Explore topic-wise InterviewSolutions in Current Affairs.

═══════════════╕☬‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌‌‌ ‌ ‌ ‌ ‌ ‌ YOUR ANSWER ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ ☬╘═════════════════════╛Solution is in the image****Graphs as GIVEN 1) 2) 3)HOPE it helps :)╒══════════════════════╕☬‌ ‌ ‌ ‌ ‌ THANKS! ‌@Brainlyconquerer ‌ ‌ ‌ ‌☬╘══════════════════════╛

avigation and ranging is very simple. You send out a PULSE of sound at a certain frequency and wait for the return. You measure the time that takes. DIVIDE it by TWO and you find out the distance it traveled at the speed of sound in the MEDIUM being probed. It is done at VARIOUS frequencies and intensities in water and air

e d = (30) 4 = 120 kmtime t = 8 hspeed V = d/t = 120/8 =15m/s

conservation of monemtum STATES that it can neither be created nor destroyed.Conservation of momentum is derived from Newton's laws of MOTION. Newton's third law states that every action has an equal but opposite reaction; the force that one OBJECT A exerts on object B is equal but opposite to the force that object B exerts on object A. By Newton's second law, this force is equal to the PRODUCT of the mass and the acceleration of the objects, so the product of the mass and acceleration of object A is equal but opposite to the product between the mass and acceleration of object B.hope it helps!please mark me as brainliest if you are satisfied

sted you app in which you get proper NOTES related to any subject the app is " CLASS 10 CBSE board "

t ADDED resistance be xCurrent in the circuit if x is added=>I = V/RtotalI = 120 / 75+xPower accros 75 ohm resistorP = I2R90 = (120/75+x)2 x 75x2 + 5625 + 150X= 12000x2 + 150x - 6375 = 0x=34.6

hat are added to petrol or other fuels, to increase fuel efficiency, are known as Fuel Additives.   Common Fuel Additives are ALCOHOLS such as Methanol (MeOH), Ethanol (EtOH), Isopropyl Alcohol etc. They increase the Octane Rating of petrol, by inhibiting CORROSION and providing a lubricating effect. It thus allows the use of HIGHER compression RATIOS, for greater power and efficiency, in internal combustion ENGINES.

potential at point B becomes equal to the potential at point D HENCE the potential difference between these two points is ZERO hence currentthrough galvanometer is NIL. The deflection in the galvanometer is nilwhen the switch K2 is closed.‌✌️ ✨I THINK IT HELPED YOU✨ ✌️

ye IMPAIRMENT.. this happens DUE to the LOSS of VITAMIN C , and e

s of the FALLING object.g = the acceleration DUE to gravity. ...ρ = the density of the fluid the object is falling through.A = the PROJECTED area of the object. ...C = the drag coefficient.same medium MEANS value p ,A and C will besameso Ratio of Vt1:Vt2= 2*m*g : 2*m*g = 2×1k×g : 2×2k:g= 2 :4=1:2

he QUESTION WELL PLEASE

arge = Qcurrent = Iresistance = Rpotential difference= Vterminal voltage = Vvoltage drop = vinternal resistance = relectro motive force = etime = twork = Wheat energy = Hpower = PEnergy= ERe = EQUIVALENT resistance FORMULAS 1) Q=It2) V=IR3) v=Ir4) Q = no. of electrons × charge of an electron 5) e = I( R+r)6) e= V+v7) W = QV8) W= VIt9) W= I^2 Rt10) I = P/V11) P= W/t12) P= QV/t13) P = I^2R14) P = V^2 / R15) H = I^2Rt16) E= Pt17) Re for series= R1+R2+R3.......18) Re for parallel = 1/ R1 + 1/ R2 + 1/ R3......HOPE THIS HELPED U !!!!!

uency of LIGHT is doubled the STOPPING POTENTIAL will ALSO be doubled because stopping potential is equal to ke max.

Mass of object = m = 0.5 KG, Radius of circle = r = 0.5 m, Tension in the string = T = 5 kg wt = 5 X 9.8 N.To Find: SPEED of the object = ?, the maximum number of revolutions per MINUTE = N =?Solution:In vertical circle maximum tension is at the lowermost pointAns: Speed of object = 6.64 m/s and maximum number of revolutions = 126.7 r.p.m.

a needle with the north end of a bar magnet around 50 times, lifting between strokes. Stroking from tail to TIP pulls the needle's iron atoms into alignment, turning the needle into a weak TEMPORARY magnet. Attach the magnetised needle to your circle of cork with TACK and CAREFULLY place it into a dish of WATERI hope this will help youIf not then comment me

adical 3 [R is the BASE of the isosceles TRIANGLE FORMED by the two force-vectors and half the base is R/2 = [V radical 3] / 2 ; and the BISECTOR of the VERTICAL ANGLE is the perpendicular bisector of the base.Let k be the vertical angle of the isosceles triangle.Sine (k/2) = [R / 2] / R → {[V radical 3] / 2} / V → radical 3 /2k/2 = 60 → k = 120The angles between the two forces is: 180 - 120 = 60 degrees.I hope this will help youIf not then comment me

the ANSWERI HOPE this will HELP YOUIF not then COMMENT me

e in KINETIC ENERGY = increase in GRAVITATIONAL POTENTIAL energy of particle12mv2=mgh1+hR∴h=v22g−v2R=(4.0×103)22×9.8−(4.0×103)26.4×106=9.35×105mwe get h=935kmI hope this will help youIf not then comment me

ong time, The capacitor will be CHARGED to the voltage of the DC battery, the positive side of the capacitor touching the positive terminal of the battery. Not much DC current will CONDUCT, EXCEPT for some tiny leakage current DUE to imperfection of the cap. The battery will be drained eventually.I hope this will help youIf not then comment me

the amount of power in the electrical GRID. Base load is the level that it typically does not go below, that is, the basic amount of electricity that is always required. PEAK load is the daily fluctuation of electricity use. It is usually lowest in the wee hours of the morning and HIGHEST in the early evening.I hope this will help YOUIF not then COMMENT me

0metres10sec=3300metres=3.30kmI HOPE this will HELP YOUIF not then COMMENT me

l displacement of a PLANK with a body of mass m on it is VARYING according to law . The minimum value of for which the mass just breaks off the plank and the moment it occurs FIRST after , are given by: ( y is positive vertically UPWARDS)I hope this will help youIf not then comment me

having the same mass and radius as the disc , is now gently placed on the disc. After some time, the two rotate with a common ANGULAR VELOCITY. 1) some friction exitsI hope this will help YOUIF not then comment me

re, transmission of signals using ground wave is RESTRICTED to frequencies UPTO 1500 KHZ as higher frequencies are dampedI hope this will help youIf not then comment me

ths BLOW the outermost ZONE of a flame with a metallic blow pipe for MELTING gold and SILVER. Hence GOLDSMITH use outermost part of the flame because it is the hottest part of flame.I hope this will help youIf not then comment me

of Work and KINETIC EnergyThe work-energy theorem states that the change in kinetic energy of a particle is equal to the work done on it by the NET force.Kf - Ki = WWe KNOW the equation in 3D : v2 – u2 = 2a.d (where u-initial velocity, v-final velocity, a-acceleration, d-displacement) now MULTIPLYING the equation by m/2 we have, ½ mv2 – ½ mu2 = ma.d = F.d (Since ma = F)I hope this will help YOUIF not then comment me

here are no non-conservative forces,So using CONSERVATION of mechanical ENERGY,(1/2)mv^2 = (1/2)kx^2Now on putting all the values in the above equation, you will GET the maximum compression of the spring and that will be = 2.HOPE it helps ☺️☺️ I hope this will HELP youIf not then comment me

y of the rocket, V = 5 km/s = 5 × 103 m/sMass of the Earth, Me = 6 × 1024 kgRadius of the Earth, Re = 6.4 × 106 mHeight reached by rocket mass, m = hAt the surface of the Earth,Total energy of the rocket = Kinetic energy + Potential energy= (1/2)mv2 + (-GMem / Re)At highest point h,v = 0And, Potential energy = -GMem / (Re + h)Total energy of the rocket = 0 + [ -GMem / (Re + h) ]= -GMem / (Re + h)From the law of conservation of energy, we haveTotal energy of the rocket at the Earth’s surface = Total energy at height h(1/2)mv2 + (-GMem / Re) = -GMem / (Re + h)(1/2)v2 = GMe [ (1/Re) – 1 / (Re + h) ]= GMe[ (Re + h – Re) / Re(Re+ h) ](1/2)v2 = gReh / (Re + h)Where g = GM / Re2 = 9.8 ms-2∴ v2 (Re + h) = 2gRehv2Re = h(2gRe – v2)h = Rev2 / (2gRe – v2)= 6.4 × 106 × (5 × 103)2 / [ 2 × 9.8 × 6.4 × 106 – (5 × 103)2h = 1.6 × 106 mHeight achieved by the rocket with respect to the CENTRE of the Earth = Re + h= 6.4 × 106 + 1.6 × 106 = 8 × 106 m.I HOPE this will HELP youIf not then comment me

g-puzzling, unphysical result that linear stability analyses lead to no transition in pipe flow, even at infinite Reynolds number, is ascribed to the use of STICK boundary conditions, because they ignore the amplitude variations associated with the roughness of the wall. Once that length scale is introduced (here, crudely, through a corrugated pipe), linear stability analyses lead to stable vortex formation at low Reynolds number above a finite amplitude of the corrugation and unsteady flow at a higher Reynolds number, where indications are that the vortex dislodges. Remarkably, extrapolation to infinite Reynolds number of both of these transitions leads to a finite and nearly identical value of the amplitude, implying that below this amplitude, the vortex cannot form because the wall is too smooth and, hence, stick boundary results prevail.This work explores the effect of wall roughness on the long known contradiction between the linear stability analysis result of infinitely stable flow in pipes with smooth boundaries and the experimental observation (1) that flows become unstable at a Reynolds number of ≈2,000 for ORDINARY pipes (2–4). A hint that wall roughness may be important can be gathered from experiments which show that for smoothed pipes, the onset of the instability can greatly exceed 2,000 (5). The present work should serve as a general warning that stick boundary conditions—for example, in narrow biological channels where the surface roughness of the wall can be a significant FRACTION of the channel width, or, as ANOTHER example, in the drag reduction problem—may be inappropriate. For the smooth-wall case, there exists a rigorous proof of stability for axisymmetric disturbances (6) and strong evidence (cf. refs. 7 and 8) that all linear perturbations decay for all values of the Reynolds number and axial and azimuthal wavenumbers. Thus, there remains much interest in the cause of this transition and how one may affect the Reynolds number at which transition occurs.I hope this will help youIf not then comment me

to the road map given in Figure 11.1. We have now finished our DISCUSSION of linearized flow and the associated compressibility corrections. Keep in mind that such linearized theory does not APPLY to the TRANSONIC flow regime, 0.8 < Мж <1.2. Transonic flow is highly nonlinear, and theoretical transonic aerodynamics is a challenging and sophisticated subject. For the remainder of this chapter, we deal with several aspects of transonic flow from a qualitative point of view. The theory of transonic aerodynamics is beyond the scope of this book.Consider an airfoil in a low-speed flow, say, with = 0.3, as sketched in Figure 11.5a. In the expansion over the top surface of the airfoil, the local flow Mach number M increases. LET point A represent the location on the airfoil SURFACEI hope this will help youIf not then comment me

investigatory project for class 12 on the topic " to estimate CHARGE. To estimate the charge induced on each one of the TWO identical styro foam balls or pith balls suspended in a vertical plane by making USE of Coulomb’s law. The BALL was charged with a known charge I hope this will HELP youIf not then comment me

easing the number of turns of the coil, so that it will PRODUCE more magnetic field2) Also decreasing the resistance of the circuit.I HOPE this will HELP YOUIF not then comment me

s the QUESTIONTHIS is WRONG questionPlease TELL properly Hope you understand

en "base units" of the SI measuring system are: metre (length), kilogram (mass), second (time), ampere (electric current), kelvin (thermodynamic temperature), mole (amount of substance), and candela (luminous INTENSITY)In actuality, the fundamental quantities are (1) the second, defined as exactly 9,192,631,770 times the period of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the cesium-133 atom. We can't accurately define the second by the rotation of the earth because it is not completely constant. For example, a large earthquake changes the length of the day, and the axis of the earth's rotation precesses every 23,000 years, etc.(2) The speed of light has an exact value of 299,792,458 meters per second. It is exact because by international agreement (3) a meter is defined to be the length of the path travelled by light in vacuum during a time INTERVAL of 1/299792458 second, and the second is extremely precisely defined as above.The standard for the kilogram is a platinum-iridium cylinder manufactured in 1889 and kept under carefully controlled conditions; however, its mass fluctuates by about 50 micrograms, and cannot be perfectly accurately defined. Thus the definition of the kilogram is a secondary measure.The standard of the gravitational constant G is notoriously uncertain, as physical constants and standards go, and since its first measurement by Cavendish in 1798 has only slightly improved in accuracy. However, the more useful number is the GM product for the Earth, GM⊕ = ( 398,600.4415 ± 0.0008 ) km^3/sec^2 ; and for the Sun, GM⊙ = ( 1.327124400 × 10^11 ) km^3/sec^2, which is known far more accurately.An important goal is to find absolute and unchanging standards and definitions for all the physical constants and units. There are 19 fundamental independent physical constants, and 5 "natural units" not based on any human construct: the Planck length, Planck mass, Planck time, Planck CHARGE, and Planck temperature.I hope this will help youIf not then COMMENT me

E Q IS PLACED AT THE CENTRE OF THE LINE JOINING TWO EQUAL CHARGES Q. SHOW THAT THE SYSTEM OF THREE CHARGES WILL BE IN EQUILIBRIUM IF Q = -Q/4. If the NET force on each CHARGE will be zero, the system will be in equilibrium.I HOPE this will help youIf not then comment me