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WAEC GCE NOV/DEC 2017/2018 QUESTIONS AND ANSWERS

Discussion in 'Waec, Weac, Neco, Waec GCE And Neco GCE' started by Peptidebond, Jul 2, 2017.

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    Chiamaka1 Member

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    Amaka, same way you subscribed for maths
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    MATHS ESSAY

    4)
    area of shaded portion=area of sector pqs-area of triangle pqs
    =(θ/360*pieR^2)-(1/2r^2sinθ)
    (60/360*22/7*14^2)-(1/2*14^2sin60)
    102.2667-84.8705
    =17.796
    =17.8cm^2
    ================================
    7a)
    log6Y+2log6x=3
    log6y+log6x^2=3
    log6(x^y)=3
    x^y=6^3
    x^y=216
    y=216/x^2
    7bi)
    draw
    7bii)
    percentage who liked football and volleyball but not boxing=20
    ii)percentage who liked exactly two part=40+20+10=70
    iii)summing
    5+5+10+40+20+10+5+x=100%
    x+95=100
    x=100-95=5%
    percentage who like none=5%
    ================================
    9a)
    diagrams
    9b)
    |CA|^2=300^2+100^2-2(300)(100)cos 120degree
    =9000+10000-60000(-cos60)
    =100000+30000
    KA|^2=130000
    KA|=√130000
    KA|=360.56KM
    9bii)
    Using sine rule
    |AB|/sinC=|AC|/sin120
    100/sinC=360.56
    sinC=100sin60/360.56
    sinC=0.2402
    C=sin^-1(0.2402)=13.90degree
    bearing=270+13.90=283.9degree = 284degree
    ================================
    1a)
    [2(1/2)+1(3/4)/1(2/5)]/[2(1/4)-1(1/2)]
    [5/2+7/4/7/5]/[9/4-3/2]
    =[5/2+7/4*5/7]/[9/4-3/2]
    [5/2+5/4]/[9/4-3/2]
    =[(10+5/4)/(9-6/4)]
    =(15/4)/(3/4)
    =15/4*4/3
    =5
    1b)
    From tan60/1*PR/3root2
    PR=3root2*1/root3
    =3root2/root3*root3/root3
    =3root6/3=root6
    From sin45/1*root6/x
    x=root6*root2
    =root12
    =root4*3
    =2root3
    ================================
    5a)
    Daigrams
    5b)
    Median = ((N + 1) / 2)th item Where: N = 31
    40, 40, 40, 40, 40, 40, 40, 41, 41, 41, 41, 42, 42, 42, 42, 42, 42, 43, 43, 44, 44, 44, 44, 45, 45, 46, 46, 46, 46, 46, 46
    Data table
    F 7 4 6 2 4 2 6
    X 40 41 42 43 44 45 46
    Median = ((31 + 1) / 2)th item
    Median = (32 / 2)th item
    Median = 16th item
    Arranging the discrete data in ascending order, We go get:
    40, 40, 40, 40, 40, 40, 40, 41, 41, 41, 41, 42, 42, 42, 42, 42, 42, 43, 43, 44, 44, 44, 44, 45, 45, 46, 46, 46, 46, 46, 46
    Median = 42
    ================================
    3)
    2x/1 - 2/5 y = 2y/1
    LCM = 5
    10x - 2y = 10y
    10x = 10y + 2y
    10x = 12y
    X = 12y/10 -------------1
    2x - 2/5y + 2 1/4x + 1/2y + 2y = 180
    LCM = 20
    40x - 8y + 45x + 10y + 40y = 3600
    40x + 45x + 10y - 8y + 50y = 3600
    95x + 42y = 3600
    95 (6y/5) + 42y =3600
    19(6y) + 42y = 3600
    114y + 42y = 3600
    156y = 3600
    Y = 3600/156
    Y = 23.08
    Sub for y in eq 1
    X = 6(23.08)/5
    X = 27.69
    ================================
    11)
    A={4,5,6,7,8,9,10,11,12}
    B={10,11,12,13,14,15,16,17,18,19,20,21,22,23,24}
    M={4,8,12,11,6,20,24,28}
    a)prob (A)=3/7
    b)prob( B)=7-4/7=3/7
    c)prob(AuB)=n(AuB)/n(m)
    but (AuB)={4,8,12,11,6,20,24,}
    prob(AuB)=6/7
    ==============
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    MATHS ESSAY

    4)
    area of shaded portion=area of sector pqs-area of triangle pqs
    =(θ/360*pieR^2)-(1/2r^2sinθ)
    (60/360*22/7*14^2)-(1/2*14^2sin60)
    102.2667-84.8705
    =17.796
    =17.8cm^2
    ================================
    7a)
    log6Y+2log6x=3
    log6y+log6x^2=3
    log6(x^y)=3
    x^y=6^3
    x^y=216
    y=216/x^2
    7bi)
    draw
    7bii)
    percentage who liked football and volleyball but not boxing=20
    ii)percentage who liked exactly two part=40+20+10=70
    iii)summing
    5+5+10+40+20+10+5+x=100%
    x+95=100
    x=100-95=5%
    percentage who like none=5%
    ================================
    9a)
    diagrams
    9b)
    |CA|^2=300^2+100^2-2(300)(100)cos 120degree
    =9000+10000-60000(-cos60)
    =100000+30000
    KA|^2=130000
    KA|=√130000
    KA|=360.56KM
    9bii)
    Using sine rule
    |AB|/sinC=|AC|/sin120
    100/sinC=360.56
    sinC=100sin60/360.56
    sinC=0.2402
    C=sin^-1(0.2402)=13.90degree
    bearing=270+13.90=283.9degree = 284degree
    ================================
    1a)
    [2(1/2)+1(3/4)/1(2/5)]/[2(1/4)-1(1/2)]
    [5/2+7/4/7/5]/[9/4-3/2]
    =[5/2+7/4*5/7]/[9/4-3/2]
    [5/2+5/4]/[9/4-3/2]
    =[(10+5/4)/(9-6/4)]
    =(15/4)/(3/4)
    =15/4*4/3
    =5
    1b)
    From tan60/1*PR/3root2
    PR=3root2*1/root3
    =3root2/root3*root3/root3
    =3root6/3=root6
    From sin45/1*root6/x
    x=root6*root2
    =root12
    =root4*3
    =2root3
    ================================
    5a)
    Daigrams
    5b)
    Median = ((N + 1) / 2)th item Where: N = 31
    40, 40, 40, 40, 40, 40, 40, 41, 41, 41, 41, 42, 42, 42, 42, 42, 42, 43, 43, 44, 44, 44, 44, 45, 45, 46, 46, 46, 46, 46, 46
    Data table
    F 7 4 6 2 4 2 6
    X 40 41 42 43 44 45 46
    Median = ((31 + 1) / 2)th item
    Median = (32 / 2)th item
    Median = 16th item
    Arranging the discrete data in ascending order, We go get:
    40, 40, 40, 40, 40, 40, 40, 41, 41, 41, 41, 42, 42, 42, 42, 42, 42, 43, 43, 44, 44, 44, 44, 45, 45, 46, 46, 46, 46, 46, 46
    Median = 42
    ================================
    3)
    2x/1 - 2/5 y = 2y/1
    LCM = 5
    10x - 2y = 10y
    10x = 10y + 2y
    10x = 12y
    X = 12y/10 -------------1
    2x - 2/5y + 2 1/4x + 1/2y + 2y = 180
    LCM = 20
    40x - 8y + 45x + 10y + 40y = 3600
    40x + 45x + 10y - 8y + 50y = 3600
    95x + 42y = 3600
    95 (6y/5) + 42y =3600
    19(6y) + 42y = 3600
    114y + 42y = 3600
    156y = 3600
    Y = 3600/156
    Y = 23.08
    Sub for y in eq 1
    X = 6(23.08)/5
    X = 27.69
    ================================
    11)
    A={4,5,6,7,8,9,10,11,12}
    B={10,11,12,13,14,15,16,17,18,19,20,21,22,23,24}
    M={4,8,12,11,6,20,24,28}
    a)prob (A)=3/7
    b)prob( B)=7-4/7=3/7
    c)prob(AuB)=n(AuB)/n(m)
    but (AuB)={4,8,12,11,6,20,24,}
    prob(AuB)=6/7
    ==============

    12a)
    9x+10=28
    9x=28-10
    9x=18
    x=18/9
    x=2
    -6x+4y=4
    -6(2)+4y=4
    4y=4+12
    4y=16
    y=16/4=4
    ================================
    13ai)
    x*y=x+y+2xy
    2*3=2+3+2*2*3
    =5+12=17
    (2*3)*5=17*5
    17+5+2*17*5
    =22+170=192
    ii)x*7=x+7+2(x)(7)
    =x+7+14x
    =15x+7
    x*5=x+5+2(x)(5)
    =x+5+10x
    =11x+5
    (x*5)*2=(11x+5)*2
    =11x+5+2+2(11x+5)^2
    =11x+5+2+44x+20
    =55x+27
    15x+7=55x+27
    15x-55x=22-7
    -40x=20
    x=-20/40=-1/2
    ================================
    2a)
    A=P(1+r/100)^n
    A=2,205,000
    n=2
    r=5%
    p=?
    2205000=p(1+5/100)^2
    2205000m=p(1.05)^2
    p=2205000/(1.05)^2=N2,000,000
    2b)
    area=30cm^2 h/b=3/2
    area=1/2bh
    b=2h/3
    30=1/2*2h/3*h
    2h^2=30*2*3
    h^2=30*2*3/2
    h^2=90
    h=√90=9.49m

    Ama101 Member

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    Pls help me with maths obj,pls.

    Ama101 Member

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    God bless you for the theory post.
    Peptidebond likes this.
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    OBJ LOADING

    Ama101 Member

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    a friend just sent this to me i dont know if its legit OBJ:
    1–10 CBACDBBDCA

    11–20 ACCBDADBDC

    21–30 BCADDABDAC

    31–40 DACBBDACCD

    41–50 ACCBDABBCD

    Check before you shade
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    it's for Ghana
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    Maths obj

    1-10ccaacbadab
    11-20
    21-30

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    Ok
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    Keep Refreshing this page for the remaining

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    The invigilators are looking at me with evil eyes; phones are not allowed in .
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    NIG Maths obj

    1-10.ccaacbadab
    11-20.b-ccb-----
    21-30.ddbdbcccba
    31-40.
    41-50.
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    ♧NIG Maths-Obj♧
    1CCAACBADAB
    11BCCCBBBBBB
    21DDBDBCCCBA
    31BDDDCCCCCD
    41ACADADCDDC

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    Geo- things on point......... Solution thins na
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    Geo theory answers

    1a)
    i)Availability of raw materials:In determining the location of an industry, nearness to sources of raw material is of vital importance. Nearness to the sources of raw materials would reduce the cost of production of the industry.
    ii)Availability of Labour:Adequate supply of cheap and skilled labour is necessary for and industry. The attraction of an industry towards labour centres depends on the ratio of labour cost to the total cost of production which Weber calls ‘Labour cost of Index’.
    iii)Proximity to Markets:Access to markets is an important factor which the entrepreneur must take into consideration. Industries producing perishable or bulky commodities which cannot be transported over long distance are generally located in close proximity to markets.
    iv)Transport Facilities:Transport facilities, generally, influence the location of industry. The transportation with its three modes, i.e., water, road, and rail collectively plays an important role.
    v)Power:Another factor influencing the location of an industry is the availability of cheap power. Water, wind, coal, gas, oil and electricity are the chief source.

    1b)
    i)Economic Growth: This is ensured through Government ownership of some organizations&Industries Are Required
    ii)Protection of the Country’s Sovereignty: industries is sometimes meant to protect the integrity and sovereignty of the country
    iii) Distribution of resources: industries is to ensure equitable distribution of the country’s resources to make profit
    iv)Social Cost: The establishment of some industries may involve a social cost, thus ownership of such organizations becomes imperative
    v)Prevention of competition: establishing industries is to make the role and duty to ensure the success of such corporations in the interest of the public

    2a)
    -Rural-Urban migration
    -Urban-Rural migration
    -Rural-Rural migration
    -Urban-Rural migration
    2b)
    -It brings social voices
    -It brings decline in production
    -It leads to high cost of living
    -It leads to congestion.

    2c)
    -Provisions of social amenities in rural areas
    - Provision of food
    -Provision of job
    -Provision of education.

    3a)
    The Economic Benefits is the improving of time performance and reduces loss and damage, thus reducing economic drag. In other words this is reducing the cost and time for existing passenger and freight movements increase transport’s contribution to economic growth

    3b)
    i)Shortage of capitalto maintain and establish new railways.
    ii)Old fashioned technologywhich make the railways carry less cargo compared to modern standard gauge
    iii)Poor government policieswhich favour development of roads instead of railways.

    3c)
    i)availability of capital to maintain and establish new railways.
    ii)New fashioned technology which make the railways carry less car go transport
    iii)availability of government policies to development railways..

    5b) -Low rainfall supports growth of short deciduous trees in the north
    - High and continuous rainfall give rise to evergreen vegetation
    - As rain decreases northwards, density of vegetation decreases

    6a)

    IMG-20170911-WA0011.jpg
    6b)
    - Air transport has indirectly helped the business to grow at and around the various airport terminals.
    - Another contribution of air transport which is closely related to job creation is that it aids mobility of labour.
    - Air transport promotes trade and commerce:- The improvement in air transport and communication worldwide has reduced the world to nothing more than a global village and it is bringing about a continuous increase in specialization.

    6c)
    -Uncertain and Unreliable:
    Air transport is uncertain and unreliable as it is controlled to a great extent by weather conditions. Unfavourable weather such as fog, snow or heavy rain etc. may cause cancellation of scheduled flights and suspension of air service.
    - Breakdowns and Accidents:
    The chances of breakdowns and accidents are high as compared to other modes of transport. Hence, it involves comparatively greater risk.
    -Large Investment. It requires a large amount of capital investment in the construction and maintenance of aeroplanes. Further, very trained and skilled persons are required for operating air service.
    -Unsuitable for Cheap and Bulky Goods:
    Air transport is unsuitable for carrying cheap, bulky and heavy goods because of its limited capacity and high cost.


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    Geo-Obj
    1ACBCCDBBCD
    11BBCCBAAADA
    21ACDCBCABCA
    31BCADCCBBCA
    41CCCCBADDCB
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    Biology 3 (Alternative to Practical Work) – 13.00 hrs. – 15.00 hrs.
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    BIOLOGY PRACTICALS
    (1ai)
    Pectoral Girdle(Scapula
    (1aii)
    I-Scapula
    II-Spine of scapula
    III-Coracoid
    IV -Glenoid cavity
    V-Acronium
    VI-Metacromion
    (1b)
    (i) Humerus
    (ii) Ball and socket Joint
    (1c)
    (i) It holds the upper limb to the axial skeleton
    (ii) Attachment of muscles
    2(a) A- symbiotic relationship
    B- predation
    C- symbiotic relationship
    D-epiphytsm
    E-parasitism
    (2b) beneficial | harmful
    A. B
    C. E
    D
    4a)
    I)Operculum
    II)Hind limb
    III)Tympanum
    IV)Trunk
    V-eye
    VI-lateral line
    VII-dorsal fin
    VIII-caudal fin
    4bi)
    Tabulate
    Vertebra
    i)they are poikilothermic or cold-blooded animal
    ii)they have fins which are used for movement in water
    iii)they have naked or moist and glandular skin with no external scales
    iv)they do not show parental care
    v)they have a two chambered heart
    Vertebrate B
    i)they are aquatic animal
    ii)they have gills which used for gaseous exchange
    iii)the skin is covered by scales but few are with out scale
    iv)they show parental care for young onces
    v)they have three chambered heart
    4bii)
    i)they both posses scale
    ii)they both posses eyes
    iii)they both posses respiratory organs
    4ci)
    A-amphibian
    B-pisces
    4cii)
    i)they posses fins which are used for movement
    ii)they posses swim bladder which posses swim bladder which enables them to maintain buoyancy in water
    iii)they posses lateral line system for detectin of vibration
    4a)
    I)Operculum
    II)Hind limb
    III)Tympanum
    IV)Trunk
    V-eye
    VI-lateral line
    VII-dorsal fin
    VIII-caudal fin
    4bi)
    Tabulate
    Vertebra
    i)they are poikilothermic or cold-blooded animal
    ii)they have fins which are used for movement in water
    iii)they have naked or moist and glandular skin with no external scales
    iv)they do not show parental care
    v)they have a two chambered heart
    Vertebrate B
    i)they are aquatic animal
    ii)they have gills which used for gaseous exchange
    iii)the skin is covered by scales but few are with out scale
    iv)they show parental care for young onces
    v)they have three chambered heart
    4bii)
    i)they both posses scale
    ii)they both posses eyes
    iii)they both posses respiratory organs
    4ci)
    A-amphibian
    B-pisces
    4cii)
    i)they posses fins which are used for movement
    ii)they posses swim bladder which posses swim bladder which enables them to maintain buoyancy in water
    iii)they posses lateral line system for detectin of vibration
    =====
    4di)
    i)they posses pigment on their skin which act as a camourplage against it predators
    ii)they posses scales on the skin which reduce water loss from the body against its predators
    iii)they posses digits or claws use for self defence
    4dii)
    i)water
    ii)land
    4diii)
    i)they are use as food
    ii)they are use for the preparation of local medication
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    BIOLOGY
    1a) i).white blood cells
    lipids

    aii)
    i.responsible for the removal of interstitial fluid from tissues

    ii.It transports white blood cells to and from the lymph nodes into the bones

    1b)
    i).oxygen
    ii.excretory products, such as carbon dioxide, urea, etc
    iii.Hormones

    ii)Plants absorb water from the soil through their root hairs which are surrounded by soil water in the soil. The soil water is a more dilute solution than the soil sap inside the root hair hence water molecules move from soil water into the root hairs by the process of osmosis. Mineral salts are absorbed through the cell membrane often from a zone of their low concentration in the soil to a zone of their higher concentration in the cells of the roots through active absorption.

    1c)
    i).Placentation is the arrangement of ovules within an ovary. The part of the ovary where the ovules are attached is called a PLACENTA.

    ii).
    Marginal placentation
    Parietal placentation
    Axile placentation
    Free central placentation

    6bi)
    Irritability is the Abnormal sensitivity or excessive responsiveness, as of an organ or body part, to a stimulus.

    6ci) Tillage is the agricultural preparation of soil by mechanical agitation of various types, such as digging, stirring, and overturning. Examples of human-powered tilling methods using hand tools include shovelling, picking, mattock work, hoeing, and raking
    ===============
    6ei)Oviparity is the producing eggs that mature and hatch after being expelled from the body, as birds, most reptiles and fishes, and the monotremes.

    ii) Viviparity is the retention and growth of the fertilized egg within the maternal body until the young animal, as a larva or newborn, is capable of independent existence. The growing embryo derives continuous nourishment from the mother, usually through a placenta or similar structure.

    6f)

    Rhizobium turns the free Nitrogen in air into

    - Ammonia
    - Nitrate
    4ai)Sickle cell anaemia; this is a group of disorders that affects hemoglobin , the molecule in red blood cells that delivers oxygen to cells throughout the body. People with this disorder have atypical hemoglobin molecules called hemoglobin S, which can distort red blood cells into a sickle , or crescent, shape.
    Signs and symptoms of sickle cell disease usually begin in early childhood. Characteristic features of this disorder(sickle cell anaemia) include a low number of red blood cells ( anemia ), repeated infections, and periodic episodes of pain.


    4d)
    i)The presence of fossils;
    preserved impressions of organisms/ ancestors that lived in the past
    ii)The vestigial structures;
    an organ that serves no purpose anymore but did with a common ancestor
    iii)The presence of homologous structures;how the bones or parts of an animal are the same but in different proportions
    iv)The comparative embryology; in development, embryos look similar until they differentiate later on
    V)The molecular evidence ;comparing DNA; to trace differences and similarities in DNA and other molecules, making the species more or less related.

    3a)
    i)community is an interacting group of various species in a common location. For example, a forest of trees and undergrowth plants, inhabited by animals and rooted in soil containing bacteria and fungi, constitutes
    ii)populationis all the organismsof the same group or species, which live in a particular geographical area, and have the capability of interbreeding.
    iii)ecological niche describes how an organism or population responds to the distribution of resourcesand competitors (for example, by growing when resources are abundant
    3b)
    i)All living organism need energy and nutrient in an environment in other to sustainlife
    ii)All living organisms need energy to maintain homeostasis
    3ci)
    i)Lack of enough food
    ii)out break of disease
    iii)lack of money
    3cii)
    i)Exposed the soil into erosion
    ii)Cause loss of nutrient into the plant
    iii)Cause low yield
    3d)
    To protect against soil erosion; improve the soil by adding organic matter, nutrients, and stability; and scavenge leftover nutrients. In other words Cover crops are used as ground cover, mulches, green manure, nurse crops, smother crops, and forage and food for animals or human

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    PHYSICS

    4a)
    Time (t)=4secs
    V=0ms^-1
    pie call,V^2=U^2+2gh
    O^2=20^2+2(-10)h
    0=400-20h
    20h/20=400/20
    h=20m
    4b)
    the magnitude of the initial speed (u) is equal to the magnitude of the height attained
    ================================
    6)
    P-type semi conductors are created by dopung an intrisic semi conductor with acceptor impurities.
    ================================
    2)
    i)used ineye surgery
    ii)used for the precise cutting of flat material
    iii)used for measuring distances
    ================================
    3)
    i)iron
    ii)steel
    iii)nickel
    ================================
    5)
    i)availability of constant sunlight
    ii)nearness to power grid
    iii)must be closer to both skilled and unskilled labour

    8a)
    force is a push or pull which changes a body state of rest or of uniform motion in a straight time
    8aii)
    i)constant force
    ii)field force
    8bi)
    bodies are streamline to make sure that the surface are in head on contact with the fluid during motion
    8bii)
    i)aircrafts
    ii)ship
    8ci)
    diagrams
    8cii)
    mass of bucket =1.3kg
    speed of loop(v)=7.0ms^-1
    radius(s)=1.2m
    (x)acceleration (a)
    a=rw^2
    v=rw
    w=v/r
    a=r(v/r)^2
    a=rv^2/r^2
    a=v^2/r
    a=(2.0)^2/1.2=49/1.2
    a=40.8ms^-2
    b)net force (F)
    F=F1-Fg
    F=Ma-Mg
    F=(1.3*40.8)-1.3*19i
    F=53-04-13
    F=40N

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    11a)
    i)hammering method
    ii)electrical method
    11b)
    electrical line of force- this is the line drawn such that the tangent to it at a given point is in the direction of the force acting on a small positive change at the point
    11c)
    diagrams
    at beaker point
    potential different accross Q=p d are p
    IiQ=I2------eq1
    p d across = p d across R
    IiS=I2R------eq2
    divide eq 1by2
    IiQ/IiS=I2P/I2R
    R=SP/Q
    11d)
    resistivity- this is the resistance of a wire unit length and unit cross-sectional area
    i;e P=RA/L
    11e)
    Given: F = 50Hz
    L = 10 mH (10×10^-³H)
    Inductive reactance, XL = 2πFL
    = 2×3.14×50×10×10^-³
    = 3..14 ohms
    11f)
    If voltmeter reads 12V, over the 500 ohms, then using I = V/R
    current in the circuit is I = 12/500
    I = 0.024A
    Hence over the 200 ohms, V = IR
    V = 0.024×200
    V = 4.8 VOLTS
    Voltmeter will read 4.8 volts
    ================================
    9a)
    i)Poor conductors of heat application
    ii)Coat (Trapped air in between the fabric
    9b)
    Renewable: Water, corn stalk
    Non-renewable: coal, natural gasgas
    9c)
    A closed is a physical system so far removed from other systems that it does not interact with them.
    9d)
    heat required to melt = mc∆ + ml
    Not sure of the temperature given, if it is -3°C or -5°C, However, *Using -3°C*
    Heat = (0.1×2200×3)+(0.1×2.26×10^6)
    = 226660 J
    9ei)
    Given: 75 heart beats in 1 min
    Therefore in 1 hour, we have (75×60) heart beats = 4500 heart beats
    Also given: 2J of energy is expended in 1 heart beat
    Therefore for 4500 heart beats, energy expended = (4500×2) J = 9000J
    9eii)
    Energy = mc∆
    9000 = (0.25×4200×∆)
    9000 = 1050∆
    ∆ = 9000/1050
    Temperature rise = 8.57°C
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