Acurate-Phy-Obj

1ABABDBBBBC

11CCCDADCABA

21BCBCCBDACD

31ADBCDACBBB

41CCBAACACCA

SECTION-B ANS 3 ONLY

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

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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

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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SECTION-A ANS 5 ONLY

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SECTION-A ANS 5 ONLY

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(1)

(a)

Horizontal speed just before it strikes floor = ucosθ

= 10cosθ * 1.0

= 10 m/s

(b)

t = √(2g/h)

= √(2*10*1)

= √20

= 4.47s

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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

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6)

P-type semi conductors are created by dopung an intrisic semi conductor with acceptor impurities.

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2)

i)used ineye surgery

ii)used for the precise cutting of flat material

iii)used for measuring distances

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3)

i)iron

ii)steel

iii)nickel

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5)

i)availability of constant sunlight

ii)nearness to power grid

iii)must be closer to both skilled and unskilled labour

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Keep Refreshing

Typing/Solving

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Completed
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ALWAYS REFRESH THE PAGE BY UP DRAGGING

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