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Nairaland Forum / Nairaland / General / Education / 2016 NAIRALAND JAMB TUTORIAL {the Physics Thread } (44404 Views)
Nairaland 2016 Jamb Tutorial Classroon [use Of English Thread] / Nairaland 2016 Jamb Tutorial Classroom [chemistry Thread] / Nairaland 2016 Jamb Tutorial Classroon [mathematics Thread] (2) (3) (4)
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Re: 2016 NAIRALAND JAMB TUTORIAL {the Physics Thread } by Geofavor(m): 1:39pm On Dec 30, 2015 |
1 Like |
Re: 2016 NAIRALAND JAMB TUTORIAL {the Physics Thread } by thankyouJesus(m): 2:01pm On Dec 30, 2015 |
[b]Measurement of Electrical Power. Electrical power consumed by an electrical appliance is measured in watt. The unit of electrical energy is kilowatt-hour. Mathematically; W = IVt. Recall that P = IV, W = Pt. Where w is energy (kwh), P is power (kw), t is time (h). Note; 1. To convert from watt to kilo-watt, divide by 1000. 2. The above formula is used when cost/money is involved.. Example; A lamp is rated 12V, 24w. How many Joules does it consume in an hour? Solution; V = 12V. T = 1hr = 3600s (money is not involved). P = 24w. Using P = IV. I = 2A. Energy = IVt. Energy = 86400J. Example; A household refrigerator is rated 200w. If electricity costs 5K per Kwh, what is the cost of operating it for 20days? Solution; P=200w=0.2kw (money is involved). T=20days=20X24=480hrs. Energy = power X time E=96Kwh. If 1kwh=5k, therefore, 96kwh=480k. Example; Find the cost of running five 50w lamps and three 100w lamps for 6hrs if electric energy costs #20 per unit. Solution; 5lamps=5X50=250w. 3lamps=3X100=300w. Total power=550w=0.55kw. Time=6hrs. Energy=0.55X6. E=3.3kwh. If #20 per unit, then, 3.3kwh is #66. Electrical Installation. Electrical installation is the branch of physics that leads to electrical engineering. It deals with how two wires, one live and the other neutral can be connected in a building. Live wires are usually insulated with red colour insulator, while neutral wires are insulated with black colour insulator[/b] 1 Like 1 Share |
Re: 2016 NAIRALAND JAMB TUTORIAL {the Physics Thread } by thankyouJesus(m): 2:18pm On Dec 30, 2015 |
Classwork. 1. An electric lamp rated at 48w, 12v supply. Calculate the current flowing in the lamp and the resistance. 2. It takes 4minutes to boil a quantity of water electrically. How long will it take to boil same quantity of water using the same heating coil, but with the current doubled? 3. Calculate the amount of of heat generated in an external load of resistance 8 Ohm if an alternating current of peak value 5A is passed through it for 1min 40sec. 4. An electric heater takes 4A when operated on a 250V supply. What is the cost of the electricity consumed at 10k per kwh when the heater is used for 5hours? 5. A man has a 40w and two 60w bulbs in a room. How much will it cost him to keep them lit for 8hours if the cost of a unit is 20kobo? Cc Geofavor Iceberryose Debbychris Oxytocin |
Re: 2016 NAIRALAND JAMB TUTORIAL {the Physics Thread } by Dohyn3(f): 3:00pm On Dec 30, 2015 |
Can anyone direct me to the literature thread. please? ![]() Been looking since...couldn't find it. Anyone? Everyone? |
Re: 2016 NAIRALAND JAMB TUTORIAL {the Physics Thread } by Geofavor(m): 3:07pm On Dec 30, 2015 |
thankyouJesus:I = p/V I = 48/12 = 4A R = V/I = 12/4 = 3ohms 2minutes w = I2Rt w = 25 x 8 x 100 w = 20000J p = IV p = 4 x 250 = 1000W = 1kw W = pt = 1 x 5 = 5kwh 10k/kwh = $/5kwh $ = 50k ( where $ is the cost) total power = 160w = 0.16kw w = pt = 0.16 x 8 = 1.28kwh (energy swallowed in 8hrs) if a unit = 20kobo, then 1.28kwh = 25.6kobo corrected* 1 Like |
Re: 2016 NAIRALAND JAMB TUTORIAL {the Physics Thread } by Geofavor(m): 3:15pm On Dec 30, 2015 |
Dohyn3:eeyah. I don't think i've seen any one around. You can go to the jamb thread and ask. If there isn't, maybe someone would create one upon your request. |
Re: 2016 NAIRALAND JAMB TUTORIAL {the Physics Thread } by Geofavor(m): 3:41pm On Dec 30, 2015 |
Orezy5, hop in here na. Can't you see it's getting cool and hot in here? ![]() |
Re: 2016 NAIRALAND JAMB TUTORIAL {the Physics Thread } by Orezy5(m): 3:51pm On Dec 30, 2015 |
Geofavor: |
Re: 2016 NAIRALAND JAMB TUTORIAL {the Physics Thread } by thankyouJesus(m): 3:53pm On Dec 30, 2015 |
Cc Geofavor. 1. Use this logic to correct question 4. If, #100 = $1, similarly will, #200 = $2, and also will, #300 = $3. Let me work with the first two, lets assume I am not aware of the conversion for #200. Let the conversion be x, therefore, #100 = $1 #200 = x cross multiply 100x = 200 divide both sides by 100, automatically, x = $2. Hint; if a unit is 10kwh, 1 = 10kwh. 2. There is a salient "two" in the question 5, check again. I am loving your solution. |
Re: 2016 NAIRALAND JAMB TUTORIAL {the Physics Thread } by Dohyn3(f): 5:04pm On Dec 30, 2015 |
Geofavor: Thanks, i will... |
Re: 2016 NAIRALAND JAMB TUTORIAL {the Physics Thread } by Geofavor(m): 5:09pm On Dec 30, 2015 |
thankyouJesus:thanks. I understand this logic. the mistake i made was take 5 x 10 to be 15 instead of 50. Smh. I should be caned ![]() 1 Like |
Re: 2016 NAIRALAND JAMB TUTORIAL {the Physics Thread } by thankyouJesus(m): 5:17pm On Dec 30, 2015 |
Okay oga mi, check question 5 again for the two. Geofavor:Okay oga mi, check question 5 again for the two. 1 Like |
Re: 2016 NAIRALAND JAMB TUTORIAL {the Physics Thread } by Geofavor(m): 5:23pm On Dec 30, 2015 |
thankyouJesus:lol. Yes i have corrected it. 1 Like |
Re: 2016 NAIRALAND JAMB TUTORIAL {the Physics Thread } by Geofavor(m): 5:30pm On Dec 30, 2015 |
Dohyn3:alright. |
Re: 2016 NAIRALAND JAMB TUTORIAL {the Physics Thread } by iceberryose(m): 6:03pm On Dec 30, 2015 |
Hey guys sorry oh. Am really busy but i'll be back before you know it. Am still following closely tho cc thankyoujesus geofavor 1 Like 1 Share |
Re: 2016 NAIRALAND JAMB TUTORIAL {the Physics Thread } by thankyouJesus(m): 8:40am On Dec 31, 2015 |
[b]Good Morning all, happy new year in advance. EMF. EMF simply means electro motive force, which is the total energy per current obtained from a cell. It is denoted by E and measured in volts. E=pd across external resistance + pd across internal resistance. E=IR + Ir. E=I(R + r). Where E is emf, I current, r internal resistance, R external resistance. Arrangement of cells. Cells in series. -----|¡------|¡-------|¡--------. When cells are arranged in series, they are used to obtain greater emf. E=E1+E2+E3. Similarly, the resultant of the internal resistance of the cell is also equal to the sum of internal resistance of the cell (arrangement of resistance in series). r=r1+r2+r3. Cells in parallel. When cells are arranged in parallel, they supply current for a long period. -------|¡------- -------|¡------- -------|¡-------. E=E1=E2=E3. Similarly, the reciprocal of the resultant internal resistance is equal to the sum of the reciprocal of the individual internal resistance of the cells. 1/r=1/r1+1/r2+1/r3. Bonus; Lost voltage or voltage drop is defined as the pd across the internal resistance. It is denoted by Vr and measured in volts. Mathematically; Vr=Ir. Where I is current. r is internal resistance. Example; A cell of emf 1.5v supplies a current of 0.6A through a cell of resistance 2 Ohm. Calculate the internal resistance of the cell. Solution; E=1.5v I=0.6A R=2 Ohm E=I(R+r) r=0.5 Ohm. Example; Two cells each of emf 2.0v and internal resistance 2 Ohm are connected in a series. A 4 Ohm resistor is connected in a series to the cell. Calculate (a) current flowing in the circuit (b) voltage drop. Solution; E=2+2=4v. r=2+2=4 Ohm (the cells are arranged in series). R=4 Ohm. (a) I=0.5A. (b) voltage drop =Ir=2V[/b] 1 Like 1 Share |
Re: 2016 NAIRALAND JAMB TUTORIAL {the Physics Thread } by thankyouJesus(m): 4:16pm On Dec 31, 2015 |
Classwork. I am sorry it is coming late. 1. A battery of emf 24v and internal resistance 4 Ohm is connected to a resistor of 32 Ohm. What is the terminal pd of the battery? 2. A cell of emf 1.5v is connected in series with a resistor of resistance 3 Ohm. A high resistance voltmeter connected across the cell registers only 0.9v. Calculate the internal resistance of the cell. 3. A cell can supply currents of 0.60A and 0.30A through a 2.0 Ohm and 5.0 Ohm resistor respectively. What is the internal resistance of the cell? 4. Two cells each having emf of 1.5v and internal resistance of 1 Ohm are connected to a resistance of 5 Ohm. Determine the current in the cell if they are connected in (a) parallel (b) series. Cc Geofavor Iceberryose Debbychris Oxytocin |
Re: 2016 NAIRALAND JAMB TUTORIAL {the Physics Thread } by muhamadnur: 10:25am On Jan 01, 2016 |
thanks man we gat your back |
Re: 2016 NAIRALAND JAMB TUTORIAL {the Physics Thread } by Orezy5(m): 5:02pm On Jan 01, 2016 |
1. E=24V, r=4ohms, R=32ohms, I=?
E=I(R + r)
24=I(32 + 4)
24=36I
I=24/36
I=0.667A
The terminal p.d, V=Ir
V=0.667 X 4
V=2.668V 2. E=1.5V, r=?, R=3ohms, I=V/R=0.9/3=0.3A E=I(R + r) 1.5=0.3(3 + r) 1.5=0.9 + 0.3r 0.3r= 1.5 - 0.9 0.3r=0.6 r=0.6/0.3 r=2ohms 3. E=I(R + r) lets form a simultaneous equation: E=?, I=0.60A, R=2ohms, r=? E=0.6(2 + r) E=1.2 + 0.6r---------(i) for the second equation, E=?, I=0.30A, R=5ohms, r=? E=0.30(5 + r) E=1.5 + 0.30r---------(ii) combine (i) with (ii) E=1.2 + 0.6r E=1.5 + 0.3r To eliminate E, multiply (ii) by -1: E=1.2 + 0.6r -E= -1.5 - 0.3r 0= -0.3 + 0.3r 0.3r= 0.3 r=0.3/0.3 r=1ohm 4a. when connected in parallel; E=1.5V, I=?, R=5ohms, r=0.5ohm E=I(R + r) 1.5=I(5 + 0.5) 1.5=5.5I I=1.5/5.5 I=0.27A 4b. when connected in series; E=3V, I=?, R=5ohms, r=2ohms E=I(R + r) 3=I(5 + 2) 3=7I I=3/7 I=0.428A Cc: thankyouJesus |
Re: 2016 NAIRALAND JAMB TUTORIAL {the Physics Thread } by thankyouJesus(m): 5:58pm On Jan 01, 2016 |
Orezy5:Great work sir, check your solution in 1. |
Re: 2016 NAIRALAND JAMB TUTORIAL {the Physics Thread } by thankyouJesus(m): 5:59pm On Jan 01, 2016 |
muhamadnur:You are welcome sir 1 Like |
Re: 2016 NAIRALAND JAMB TUTORIAL {the Physics Thread } by Oxytocin(m): 6:33pm On Jan 01, 2016 |
Still following ![]() 1 Like |
Re: 2016 NAIRALAND JAMB TUTORIAL {the Physics Thread } by thankyouJesus(m): 9:03am On Jan 06, 2016 |
Good morning all, I want to assume most of you know this basic topics in Physics, lets move to more complex topics. To achieve this, could you drop a topic for me to write on taking into consideration my resumption date in UI. Cc DebbyChris Oxytocin Geofavor Iceberryose Muhamadnur Orezy5 |
Re: 2016 NAIRALAND JAMB TUTORIAL {the Physics Thread } by Geofavor(m): 10:25am On Jan 06, 2016 |
thankyouJesus:welcome back sir. Here, PRESSURE a) AtmosPheric Pressure i) definition of atmosPheric Pressure ii) units of Pressure iii) measurement of Pressure iv) simPle mercury barometer, aneroid barometer and manometer v) variation of Pressure with height vi) the use of barometer as an altmeter b) Pressure in liquids i) the relationshiP between Pressure, dePth and density (P = pgh) ii) transmission of pressure in liquids ( pascal's principle) iii) application. |
Re: 2016 NAIRALAND JAMB TUTORIAL {the Physics Thread } by thankyouJesus(m): 11:04am On Jan 06, 2016 |
Geofavor:I will do justice to this after charging my battery, today or tomorrow (erratic power supply). Thank you boss. 1 Like 2 Shares |
Re: 2016 NAIRALAND JAMB TUTORIAL {the Physics Thread } by thankyouJesus(m): 8:40am On Jan 07, 2016 |
[b]Good morning all. Pressure. Pressure is defined as the force acting at right angle, normal or perpendicular per unit surface area in contact with a substance. Thus, the force applied on the surface area of a substance is called pressure. It is a scale quantity and exists in solid, liquid and gases. Mathematically; Pressure = Force/Area. Example; The weight on the narrow heel of a girl's shoe is 250N and the surface area of the heel in contact with the floor is 50mm2. Determine the pressure exerted on the heel. Solution; F = 250N.c. A = 50mm2 = 50 X 10-6m2. Pressure = Force/ Area = 5 X 106N/m2. Example; Calculate the pressure on the surface of a rectangular box of weight 100N if the base of this box has an area of 2m2. Solution; Weight (Force) = 100N. Area = 2m2. Pressure = Force/Area = 50N/m2. Pressure in Liquid. Consider a baker containing liquid at level h. The total force acting at the base of the beaker is called thrust. Thrust per unit area is called pressure of the liquid. Mathematically; Pressure = Force/Area. Pressure = mg/A. Recall that, Density = mass/volume. Therefore; Pressure = density X volume X g /A. Recall that; Volume = Area X Height. Therefore; Pressure = density X height X acceleration due to gravity. Example; A cylindrical jar of radius 7cm and height 25cm is filled with a liquid 0.80g/cm3. What is the pressure exerted at the bottom of the jar by the liquid? Solution; H = 25cm = 0.25m r = 7cm Density = 0.8g/cm3 = 8000kg/m3. Pressure = density X height X g = 2 X 103N/m2.[/b] |
Re: 2016 NAIRALAND JAMB TUTORIAL {the Physics Thread } by thankyouJesus(m): 9:40am On Jan 07, 2016 |
Pascal's Law. A scientist called Blaise Pascal propounded a law relating to pressure. According to him, Liquid transmits pressure equally in all direction. Pascal demonstrated this by putting water inside a rubber ball and ensured that the water was evenly distributed. He pressed the ball with equal pressure and water was seen out of the rubber ball at the same time. This shows that the increase in pressure or pressure applied equally on the same surface is equally distributed in all directions. |
Re: 2016 NAIRALAND JAMB TUTORIAL {the Physics Thread } by thankyouJesus(m): 11:59am On Jan 07, 2016 |
Atmospheric Pressure. The surrounding environment is full of air blowing from one part to the other. The force due to the vertical column of air on a unit area of the surface gives the atmospheric pressure. Fortin Barometer. Fortin barometer is used for more accurate measurement of atmospheric pressure than the simple or Torricellian barometer. It is made up of a tube containing mercury which is protected by enclosing it in a brass tube. Aneroid Barometer. Aneroid barometer has no liquid and is widely used in the home for showing weather changes. The essential part of the barometer is a flat cylindrical metal box or capsule, corrugated for strength, and sealed after having been partially exhausted of air. Increase in atmospheric pressure causes the box to cave in slightly, while a decrease allows it to expand. |
Re: 2016 NAIRALAND JAMB TUTORIAL {the Physics Thread } by thankyouJesus(m): 12:21pm On Jan 07, 2016 |
Cc Sexykaycee, Prot0n, Mathefaro et al, now will be a good time to come in (my schedules, power supply and other unforeseen circumstances are not favourable). 1 Like |
Re: 2016 NAIRALAND JAMB TUTORIAL {the Physics Thread } by mathefaro(m): 1:17pm On Jan 07, 2016 |
I hail thee my Oga for the wonderful with you're doing here. Although I'll suggest that questions should be asked and answered in details pertaining to the topic at hand instead of giving a kind of note. I'll be around to help henceforth when needed |
Re: 2016 NAIRALAND JAMB TUTORIAL {the Physics Thread } by thankyouJesus(m): 5:16pm On Jan 07, 2016 |
mathefaro:cc Geofavor |
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