setting P F / A F / ( π r ), ( P ) F / π r have F F gh r π r + ρ π r ( N ) π π ( π ) 3 ( m) ( 0.5 m), and solving for F, we (.4) 3 3 ( ) ( ) ( ) π ( ) kg/m 9.80 m/s.30 m m 94.9 N WebAssign Problem 8: A paperweight, when weighed in air, has a weight of. When completely immersed in water, however, it has a weight of. Find the volume of the paperweight. REASONING The paperweight weighs less in water than in air, because of the buoyant force F B of the water. The buoyant force points upward, while the weight points downward, leading to an effective weight in water of W In water W F B. There is also a buoyant force when the paperweight is weighed in air, but it is negligibly small. Thus, from the given weights, we can obtain the buoyant force, which is the weight of the displaced water, according to Archimedes principle. From the weight of the displaced water and the density of water, we can obtain the volume of the water, which is also the volume of the completely immersed paperweight. SOLUTION We have W W F or F W W In water B B In water According to Archimedes principle, the buoyant force is the weight of the displaced water, which is mg, where m is the mass of the displaced water. Using Equation., we can write the mass as the density times the volume or m ρv. Thus, for the buoyant force, we have F W W ρ Vg B In water
8 Solving for the volume and using ρ kg/m 3 for the density of water (see Table.), we find V W W ρ g In water 6. 9 N 4. 3 N c.00 0 kg / m hc9. 80 m / s h 4 m 3 WebAssign Problem 9: Concept Simulation. at reviews the concept that plays the central role in this problem. (a) The volume flow rate in an artery supplying the brain is. If the radius of the artery is 5. mm, determine the average blood speed. (b) Find the average blood speed at a constriction in the artery if the constriction reduces the radius by a factor of 3. Assume that the volume flow rate is the same as that in part (a). REASONING a. According to Equation.0, the volume flow rate Q is equal to the product of the cross-sectional area A of the artery and the speed v of the blood, Q Av. Since Q and A are known, we can determine v. b. Since the volume flow rate Q through the constriction is the same as the volume flow rate Q in the normal part of the artery, Q Q. We can use this relation to find the blood speed in the constricted region. SOLUTION a. Since the artery is assumed to have a circular cross-section, its cross-sectional area is A π, where r is the radius. Thus, the speed of the blood is r Q 6 3 Q m / s A π r v 3 ( 5. 0 m) 4. 0 m/s π (.0) b. The volume flow rate is the same in the normal and constricted parts of the artery, so Q Q. Since Q A v, the blood speed is v Q /A Q /A. We are given that the radius of the constricted part of the artery is one-third that of the normal artery, so r r. Thus, the speed of the blood at the constriction is 3 v Q 6 3 Q Q m / s A π r π r π ( ) ( 5. 0 m) 0.38 m/s
9 WebAssign Problem 0: See Multiple-Concept Example 5 to review the concepts that are pertinent to this problem. The blood speed in a normal segment of a horizontal artery is 0. m/s. An abnormal segment of the artery is narrowed down by an arteriosclerotic plaque to one-fourth the normal cross-sectional area. What is the difference in blood pressures between the normal and constricted segments of the artery? REASONING We assume that region contains the constriction and region is the normal region. The difference in blood pressures between the two points in the horizontal artery is given by Bernoulli s equation (Equation.) as P P ρ v ρ v, where v and v are the speeds at the two points. Since the volume flow rate is the same at the two points, the speed at is related to the speed at by Equation.9, the equation of continuity: A v A v, where A and A are the cross-sectional areas of the artery. By combining these two relations, we will be able to determine the pressure difference. SOLUTION Solving the equation of continuity for the blood speed in region gives v v A /A. Substituting this result into Bernoulli s equation yields Since A 4 v A ρ ρ ρ ρ A P P v v v A, the pressure difference is v A ρ ρ ρ A 4 3 ( ) ( ) ( ) ( ) P P v v kg/m 0. m/s 5 96 Pa We have taken the density ρ of blood from Table.. WebAssign Problem : Poiseuilles' law remains valid as long as the fluid flow is laminar. For sufficiently high speed, however, the flow becomes turbulent, even if the fluid is moving through a smooth pipe with no restrictions. It is found experimentally that the flow is laminar as long as the Reynolds number Re is less than about 00. Re R / Here,, and are, respectively, the average speed, density, and viscosity of the fluid, and R is the radius of the pipe. Calculate the highest average speed that blood ( 060 kg/m 3, Pa s) could have and still remain in laminar flow when it flows through the aorta (R m).
10 REASONING AND SOLUTION The Reynold's number, Re, can be written as Re v ρ R / η. To find the average speed v, v 3 ( 000) ( Pa s) 3 3 ( ) ( ) (Re) η ρ R 060 kg/m m 0.5 m/s
11 Practice conceptual problems: Note: Chapter 0 problems were included in last week s solutions. Chapter : 3. A person could not balance her entire weight on the pointed end of a single nail, because it would penetrate her skin. However, she can lie safely on a bed of nails. A bed of nails consists of many nails driven through a sheet of wood so that the pointed ends form a flat array. Why is the bed of nails trick safe? REASONING AND SOLUTION A person could not balance her entire weight on the pointed end of a single nail, because it would penetrate her skin. According to Equation.3, the pressure exerted by the nail is P F / A where F represents the weight of the person, and A is the area of the tip of the nail. Since the tip of the nail has a very small radius, its area is very small; therefore, the pressure that the nail exerts on the person is large. The reason she can safely lie on a "bed of nails" is that the effective area of the nails is very large if the nails are closely spaced. Thus, the weight of the person F is distributed over all the nails so that the pressure exerted by any one nail is small. 8. Could you use a straw to sip a drink on the moon? Explain. REASONING AND SOLUTION When you drink through a straw, you draw the air out of the straw, and the external air pressure leads to the unbalanced force that pushes the liquid up into the straw. This action requires the presence of an atmosphere. The moon has no atmosphere, so you could not use a straw to sip a drink on the moon. 3. On a distant planet the acceleration due to gravity is less than it is on earth. Would you float more easily in water on this planet than on earth? Account for your answer. REASONING AND SOLUTION According to Archimedes' principle, any fluid applies a buoyant force to an object that is partially or completely immersed in it; the magnitude of the buoyant force equals the weight of the fluid that the object displaces. Therefore, the magnitude of the buoyant force exerted on an object immersed in water is given by F ρ Vg B water, where ρ water is the density of water, V is the volume displaced by the immersed object, and g is the magnitude of the acceleration due to gravity. If the acceleration due to gravity on a distant planet is less than it is on earth, then, other factors remaining the same, the buoyant force will be less on the planet than it is on earth. However, the weight of the object will also be less than the weight of the object on earth.
12 When an object floats in water, the upward buoyant force exerted by the water must be equal in magnitude and opposite in direction to the weight of the object, as shown at the right. Hence, F m g B object. It follows that ρ water Vg m object g. Notice that the acceleration due to gravity, g, appears on both sides of this equation. Algebraically canceling the g's we have ρ water V m object. Therefore, the object will float so that it displaces a volume of water V, where V m / ρ object water. This result is independent of g. It is the same on earth as it is on the distant planet. Therefore, it would be no more difficult to float in water on this planet than it would be on earth. B u o y a n t f o r c e o n o b j e c t W e i g h t o f o b j e c t 8. In steady flow, the velocity of a fluid particle at any point is constant in time. On the other hand, a fluid accelerates when it moves into a region of smaller cross-sectional area. (a) Explain what causes the acceleration. (b) Explain why the condition of steady flow does not rule out such an acceleration. REASONING AND SOLUTION In steady flow, the velocity v of a fluid particle at any point is constant in time. On the other hand, a fluid accelerates when it moves into a region of smaller cross-sectional area, as shown in the figure below. X v X Y v Y a. A fluid particle at X with speed v X must be accelerated to the right in order to acquire the greater speed v Y at Y. From Newton's second law, this acceleration can arise only from a net force that acts in the direction XY. If there are no other external forces acting on the fluid, this force must arise from the change in pressure within the fluid. The pressure at X must be greater than the pressure at Y. b. The definition of steady flow makes no reference as to how the velocity of a fluid particle varies from point to point as the fluid flows. It simply states that the velocity of a fluid particle at any particular point is constant in time. Therefore, the condition of steady flow does not rule out the acceleration discussed in part (a). 3. Which way would you have to spin a baseball so that it curves upward on its way to the plate? In describing the spin, state how you are viewing the ball. Justify your answer. REASONING AND SOLUTION The figure below shows a baseball, as viewed from the side, moving to the right with no spin. Since the air flows with the same speed above and below the ball, the pressure is reduced by the same amount above
13 and below the ball. There is no net force to cause the ball to curve in any particular direction (except for gravity which results in the usual parabolic trajectory). up v ball right Without spin If the ball is given a spin that is counterclockwise when viewed from the side, as shown below, the air close to the surface of the ball is dragged with the ball. In accord with Bernoulli's equation, the air on the top half of the ball is "speeded up" (pressure reduced by a greater amount), while that on the lower half of the ball is also speeded up, but less so (pressure reduced by a smaller amount). Thus, the pressure on the top half of the ball is lower than on the bottom half. F a s t e r a i r, l o w e r p r e s s u r e D e f l e c t i o n f o r c e u p v b a l l r i g h t S l o w e r a i r, h i g h p r e s s u r e Because of the pressure difference, a deflection force is generated that is directed from the higher pressure side of the ball to the lower pressure side of the ball. Therefore, the ball curves upward on its way to the plate.
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Failing to get space inside the train, Eid holidaymakers climb onto its roof to reach their destinations. The photo was taken from Airport Railway Station in the city on Monday. Photo: Bayazid Akter/ UNB.
Dhaka, Aug 20 (UNB) – Monday being the last office day for most of the public and private sector job holders before the Eid-ul-Azha, the mad rush of home-goers started from this evening.
The holidaymakers were seen rushing towards the city's train, bus and launch stations in their pursuit to go to respective village homes and share the Eid joys with near and dear ones.
Photo: Bayazid Akter/UNB
The three-day official Eid holidays will begin on Tuesday (August 21) as Eid-ul-Azha will be celebrated on Wednesday (August 22).
City dwellers, especially the family members of both government and private job holders, started leaving the capital from Friday morning to celebrate Eid-ul-Azha with their relatives in the villages and district towns. Many have chosen to send their families a bit early to avoid the last moment's mad rush just a day ahead of Eid.
Thousands of city dwellers are expected to leave the capital Dhaka this year to celebrate the Eid, said AIG (Media) Police headquarters Md Sohel Rana.
Most of the holidaymakers, specially family members of government and semi-government officials and employees as well as private service holders and employees of various industries have already left the city, he said.
Photo: Zabed Hasnain Chowdhury
There was a greater crowd than usual at Kamalapur Railway Station as the home-goers were using the railway for their journey fearing traffic congestions and hassles on highways.
Officer-in-charge of Kamalapur Railway police station Yasin Faruque told UNB that a total of 84 trains were scheduled to leave the station on Monday. Passengers had to suffer much as most of the trains left the station some two to four hours behind their scheduled times.
Replying to a question he said as there was heavy tailback on the highways in last several days, passengers preferred to travel by train despite overcrowding. Return trains failed to reach the Kamalapur railway station timely as those had to run slowly to reach respective destinations carrying too many passengers on the rooftops. On the other hand, it needed more time to restrain passengers from riding on rooftops, the OC said.
Sources at the Sadarghat said though there was huge crowd but most of the launches left the terminal timely on Monday.
Photo: Zabed Hasnain Chowdhury
Officer-in-charge of Darus Salam police station Selim uz-zaman said most of the inter-city buses left the Gabtoli bus terminal normally on Monday.
Another police officer of Jatrabari police station said buses left Saidabad and Jatrabari almost timely as no tailback was prevailing on Dhaka- Chattogram highway.
Officer-in-charge of Daudkandi Highway police station Abul Kalam Azad said though there was huge pressure of vehicles on the Dhaka-Chattogram highway from Monday morning but there were no jams as vehicles moved in discipline.


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The Gloaming
A Hardback edition by Kirsty Logan (Apr 19, 2018)
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Short Description: 'The best lives leave a mark.' A bewitching tale of first love, shattering grief, and the dangerous magic that draws us home.... Read more
Product Description
'The best lives leave a mark.' A bewitching tale of first love, shattering grief, and the dangerous magic that draws us home.
Mara's island is one of stories and magic, but every story ends in the same way. She will finish her days on the cliff, turned to stone and gazing out at the horizon like all the islanders before her.
Mara's parents - a boxer and a ballerina - chose this enchanted place as a refuge from the turbulence of their previous lives; they wanted to bring up their children somewhere special and safe. But the island and the sea don't care what people want, and when they claim a price from her family, Mara's world unravels.
It takes the arrival of Pearl, mysterious and irresistible, to light a spark in Mara again, and allow her to consider a different story for herself.
The Gloaming is a gorgeous tale of love and grief, and the gap between fairy tales and real life.
The Gloaming Hardback edition by Kirsty Logan
Product Details
Product Specification
- Author
- Kirsty Logan
- 9781911215936
- Format
- Hardback,
- Publishers
- Vintage Publishing , Harvill Secker
- Publication date
- Apr 19, 2018
- Pages
- 320
- Product dimensions
- 140 x 221 x 31mm
- Weight
- 484g


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January saw us back to work in the forest garden and tackling the buttercups that were strangling some of the plants and taking over the paths. Some lovely pics courtesy of Michael Hill. We will continue to meet on the second Sunday of every month so if you want to do some volunteering in 2019 then come and join us! https://friendsofdalbyforest.wordpress.com/about/
Category Archives: forest garden
September work task in the forest garden
Thank you to all the volunteers who came along to the work task this morning. There were only 6 of us but it felt like we achieved a lot.
We laid a membrane to suppress the weeds in the area behind the cafe at the courtyard (it's very stony so that seemed the best way to kill off the weeds). We will work in this area next year as it is technically also part of the forest garden. We have lots of strawberries and currants to plant there.
In addition, we built a bug house (hotel?) with some pallets kindly donated by Mark from the maze . There are so many insects in the garden especially ladybirds and beetles at the moment, so we hope they enjoy their new accommodation!
Hedgerow work along the Ellerburn trail
Our May worktask involved removing the trees tubes along the Ellerburn trail. A few years ago we planted a range of hedgerow plants including holly and hawthorne but now they are more established, it was time to remove the tubes now that they are less likely to be nibbled by deer and rabbits. We started the task last November – thankfully it was a bit warmer today!
A couple of us also did some more work on the forest garden which has burst into life this week. If you visiting the forest do call in to see it – and catch the Helen Sear exhibition in the courtyard at the same time.