CRS OBJ:
1-10: ABBCBABDCA
11-20: CBABBABDBA
21-30: DCCABBBDBC
31-40: DCABCADDDD
41-50: AADCACCADB
===================
Irs Obj
01-10 BBDBAAABAA
11-20 BBCBDABBAC
21-30 CADAAAACCC
31-40 CCCBCCCBCC
Completed.!!!!!

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PHYSICS OBJ AND THEORY

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9bii)
Slope = ∆F/∆e
Point 1: (Extension = 10.0 cm, Force = 2.0 N)
Point 2: (Extension = 30.0 cm, Force = 6.0 N)
S = (6.0-2.0)/(30.0-10.0)
S = 4.0/20.0
S = 0.2N/cm

(9biii)
The slope represents the spring constant
k, indicating the stiffness of the spring.

(9c)
Given:
Mass (m) = 40 g = 0.04 kg
Elastic constant (k) = 350 N/m
Extension (x) = 4 cm = 0.04 m
Elastic potential energy stored in the rubber(E) = (1/2)kx²
= (1/2) × 35 × (0.04)²
(1/2) × 35 × 0.0016
= 0.28 J

This energy is converted to kinetic energy of the stone:
K = (1/2)mv² = E
(1/2)mv² = 0.28 J
(1/2) × 0.04 × v² = 0.28
v² = (2 × 0.28)/0.04
v² = 14
v = √14
v = 3.74 m/s

(9d)
(i) Bulk modulus
Bulk modulus is a measure of a material's resistance to compression under uniform pressure.
(ii) Energy gained by the wire
Given:
Mass (m) = 2.5g = 0.0025kg
Extension (x) = 2cm = 0.02m
g = 10m/s²
Force (F) = mg = 0.0025 × 10 = 0.025N
Energy gained (E) = (1/2)Fx
= (1/2) × 0.025 × 0.02
= 2.5 × 10⁻⁴ J

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PHYSICS OBJ
01-10: BBCDACBCBB
11-21: ADBBCCDCBC
21-30: CDABAAADBA
31-40: ABABCCBCDA
41-50: BBACBBCABC
*COMPLETED*✅

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Chemistry theory and obj question dropped

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WAEC UP NEXT 22ND MAY 2025

CHEMISTRY THEORY & OBJ #1500

LITERATURE PROSE & OBJ #1000

GET IT BY 11PM TONIGHT

WHATSAPP 09056863938 ONLY

1500


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CHEMISTRY OBJ ANSWERS

01-10 BCAACBBBCC
11-20 CCAAAABABA
21-30 BBABDADCAC
31-40 ABDBCDBBAA
41-50 BCBACCDDDB
51-60 ADACCCDAAD

COMPLETED ✅✅✅

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