Om die struktuur van sommige regte prismas te verstaan
[LU 3.3, 3.4]
A.
Bou houers
Daar sal aan jou ‘n vel papier met vorms gegee word. Jy benodig ‘n liniaal waarmee jy kan meet, ‘n skêr en gom of kleefband. Kleurpotlode sal ook help. Doen die volgende:
Meet al die lyne en skryf jou mate netjies neer (jy behoort tot die naaste half-millimeter te kan meet). Doen ook jou bes om die deursnee (of die radius) van die sirkel te meet. As jy ‘n gradeboog beskikbaar het, bepaal ook waar die 90° hoeke is.
Bepaal nou die oppervlaktes van al die vorms vanuit jou mates. Tel die verskillend dele bymekaar om die totale oppervlaktes van die vier verskillende vorms te bereken. Sit jou werk versigtig uiteen sodat enigeen kan begryp wat jy doen. Gebruik die regte name vir die vorms waarmee jy werk.
Byvoorbeeld, vir die laaste figuur sou dit so lyk:
Totale oppervlakte = klein reghoek + klein reghoek + groot reghoek
= (l × b) + (l × b) + (l × b)
ensovoorts . . . (Onthou om altyd geskikte eenhede te gebruik.)
Sny nou die vorms versigtig uit. Jy kan hulle inkleur as dit jou sou help om die bokant en onderkant van die sykante (met die strepe) te onderskei. Vou die vorms nou en gebruik kleefband of gom en papierstrokies om vier houertjies te maak. Hou die kante met die strepe aan die buitekant.
Skryf die
Totale Buite-Oppervlakte (TBO) van elke vorm neer. (Dis wat jy reeds bereken het!)
Werk saam in groepies van drie of vier en probeer uitwerk hoeveel 1cm × 1cm blokkies in elk van die houers sou inpas. Hierdie waarde is
volume van die houer. As julle ‘n metode of formule kan vind wat vir elkeen van die vorms sou werk, skryf dit sorgvuldig neer.
As jy klaar is met hierdie oefening behoort jy twee formules te hê.
B.
Regte prismas
Hierdie vier houers is elk ‘n
regte prisma . Hierdie vorms het ‘n basis en ‘n bokant wat presies dieselfde grootte en vorm het, met sye wat reguit boontoe loop en ‘n 90° hoek vorm met die basis. Soek vir jou items wat aan hierdie vereistes voldoen en dus regte prismas is.
Ons benoem regte prismas volgens die vorm van die basis, byvoorbeeld vierkantige prisma, reghoekige prisma, driehoekige prisma en sirkelvormige prisma (silinder).
Is hierdie twee vorms regte prismas? Beskryf die vorm van elkeen se basis en bevestig of die sye regop loop teen 90° met die basis.
C.
Formules
Om die totale buite-oppervlakte (TBO) en volume (V) van enige regte prisma te bereken, gebruik ons die volgende algemene formules: (Let op dat H na die prisma se hoogte verwys.)
TBO = 2 × basisoppervlakte + sy-oppervlakte en V = basisoppervlakte × prismahoogte
Die volgende voorbeelde is belangrik. Dit is die vier houers wat jy uitgesny en gevou het. Let op dat elke afdeling van die berekening apart gedoen word en dan uiteindelik in die formule ingestel word.
Three charges q_{1}=+3\mu C, q_{2}=+6\mu C and q_{3}=+8\mu C are located at (2,0)m (0,0)m and (0,3) coordinates respectively. Find the magnitude and direction acted upon q_{2} by the two other charges.Draw the correct graphical illustration of the problem above showing the direction of all forces.
To solve this problem, we need to first find the net force acting on charge q_{2}. The magnitude of the force exerted by q_{1} on q_{2} is given by F=\frac{kq_{1}q_{2}}{r^{2}} where k is the Coulomb constant, q_{1} and q_{2} are the charges of the particles, and r is the distance between them.
Muhammed
What is the direction and net electric force on q_{1}= 5µC located at (0,4)r due to charges q_{2}=7mu located at (0,0)m and q_{3}=3\mu C located at (4,0)m?
Capacitor is a separation of opposite charges using an insulator of very small dimension between them. Capacitor is used for allowing an AC (alternating current) to pass while a DC (direct current) is blocked.
Gautam
A motor travelling at 72km/m on sighting a stop sign applying the breaks such that under constant deaccelerate in the meters of 50 metres what is the magnitude of the accelerate
velocity can be 72 km/h in question. 72 km/h=20 m/s, v^2=2.a.x , 20^2=2.a.50, a=4 m/s^2.
Mehmet
A boat travels due east at a speed of 40meter per seconds across a river flowing due south at 30meter per seconds. what is the resultant speed of the boat
which has a higher temperature, 1cup of boiling water or 1teapot of boiling water which can transfer more heat 1cup of boiling water or 1 teapot of boiling water explain your . answer
I believe temperature being an intensive property does not change for any amount of boiling water whereas heat being an extensive property changes with amount/size of the system.
Someone
Scratch that
Someone
temperature for any amount of water to boil at ntp is 100⁰C (it is a state function and and intensive property) and it depends both will give same amount of heat because the surface available for heat transfer is greater in case of the kettle as well as the heat stored in it but if you talk.....
Someone
about the amount of heat stored in the system then in that case since the mass of water in the kettle is greater so more energy is required to raise the temperature b/c more molecules of water are present in the kettle
pratica A on solution of hydro chloric acid,B is a solution containing 0.5000 mole ofsodium chlorid per dm³,put A in the burret and titrate 20.00 or 25.00cm³ portion of B using melting orange as the indicator. record the deside of your burret tabulate the burret reading and calculate the average volume of acid used?
No. According to Isac Newtons law. this two bodies maybe you and the wall beside you.
Attracting depends on the mass och each body and distance between them.
Dlovan
Are you really asking if two bodies have to be charged to be influenced by Coulombs Law?
Robert
like charges repel while unlike charges atttact
Raymond
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