Problem 1: A Toy Grasshopper

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Claude Hawkins
6 years ago
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When air is allowed to get in between the sucker and a surface, the grasshopper will jump up from the surface, because the legs of the grasshopper act like springs.

1 Problem 1: A Toy Grasshopper In the illustration on the right you see a picture of a toy grasshopper. At the underside of the bodypart there is a sucker that will strongly attach itself to any surface when firmly pushed down. When air is allowed to get in between the sucker and a surface, the grasshopper will jump up from the surface, because the legs of the grasshopper act like springs. Claire and Joanne decide to research this type of toy grasshopper a little further. Their research question is: What speed does the grasshopper has the moment the legs are released from the surface. To get an idea of this speed, they let the grasshopper be released several times from a surface. They estimate the jump to be, on average, 1.0 m above ground. 1. Calculate at which speed the grasshopper will be released from the ground. Friction by air can be ignored in this calculation. mgh = ½ mv 2 so 9.81*1.0 = ½ v 2 and v = 4.4 m/s In the illustration below you see two situations in the release of the grasshopper. At time t 0 the sucker just releases from the surface; at time t 1 the legs are just released from the surface. Claire has attached a distance sensor to a computer and registers the height of the grasshopper as function of time. The result is shown in the graph below. The distance sensor has been set in such a way that the height h=0 corresponds to time t 1 (see the respective illustrations and graph). The graph is also depicted (enlarged) in the annex of problem Determine from the graph in the annex the speed at t 1.

2 Draw the tangent at t1 and determine the slope. v = 1.4/0.30 = 4.7 m/s At time t=0.75 s the grasshopper starts to fall. 3. Determine from the graph whether or not the grasshopper experiences any friction by air. The graph is perfectly symmetrical, the time from t1 to 0.75 s is the same as from 0.75 s to the end of the fall. There is no friction by air. Second argument: determine the slope at t = 1.25 s: this is exactly the same as at t1. Claire and Joanne have an additional research question: How much of the original energy through the spring is converted into energy by height?. For this Joanne pushes the grasshopper down several times with different forces F. Each time the force F she uses is registered. Also registered is the compression u of the sucker at those times. The graph to the right shows the relation between F and u. At u=4 cm the sucker is fully compressed. From this graph one can establish the amount of spring energy that is stored in the legs before release. Again friction by air is neglected. This graph is also shown on the annex. Using both this graph and the previous graph in the annex, the percentage of conversion of spring energy to energy of height can be calculated. The mass of the grasshopper is 6.2 g. 4. Calculate this percentage. Spring energy is ½ Cu 2 Determine C from the graph: C = 7.4/0.04 = 185 N/m. So spring energy is ½ C* = 0.15 J. Energy of height = mgh = *9.81*1.0 = J. The percentage is thus 0.061/0.15 = 41%

Problem 2: An airplane in a turn While an airplane is flying at constant height and constant speed, there are two forces in effect on the airplane: the gravitational force and the lift force.

Proof that the airplane will descend if the pilot does not correct the airplane in this situation. Do so by a drawing illustrating the forces acting on the airplane in such a situation.

3 Problem 2: An airplane in a turn While an airplane is flying at constant height and constant speed, there are two forces in effect on the airplane: the gravitational force and the lift force. For the airplane to make a (partial) turn, the airplane will tilt to make an acute angle with respect to the horizon. 5. Proof that the airplane will descend if the pilot does not correct the airplane in this situation. Do so by a drawing illustrating the forces acting on the airplane in such a situation. The vertical component of the lift is now smaller than the force of gravity. By increasing speed, the pilot prevents loosing height. As such the airplane describes a perfect circle. In the figure on the corresponding annex the gravitational force in this circumstance is shown (see arrow). Also, the direction of the lift force is indicated. 6. Reconstruct, by using the annex, the lift force and inward circular force.

4 The time T it takes the airplane to complete a full circle is given by the formula 2 tan with r being the radius of the circle the airplane is making, and α the angle (tilt) of the wings with respect to the horizon. 7. Show this formula to be correct. From the previous construction it appears that α is also the top angle so this gives tan α = F centr /F z = mv 2 /rmg = v 2 /rg. As v = r/t it follows that tan α = 4π 2 r 2 /T 2 rg = 4π 2 r/t 2 g so T 2 = 4π 2 r/gtan α and At standard landing procedure it is custom for airplanes to take circles with a diameter of 4 km and completion of one circle in 2 minutes time. 8. Calculate at which angle the wings must tilt with respect to the horizon in order for the airplane to accomplish this. so 2*60, therefore tan α = 0.56 and α = 29 Problem 3: Aquarium An aquarium is filled with 63 liters of water. A heating element regulates the temperature. Its specifications are 230 V and 30 W. The heat capacity of the aquarium without water is 15 kj/k. 9. Calculate the electrical resistance of the heating element. P = U*I = U 2 /R so R = U 2 /P = /30 = 1.8 kω 10. Calculate the heat capacity of the aquarium filled with water. Total heat capacity is 15 kj/k *63 J/K = J/K The heating element is engaged when the temperature drops below 22 C. A temperature sensor is sensitive for changes in temperature between 10 C and 30 C, and responds linearly between these temperatures by giving off a potential between 0.0 en 5.0 V. 11. Calculate the sensitivity of this heat sensor. Sensitivity is volts/temperature difference = 5 0/30 10 = 5/20 = 0.25 V/ C 12. Provide a design scheme of logical components which makes the heating element being engaged below 22 C. If one of the logical components is a comparator also provide the reference potential. Uref = (22 10)*0.25 = 3 V An 8 bits A/D convertor (ADC) is used to convert the potential provided by the temperature sensor into a digital signal. The minimum signal of the ADC corresponds to 0.0 V, whereas the

maximum signal corresponds to 5.0 V. 13. Calculate the binary signal that corresponds to an input potential of 3.0 V. 8 bits means 2 8 divisions so 5/256 = 0.

5 maximum signal corresponds to 5.0 V. 13. Calculate the binary signal that corresponds to an input potential of 3.0 V. 8 bits means 2 8 divisions so 5/256 = V/div, and 3V corresponds with the 3/ th division = 153. Binary code of 153 is The heating sensor is also used to set off a sound alarm when the water temperature drops below 15 C, or increases to temperatures 25 C and above. The alarm will continue to sound even when water temperature returns to in between 15 C and 25 C; it has to be put off manually. This extension of the design scheme is shown in the illustration below. The illustration is also given in enlarged form on the annex. 14. Finish the scheme by adding the necessary components. Again, provide reference potentials of the comparators. Problem 4: Technetium as tracer. In nuclear medicine radioactive substances are used as 'tracer': because these substances give of radiation they can be used to locate where in the body these substances end up. Preferably radioactive substances are used that radiate gamma radiation. An appropriate half life time of a tracer is a requirement. 15. Explain what the disadvantages are of a tracer with too short or too long a half life time. Too short: the radiation power expires before the tracer reaches the right spot in the body. Too long: the patient carries a radioactive source in the body for a long time. In medical research the isotope of Technetium 99m Tc is used as tracer. This isotope is an almost ideal material for nuclear medicine. It is easily obtained from the bèta decay of 99 Molybdenum. The letter m in 99m Tc indicates that the nucleus is in a metastable state: a nucleus can remain for longer time in an excited state.

6 16. Provide the decay equation from which 99m Tc results Mo 1 0 e + 99m 43 Tc Molybdenum is being produced in nuclear power plants and subsequently transported to hospitals where research is taking place. Because of decay the time between production and use should not be too long. 17. Calculate the allowed period of transport time when requirements state that at least 10% of the molybdenum is still available for use in hospitals. t 1/2 = 68.3 hours so 0.1 = 0.5 t/68.3 It follows that t = 68.3*log0.1/log0.5 = 227 hours (=29.5 days) The half life time for 99m Tc is 6.0 hours. When a nucleus returns to its groundstate, it emits a gamma photon of 140 kev. These photons are registered by the equipment used in research. 18. Calculate the wavelength of the emitted photon. λ = c/f and E = hf so λ = hc/e = * /(140*10 3 * ) = m At emission the nucleus looses mass. 19. Calculate the mass percentage a Tc nucleus looses during emission of a photon. E = mc 2 so * = m*( ) 2 and m = kg. Mass of Tc 99 is 99u = kg so percentage is / *100 = %

Problem 5: A weir in the river Lek. To keep the Lek river navigable during dryer periods, weirs are used to maintain water levels. At Hagestein (NL) such a weir is in operation.

7 Problem 5: A weir in the river Lek. To keep the Lek river navigable during dryer periods, weirs are used to maintain water levels. At Hagestein (NL) such a weir is in operation. The photo below depicts the weir at Hagestein. The width of the river at that location is indicated by the letter l. At the position of the weir the width of the river Lek is 110 m. The photo is the result of an enlargement of 12,0 times with respect to the image in the camera that recorded the photo. The lens of the camera had a focal length of 200 mm. 20. Calculate at which distance the picture, depicted in the photo above, was made. The weir at the photo measures 6.0 cm, so the image on the camera chip is 6.0/12 = 0.5 cm. The enlargement is thus /110 = This is b/v and b = f = m. So v = / = 4.40 km When the weir is closed water will accumulate at one side of the weir. At one side the level of water will be higher than on the other side. This difference in height can be used to generate electricity, by regulating water to flow downwards through turbines. As such, the maximally attained power generated is 1,8 MW. This maximally attained power is generated when, per second, 95 m 3 of water is allowed to fall down over a height difference of 3,0 m. 21. Calculate the efficiency by which electrical power is generated at this weir. Power input is m*g*h per second = 95*1000*9.81*3 = 2.8 MW. So efficiency is power output/power input = 1.8/2.8 *100% = 64% As indicated, the electrical power is generated in turbines which convert water flow through rotating blades into electrical current, providing electrical energy. This energy is transported though copper wires over a length of 4.0 km. The copper wires have a through section of 50 mm 2. Subsequently, the electrical energy is transported through a 10kV power net. See figure below.

8 At a certain time the turbines generate 0.15 ka in current. 22. Calculate the loss of electrical power due to the use of copper wire outside the 10 kv circuit. R wire = ρ*l/a = *4000/ = 1.36 Ω Power loss = I 2 *R = ( ) 2 *1.36 = 31 kw When close to urban areas the transport voltage of 10 kv is reduced to 230 V using a transformer. Users, at the secondary side of the transformer, take 100 kw of electrical power. Unfortunately, transformers are generally not ideal in transforming energy. There is a loss of 1.20 kw power through heat. 23. Calculate the current that is operating in the primary coil of this transformer. Power in primary coil = = kw = U*I = 10 4 *I so I = 10 A The water power plant at Hagenstein delivers electrical energy to the same users as the gaspowered power plant at Utrecht (NL). At the Utrecht power plant, natural gas is burned with an efficiency of 30%. Burning 1 m 3 natural gas gives an energy amount of J. During a specific month the Hagenstein power plant delivered kwh of electrical energy to users. 24. Calculate how much gas (in m 3 ) was saved through the use of the water power plant at Hagestein during this month. Saved kwh = J, this means /0,3 = J gas saved. This amounts to / = m 3 gas. End

9 Annex to questions Problem 1, questions 2 and 3: Your name:... Problem 1, questions 3 and 4

10 Problem2, question 6 Your name:...

11 Problem 3, question 14

An ion follows a circular path in a uniform magnetic field. Which single change decreases the radius of the path?

An ion follows a circular path in a uniform magnetic field. Which single change decreases the radius of the path? T5-1 [237 marks] 1. A circuit is formed by connecting a resistor between the terminals of a battery of electromotive force (emf) 6 V. The battery has internal resistance. Which statement is correct when