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are FOUR FACTORS that influence the resistance in a conductor. Thickness (cross sectional AREA of the wire), length, and temperature. The fourth factor is the CONDUCTIVITY of the material that is usedgmark as BRAINLIST

ts true that the work DONE is ZERO as there was no DISPLACEMENT done by the FORCE applied by the CHILD but the energy got stored as potential or kinetic.

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end...Here is your ANSWER..CALORIFIC value of PETROL =45.5 MJ/kgHOPE that it will help you

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The LENGTH of ROD = 8 m .A WEIGHT of 10 kg is put at A.Fulcrum is 3 m away from B .Let the weight of load at B be x .Distance of A from fulcrum = 8 m - 3 m                                              = 5 mBy principle of MOMENTS :SUM of clockwise moments = sum of anti-clockwise moments .Hence we can write that :5 m × 10 kg = x × 3 m= > 50 kg m = 3 x m= > 3 x = 50 kg= > x = 50/3 kg= > x = 16.666 kg= > x = 16.67 kgANSWER :The load at B would be 16.67 kg____________________________

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initial velocity ( u ) = 40 km/hr                               = 40 × 1000 m / 3600 s                               = 40 × 5 / 18 m/s                               = 200 / 18 m/s .FINAL velocity ( V ) = 0 m/sDistance ( S ) = 40 m .v² = u² + 2 a S= > 0 = u² + 2 a S= > ( 200 / 18 )² + 2 × 40 × a = 0= > 80 a = - 40000 / 324= > a = - 500 / 324When initial velocity = 80 km/hr                                   = 80 × 5/18 m/s                                   = 400/18 m/sv² = u² + 2 a S= > u² + 2 a S = 0= > ( 400 / 18 )² + 2 × - 500 / 324 × S = 0= > 1,60,000 / 324 - 1000 / 324 S = 0= > 1,6000 / 324 S = 40,000 / 324= >  S = 1,60,000 / 1000= > S = 160 mANSWER :The distance WOULD be 160 m

tors on which the amount of work done depends are force and DISPLACEMENT. When we push the walls of our HOUSE force is applied but no work is done because there is no displacement in the POSITION of the wall. When luggage from the 3rd floor to the GROUND floor the work done is positive because the force is applied in the for of gravity... hope it will help you.

lectric ENVISION is not EXPLAINED by HUYGENS WAVE THEORY

centage increase in DISPLACEMENT during sixth second COMPARED to that in 5th second is 44%Explanation : LET the acceleration of the particle is a m/s²From the laws of motion we know that,S = ut + at²/2since u = 0, hence S = at²/2Displacement at the end of 5 second,S₅ = a(5)²/2 = 12.5aDisplacement at the end of 6 second,S₆ = a(6)²/2 = 18aIncrease in displacement,S = 18A - 12.5a = 5.5aHence percentage increase in displacement= (5.5/12.5) x 100= 44%Hence The percentage increase in displacement during sixth second compared to that in 5th second is 44%

re! First of all such a good question! Had to use so many concepts to get here…Here goes,This is a diagram of the situation. Now, NEXT will be a free body diagram of the SOLID Cylinder.Now, by LOOKING at the situation, we can say that the normal force will be balanced by Mgcosβ, while friction will APPLY in the opposite direction of the motion. We know that rolling is a motion of the body made possible due to torque, so let's calculate the torque due to all the forces here.(consider radius of the cylinder to be ‘r’ and point of rotation to be the Centre of the mass of the sphere)Torque due to Normal force and Mg (and all its components) will be zero (because they pass through the point of rotation). But, torque due to friction would be =(f*r) (force by perpendicular distance).Also, we know thatTorque = Moment of inertia*Angular accelerationThus we getτ=IαBut τ=f*r and I(for solid cylinder is) = 1/2 M*r² and also α=a/rThus, we get, after substituting the values for f, I, and a;μ(Mgcosβ) * r = 1/2 Mr² * gsinβ/rSolving which gives usμ=1/2 tanβ

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re!The value of one main SCALE division = 1/20 cmThe NUMBER of divisions on VERNIER scale = 25L.C = Value of one main scale division / Number of divisions on vernier scale       = (1/20) / 25 = 1/500 cm = 0.002 cmHope It Helps You!

erberation — the PERSISTENCE of sound due to REPEATED reflection is CALLED REVERBERATION.

trifugal FORSE is GREATEST on this EARTH and minimimum NEAR POLES.

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E HERE IS YOUR ANSWER.AS WE KNOW IF THERE IS ANOTHER CHARGES THEN THEY ATTRACT EACH OTHER . BUT WHEN THE CHARGES ARE THEY REPEL EACH OTHER . FOR EX. WE RUB TWO BALLONS IF THEY HAVE SAME CHARGE THEY REPEL EACH PTHER BUT IF THE THE CHARGE IS DIFFRENT THE ATTRACT EACH OTHER.HOPE IT WILL HELP YOU . I HAVE MANY OTHER EX.

re!Given,Size of bacteria = 1 mewwe know,1 MEW = 10^-6 MNO. of bacteria = 1 m / 10^-6 mNo. of bacteria = 10^6 HOPE It HELPS You!

e the HEIGHT of the object.we knowPE = MGHH = PE / MG= 900/ (100x10)= 0.9 m

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the previous series resistor circuit, in a parallel resistor network the circuit current can take more than one path as there are MULTIPLE paths for the current. Then parallel circuits are classed as current dividers.Since there are multiple paths for the supply current to flow through, the current may not be the same through all the branches in the parallel network. However, the voltage drop across all of the RESISTORS in a parallel resistive network IS the same. Then, Resistors in Parallel have a Common Voltage across them and this is true for all parallel connected elements.So we can define a parallel resistive circuit as one where the resistors are connected to the same two points (or NODES) and is identified by the fact that it has more than one current path connected to a common voltage source. Then in our parallel resistor example below the voltage across resistor R1 equals the voltage across resistor R2 which equals the voltage across R3 and which equals the supply voltage. Therefore, for a parallel resistor network this is given as:In the following resistors in parallel circuit the resistors R1, R2 and R3 are all connected together in parallel between the two points Aand B as shown.Parallel Resistor CircuitIn the previous series resistor network we saw that the total resistance, RT of the circuit was equal to the sum of all the individual resistors added together. For resistors in parallel the equivalent circuit resistance RT is calculated differently.Here, the reciprocal ( 1/R ) value of the individual resistances are all added together instead of the resistances themselves with the inverse of the algebraic sum giving the equivalent resistance as shown.Parallel Resistor EquationThen the inverse of the equivalent resistance of two or more resistors connected in parallel is the algebraic sum of the inverses of the individual resistances.If the two resistances or impedances in parallel are equal and of the same value, then the total or equivalent resistance, RT is equal to half the value of one resistor. That is equal to R/2 and for THREE equal resistors in parallel, R/3, etc.

________________Length - m ( metre ) Mass - kg ( kilogram ) TIME - s ( second ) Temperature - k ( kelvin ) Electric Current - A ( Ampere ) Luminous Intensity - cd ( Candela ) Quantity of matter - mol ( mole ) Supplementary Quantity, ______________________Plane angle - rad ( radian ) SOLID angle - sr ( STERADIAN )

re!The three REQUIREMENTS for selecting a UNIT of a physical quantity are:1. It should be POSSIBLE to DEFINE the unit without ambiguity.2 . The unit should be reproducible.3. The unit should be of convenient size.Hope it Helps YoU!

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y farming research PROJECTS, it is highly desirable to initiate on-farm research activities as early as possible. Naturally, before alley farming experiments can move on-farm, a certain amount of time will be devoted to purely on-station research. This is necessary for screening and evaluation of multipurpose trees, and for EXPERIMENTATION with management practices. However, AFNETA's strategy is for collaborating institutions to begin on-farm work as soon as researchers have found a "best-bet" prototype for local agroecological conditions.This unit begins by defining on-farm research (OFR) and explaining the important role of OFR in the testing and development of alley farming systems. It then describes the various types of OFR which alley farming research projects may entail. Detailed methodological guidelines and experimental designs are not provided here, but may be found in the Annex and in Volume 2.As part of AFNETA's Core Course in Alley Farming, this unit restricts its coverage to experimental and developmental OFR. These are the types of OFR that are directly involved in alley farming research projects such as those supported by AFNETA. However, it is assumed that researchers have conducted on-farm surveys at an EARLIER stage. Such diagnostic on-farm work is necessary to demonstrate the potential relevance of alley farming in the local area. For example, a station or a ministry may have found through on-farm surveys that local farmers face constraints such as declining soil fertility, soil erosion, POOR livestock nutrition, and/or land scarcity. Such findings would justify research into alley farming or other TECHNOLOGIES that could address the constraints.The pre-experimental type of on-farm activity is covered in the technical papers on Diagnosis and Design methods and socio-economic surveying Hope it helps u plz plz plz plz plz plz plz plz plz mark me as brainlist

re!Given,Mass of ONE atom of URANIUM = 3.9 * 10^-25 KG = 3.9 * 10^-22 g[ 1 kg = 1000 g]NUMBER of atoms in 1  g of uranium = 1 / 3.9*10^-22 = 2.56 * 10^21Hope It HELPS You!

re!GIVEN,The SIZE of the PARTICLE = 4.6Since 1 = 10^-6 m∴ The size of the particle = 4.6 * 10^-6 mHope It HELPS You!

Formula. The tension on an object is equal to the mass of the object X gravitational FORCE plus/minusthe mass x ACCELERATION. Tension Formula QUESTIONS: 1) There is a 5 kg mass hanging from a rope.