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Here m = 4 m = v/u = 4 v = 4u By lens formula 1/v - 1/u = 1/f 1/4u - 1/u = 1/20 (1-4)/4u = 1/20 -3 / 4u = 1/ 20 u = -15 So object should be placed at a distance 15cm from lens

Given information is,R1=10cm;R2=∞;f=80cmandμwater=43So, Focal length of the plano - convex lens, when it is immersed in water is,1fw=(μrel−1)[1R1−1R2]⇒180=(μmμwater−1)[110−1∞]⇒180=(μm43/−1)[110]⇒18=(3μm4−1)⇒3μm4=98⇒μm=32Therefore, refractive index of the material is1.5

Train accelerate from start for 10 sso velocity reached at 10s is a×t= 2×10=20m/s1) 20m/s is max velocity as after this the train runs on constant speed2) the train is retarded for 50s from this speed to rest so retardation is velocity /time=20/50=-0.4m/s23) distance travelled is total of distance travelled with accelaration+distance with constant speed+ distance with retardationdistance with acceleration is ut+1/2at2=1/2*2*10*10=100mdistance with constant speed is s×t=20×200=4000mdistance with retardation is ut+1/2at220×50-1/2×.4×50×50=1000-500=500mtotal distance =100+4000+500=4600m4) average velocity= total distance/total time= 4600/260=17.69m/s

When the bus is not moving the passengers are in the state of rest and they have inertia of rest. When the bus starts moving suddenly, the lower part of the body of passengers, which is in contact with the bus, come in motion, but upper part of their body tends to be in the state of rest and the passengers fall backward.

pls answer in one word

A very easy solution is to take a sample of colourless liquid and put on stove if it starts boiling exactly at 100 ºC then it is pure water. Any other colourless liquid such as vinegar always have different boiling point.

A very easy solution is to take a sample ofcolourless liquidand put on stove if it starts boiling exactly at 100 ºC then it ispure water. Any othercolourless liquidsuch as vinegar always have different boiling point.

option b is correct pgh.

P=force/area P=mg/area.P=density ×V×g/A. P=density×A×h×g/A.P=density×g×h

1/f={n-1)(1/R)...from lens maker formula1/f=(1.5-1)(1/15)1/f=0.5/15f=15/0.5=150/5=30cm

Focal length = 2m

Power = ?

so by using ,

P=1/f

v get... P = 1/-2= -0.5 D

Could u post the question once again completely, as in some words seem to be missing?

0°c= 273know 300k = 300-273= 27° c

2) 573k= 573-273= 300°c

As there is a balance bridgeso middle resistances will be truncatedso there are only 4 resistance of value 3 ohm will be therenow3+3//3+36//6= 6*6/12= 3 ohmanswer

sini = v1t/ACsinr = v2t/ACthus refractive indexnunu= sini/sinr = v1/v2*Huygens principle: Every point on a wavefront acts as a source of secondary wavelets. The geometric envelope of the waves gives the shape of the wavefront at a later time.

Its temperature increases because liquid moves against the viscous forces. Work done againstviscous forces is converted into heat

When liquid is shaken in a thermos flask, work is done on the liquid which gets converted into internal energy of the liquid. Hence, temperature of liquid increases.

option(a) is correct

AB=root((3-2)^2+(4-3)^2+(7-4)^2) =root(1+1+9)=root(11)BC=root((-2-3)^2+(-5-4)^2+(10-7)^2) =root(25+81+9)=root(115)3AB+2BC=3root(11)+2root(115)

the square root of the number 45369 is 213

213 is the right answer

the square root is 213

Given In Wheatstone network the resistance in cyclic order are p=10,Q=5 ,s=4 ,R=4 for the bridge to balance

We know that in a balanced Wheatstone bridge or network

p/q = r/s

10 / 5 = 4/4

2 = 1 is not balanced so we need to add 5 to q then the circuit will be balanced.

So 5 ohm should be connected in series with q for the bridge to be balanced.

option 3 is the correct answer .

as , when the magnitude of P will be greater than Q , the resultant will shift towards P.due to more contribution by P in the resultant.

wrong answer

upon moving The horizontal 5N force to connect with the tail

we get the angle as 180°-60° = 120°angle( __/ )

so, resultant is √5²+5²-2*5*5cos(120) = √5²+5²-5² = √5² = 5N

latent heat of ice=80 cal should be use

we know that

pressure = force /area

here force is nothing but the thrust exerted

so,

thrust = pressure x area

or

F = 1.013 x 10^5* 1.5

thus, thrust will be

F = 1.5195 x 10^5N

three forces of magnitude 7 PSP and 80 are acting at a particle and maintaining and barium the last two forces angle is 102

solve do

forces are (P+Q) and (P-Q)Resultant R is (3P^2 +Q^2)^0.5resultant for A,B : R^2 = A^2 +B^2 +2ABcosxR^2=(P+Q)^2 +(P-Q)^2 +2 (P+Q)(P-Q)cosx3P^2 +Q^2 =P^2 +Q^2 +2PQ+P^2 +Q^2-2PQ+2(P-Q)(P+Q)cosx3P^2 +Q^2=2P^2 +2Q^2+2(P-Q)(P+Q)cosxP^2 -Q^2 =(P^2 -Q^2)cosxcosx=1x=0vectors are parallel

Let R , Q are two vector which magnitude is equal also resultant of it also equal to either e.g | R | = | Q |

and resultant = R + Q magnitude of resulatant = | R | = | Q|

so, | R | = | R + Q |

| R | = √( | R|² + | Q|² +2|R | |Q| cos∅)

take square of both sides

R² = R² + Q² + 2|R| |Q| cos∅

put | R| =|Q|

R² = R² + R² +2R² cos∅

cos∅ = -1/2

∅ = 2π/3

2¥/3 is the right answer

Let R , Q are two vector which magnitude is equal also resultant of it also equal to either e.g | R | = | Q |

and resultant = R + Q magnitude of resulatant = | R | = | Q|

so, | R | = | R + Q |

| R | = √( | R|² + | Q|² +2|R | |Q| cos∅)

take square of both sides

R² = R² + Q² + 2|R| |Q| cos∅

put | R| =|Q|

R² = R² + R² +2R² cos∅

cos∅ = -1/2

∅ = 2π/3

The magnitude of each force is 40 N

thank you