PHYSICS PRACTICALS AREA OF CONCENTRATIONπŸ’―πŸ’―

🚬πŸŒͺ️πŸŒͺ️

NUMBER 2

Make sure you use almost that of your teacher value rapport with him or her to find out .

Note yourπŸ‘‡πŸ‘‡

**Observation W = ????
**IΒ°,d(cm),L(cm),e/Β°,Ξ¦/Β°,Ξ¦ = I + e/Β° any one asked 🚬πŸŒͺ️πŸŒͺοΈπŸ’―
**Scale eg. Let 2cm represent 0.1unit on cos Ξ¦ axis
Let 4cm represent 0.1unit on sin I axis
Depending on what you are ask to place on the graph πŸ‘†πŸ‘†
**Slope
**Intercept on the vertical/horizontal axis if asked.πŸ’―
*Make sure your graph, scale, and, intercept are well drawn and taken

#PRECAUTIONS:

I. I ensured neat traces by using sharp pencil.
II. I ensure that the pins are vertically erect.
III. I ensure that the pins are well spaced which is 4cm apart.
IV. I avoided parallax errors when reading protractor.
V. I avoided error due to parallax when reading the metre rule.
Choose any asked

Study this incase if asked ..πŸ‘‡πŸ‘‡ 🚬πŸŒͺ️πŸŒͺ️

A. State the conditions necessary for total internal reflection of light to occur.

B. The Critical angle for a transparent substance is 39Β°. Calculate the refractive index of the substance.
C. State snell’s law
D. Calculator the critical angle for a water – air interface [ refractive index of water = 4/3 ]
WORKINGS ✍🏿✍🏿

A. I. Light must be traveling from a denser medium to a less dense medium.
II. The angle of incidence in the dense medium must be greater than the critical angle.

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B. Refractive index πŸ‘Ž = 1/sinC = 1/Sin39Β° = 1/0.6293 Therefore, n = 1.59

C. Snell’s law state that the ratio of the angle of refraction is a constant for a given pair of media.

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D. 1/SinC = a ^N w Therefore, SinC = 3/4 = 0.75 Hence C = Sin inverse 0.75 = 48.6

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MT πŸ’―πŸ’―

Study It area of concentration no. 2

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NUMBER 3. STUDY WELL πŸ’―πŸ’―πŸ’― OJCA✍🏿✍🏿✍🏿

Same as num 2 rapport with your teacher to get almost the value of his πŸ’―πŸ’―

## Your Observation

Make sure your readings, graph, scale and intercept are well drawn and taken

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Precautions _OJCA

I. I ensured Tight connections.
II. I noted the zero error on the ammeter and voltmeter.
III. I avoided error due to parallax when reading the ammeter and voltmeter.
IV. I ensured that I remove the key when readings are not been taken to prevent the cells from running down.

Study this incase if askedπŸ‘‡πŸ‘‡

A. Two method by which electric current can be produced

B. Explain why E. M. F. of a cell is greater than the PD. across the cell when it is supplying current though an external resistance

C. Define potential difference between two points in an electric circuit.

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#WORKINGS ✍🏿✍🏿

A. Electric current can be provided using:
I. Chemical Cells
II. Generator
III. Solar Cells
IV. Gas Turbines
V. Hydro Power
VI. Wind Mills
Vii. Geothermal
Viii. Nuclear Power

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B. E. M. F. (E) is the Work done across external resistance (R) and internal resistance of the cell (r) While (PD) [v] is the work done across external resistance (R) only, hence, emf of a cell is greater than the PD across the cell.
Therefore, E = V + Ir

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C. The potential difference between two points in an electric circuit is the Work done in Joules (J) in moving a charge of one Coulomb (c) from one point to the other..

Make sure you study thisπŸ’―πŸ’―πŸ‘†πŸ‘†

Goodluck

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