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WAEC 2018/19 Physics Obj And Theory Answers Out

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WAEC 2018/19 Physics Obj And Theory Answers Out Online -WAEC 2018/19 Physics Obj And Theory Answers Out

1a

Strain is a force (a part of one’s body or oneself) to make an unusually great effort.

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2 )

i ) quantity of charge

ii) nature of element

iii ) Time

2bi

The boiling point of a liquid Is the point at which The vapour pressure is equal To atmospheric pressure

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3)

-Diamagnetic material

-Paramagnetic material

-Ferromagnetic material

4a)

Satellite are used for monitoring weather conditions 

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4b)

i)Rockets are used for space travel

ii)military uses, 

iii)launching satellites into space

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7(a). LASER stands for Light Amplification by Stimulated Emission of Radiation. 

  (b). A laser is a device that emits a beam of coherent light through an optical amplification process.

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10a) Diffraction refers to various phenomena that occur when a wave encounters an obstacle or a slit. It is defined as the bending of light around the corners of an obstacle or aperture into the region of geometrical shadow of the obstacle.

10bi

the angle of incidence beyond which rays of light passing through a denser medium to the surface of a less dense medium are no longer refracted but totally reflected.

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12a)

binding energy is the minimum energy that would be required to disassemble the nucleus of an atom into its component parts. These component parts are neutrons and protons, which are collectively called nucleons.

(12b) They have short wire length and high frequency.

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-They are highly penetrating

-They travel in straight line

-They don’t require material medium for their propagation

(12c) 

-It is used in production of electricity. 

-It is used to study and detect charges in genetic engineering. 

-It is used in agriculture. 

-It is used in treatment of cancer. 

(12di)

E = hf-hfo

but f = v/landa

E= v/landa.h – wo

Where wo = hfo = work function 

f= frequency 

landa = wavelength 

Hence 

hf = hfo – E

f = hfo – E/h

f = wo – E/h

Recall; that v = f landa

Therefore f = v/landa = 3×10^8/4.5×10-7

=3/4.5 × 10^8+7

=6.6×10^14Hz

f = 6.6×10^14Hz

(12dii) 

E = hf

=6.6×10^-34 × 6.6×10^14Hz

=43.56×10^-20J

(12diii) 

Energy of the photoelectron E = hf – vo

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=Energy of incident electron – work function 

=4.356×10^-19J – 3.0×10^-19J

=1.356×10^-19J

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