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Explanation:3.Let the charge Q be placed at distance x from charge 2q. Now for the charge to be in equilibrium the forces due to both charges must cancel each other or the forces must be equal and opposite. Now the forces due to each charges is given by F  2q ​   =2kQq/x  2    and F  q ​   =kQq/(l−x)  2   for equilibrium we must have F  2q ​   =F  q ​      or, 2kQq/x  2 =kQq/(l−x)  2  or, (l−x)  2  x  2  ​   =2   or,    l−x x ​   =  2 ​     or,  x=  1+  2 ​    2 ​   l ​    l−x x ​   =−  2 ​       or,  x=  2 ​   −1 2 ​   l ​    Since second solution of x lies outside the l we will neglect it. Hence the answer is- $x=  2 ​   +1 2 ​   l ​   $ Now if the charge is displaced longitudinally towards charge q then x will increase hence F  2q ​    will decrease and F  q ​    will increase. Hence the charge will be forced to move towards  charge q. Similarly if it is moved towards charge 2q, F  2q ​    will increase and F  q ​    will decrease thus the charge Q will be moved towards charge 2q. Hence the charge Q is in unstable equilibrium with respect to longitudinal motions.2.Let"R" be the distance from the charge q where Q is in equilibrium .Total Force acting on Charge q and 4q:F=kqQ/r² + k4qQ/(l-r)²For Q to be in equilibrium , F should be equated to zero.kqQ/r² + k4qQ/(l-r)²=0(l-r)²=4r²⇒l-r=2r⇒l=3r⇒r=l/3Taking the third charge to be -Q (say) and then on applying the condition of equilibrium on + q chargekQ/(L/3)² =k(4q)/L²9kQ/L²=4kq/L²9Q=4qQ=4q/9Therefore a point charge -4q/9 should be placed at a distance of L/3 RIGHTWARDS from the point charge +q on the line joining the 2 charges.1.Given,qA = +4eqB = +eAB = xLet, Q be the third point charge located at some point P.AP = xtherefore, PB = (a-x)Force on Q due to qA,F1 = kqAQ/x2Force on Q due to qBF2= kQqB/(a-x)2Since the charge is in equilibrium,F1 = F2kqAQ/x2 = k qBQ/(a-x)24e/x2 = e/(a-x)24/x2 = 1/(a-x)2 (Taking square ROOT on both sides)2/x = 1/(a-x)2(a-x) = x(1)2a - 2x = x2a = 3xx = 2/3aTherefore, the third point charge Q should be placed 2/3a from A.if helps mark as brainlist answer

teExplanation:Here, the point at which the BEAM would have converged in absence of the lens acts as VIRTUAL object. So,u=+20cm.Again since the image is INVERTED and of same size, m=−1.∴m=−1=uv⇒v=−20cm∴f1=v1−u1=−201−201=20−2=10−1.I hope this will HELP you❤️❤️

bro cube ki SIDE toh de diExplanation:YE toh BATA do KYA KARNA hai volume likalni hai ya area

For the first CIRCUIT the answer is 5/7 ohm.RESISTANCES 3 and 2 ohm are in series while the resultant resistance of these and 1 ohm are in parallelFor the second circuit the answer is 24/5 ohmExplanation:resistances 8,9,7 ohm are in series while it's resultant and 6 ohm are in parellel

1 V/mExplanation:GIVEN V= log XAS we know E = -(dv/dx,)DERIVATIVE of log x is 1/x HENCE the value of derivative @ x= -1 is 1 V/m or 1 N/C

∆x = 18 mx-t graph will STRAIGHT line straight line from originExplanation:Area under v-t CURVE GIVES XA straight line from origin in x-t curve DENOTES constant velocity

The man who CLIMBS the slope.Explanation:b'coz work equals to FORCE x distanceand as he climbs up for E of gravity is exerted more.Mark as Brainliest

The concept of "relative rest" is closely linked to that of inertial observers and the statement that "nothing is at ABSOLUTE rest" is loosely equivalent to stating that there are no frames of reference which are TRULY inertial. So-called non-inertial observers are ADDRESSED by the theory of general relativity.Explanation:FOLLOW me..

the question of whether you have any questions or need any further information PLEASE contact me at the end of the MONEY to pay 57 CDS

the VELOCITY of v.t GRAPH is 5

here's your ANSWEREXPLANATION:HOPE it works PLEASE mark it as BRAINLIST

A is the CORRECT ANSWER

effctive RESISTANCE 2 is a RIGHT ANSWER

y the VECTOR LAW of ADDITION of FORCES,     144² = F² + F²   F = 144/√2 = 72√2 Newtons

aulic lift                      a) charge is zero2) PURITY of gold                    b) net charge is zero3) acceleration                     c) pascal's law4) energy                            d) Archimedes principle5) neutral charge                  E) scalar QUANTITY                                           f) vector quantity2 - d         Archemedes discovered the law, to find the purity of gold in the crown.1 - c         Pascal's law - pressure is same at the same height in a liquid.                 this is used to lift very heavy loads by hydraulic lift.3 -  f           5 -  b       uncharged particle means charge is 0.  neutral charge                 means that there are EQUAL and opposite charges present on a                body  or at a location.4   -   e

s = 80 m = displacement       u = 0  = initial velocity           g = 10 m/sec/sec  acceleration of the STONE          equation of motion :  v² = u ² + 2 a s         v² = 0 + 2  * 10 * 80 = 1600        v = final velocity = 40 m/sec2)    u = 0        a = 0.50 m/sec/sec       s = 25 m         v² = u² + 2 a s             = 0 + 2 * 0.4 * 25  = 20           v = √20 m /sec = 2 √5 m/s         v = u + a t           t = (v - u) / a = 2√5 / 0.50  = 4 √5  seconds.

be a PARABOLA.let speed = xkinetic energy = y = 1/2 m x²the shape of this CURVE is a parabola, as it is in the form:  x² = 4 a y   The curve is ALWAYS above x axis. The axis of the parabola is x = 0  or the y axis. the directrix of parabola is  at  y = - 1/(2m)

string is rigid , tight, inextensible with tension > 0, then the two bodies have the same (magnitude of)  ACCELERATION, along the direction of the string on either side.  If the string is loose or folded, then the accelerations cannot be inter-related easily.I SUPPOSE there is no rotation of the two bodies..Is this question / DOUBT from the  TOPIC  pulleys, bodies attached to strings rolling over pulleys?    Then yes, certainly.

related to bipolar junction transistor in the common EMITTER  configuration.  The CURRENT GAIN is defined as α, which is usually around 50.  A small DC current in  the base of the transistor triggers a high current in the collector region which flows TOWARDS the emitter.  The base is less dop ed and collector is heavily dop ed with impurities.Current gain = collector current / base current                   =  [ emitter current - base current ] / base current                   = emitter current / base current - 1base current =  emitter current / [ current gain + 1 ]

T Cos 30 =  1 kg * a       30 N - T cos 30 = 2 kg * a       =>  3 a = 30      =>  a = 10 m/sec/sec       =>  T  = 20/√3  N       Normal reaction:    mg - T sin 30 = 1 * 10 - 20/√3 * 1/2 = 10 - 10/√3  N===========15.           m1 = 2 kg        m2 = 5 kg         F - (m1 + m2) g = (m1 + m2 ) a      =>  F - 70 = 7 a           So the pulley moves up only if  F > 70 NEWTONS.     F = 70 N,  then accelerations are both 0, as the pulley still does not move up.=======================16.         m2 g = tension    as m2 is at rest.       m1 g Sin Ф - tension        as m1 is at rest.         m1 Sin Ф = m2      verify the options by substitution.=======17       F1 = 10 N,  F2 = 20 N          F1 Cos 60  + F2 = 25 N     ---  in x direction    F1 Sin 60 = 5√3 N         ---  in the y direction,  as block is at rest.     resultant  =  √ [25² + (5√3)² ] = 26.46  N             the result is supposed to be an integer???/    Are you expected to write only the force in the x direction ???==============18.   m1 = 3 kg,        m2 = 2kg            F cos 60 =  m1 a - T        T =  m2  a       a = F * Cos 60  /(m1+m2)  = 1 m/s/s     T = 2 N===============19.           F - mg = ma            F - 2 * 10 = 2 * 5            F = 30 N==================20.             F - mg  = m a = - m g          as  F is upwards, and a = -g  downwards             F = 0 N

Normal force on B by the floor = F       Force exerted by block A on B:  F1         F - m_B g -  F1 = - m_B * 2        =>  F = F1 + m_B g - 2 m_B         F1 - m_A  g = - m_A * 2            =>  F1 = g/2 - 1  = 4  Newtons             =======================9.  component of weight/acceleration due to gravity along the inclined slope             = g Sin 30 = g/2      Component of weight/acceleration along the groove OA = g/2 Sin 30 = g/4     s = u t + 1/2 a t²    => t² = 2 s / a               t² = 2 * 5 /  (g/4) = 4              =>           t = 2 sec.=================10.      m1 = 100 kg          M2 = 60 kg  a1 = acceleration of 100 kg upwards.  Rope is also moving with the same acceleration.       T - 100 kg * g = 100 kg * a1        => T = 1000+ 100 a1          ---- (1)          Acceleration of the person = a     relative acceleration of the man wrt rope  = a + a1  = 5g/4              => a = 5g/4 - a1  = 12.5 - a1      upwards     T - 60 kg * g = 60 kg * a       T = 600 + 60 a = 600 + 60 (12.5 - a1) = 1350 - 60 a1      ----- (2)       Solving the two equations          a1 = 350/160  m/sec/sec       T = 1000 + 100 * 35/16 = 1218  N======================11         F = buoyancy force upwards  remains same         F - M g  = M a        => F = M (g+a)         F - (M-m) g = (M-m) a'          =>  a' = [ M g + Ma - Mg + mg ] / (M-m)                 a' = (M a + m g)/(M-m)===============================12.        T - 5 g = 5 a             10 g - T = 10 a            a = g/3=================   13.      F - m g Sin 37  = m a          along the slope           5.5 - 0.50 * g * 0.6  = 0.50 a      =>  a = 5 m/sec/sec                         v^2 = 2 a s = 2 * 5 * 10  = 100              v = 10 m/s    at the TOP of the incline.         ANGLE of projection = 37 deg.             y initial = 10 Sin 37 = 6 meters           Time to REACH the top point :    u Sin 37 / g = 10 * 0.6 / 10 = 0.60 sec           height reached = s  =  u sin 37 * 0.60 - 1/2 * 10 * 0.60^2                                         = 10  * 0.6 * 0.60 - 5 * 0.60^2  = 1.80 meters           total height at the top of the flight = 6 + 1.80 = 7.80 m           time to reach the ground = t  => t^2 = 2s / g =>        t = 2 * 7.80 / 10 = 1.56 sec       total time of the flight = 2.16 sec.          total  horizontal distance travelled after projection from the ramp                   =  u cos 37 * 1.56 sec                    = 12.48 meters

ball=1/2mv²when speed or velocity is DOUBLED i.e. 2vtherefore NEW K.E=1/2m(2v)²                          =4 × 1/2mv²Hence, the new K.E becomes 4 TIMES the EARLIER K.E.

are GOING on an inclined line on a slope on a ROLLER coaster, the chair or the floor exert a force on us to balance the weight due to gravitation of Earth.  When the reaction force acts on our feet or our buttocks or back, that pressure and sensation makes us feel the our weight.    When we go through a vertical loop we are going in a circular arc.  At the top of the roller coaster (inverted U arc), a force called centripetal force is required to make us go in a circle.  This force depends on the speed, the radius of curvature and our mass.         our weight = Normal reaction of the floor/chair + centripetal force     normal reaction of the floor or chair = our weight - centripetal forceSo the apparent weight is reduced by the centripetal force.  So the contact between our feet or BODY with the roller coaster becomes very LIGHT.  So we feel weightlessness.====================Also at some points, there is a near free-fall of the persons on a roller coaster.  Then the apparent weight we feel = our weight - m * acceleration of the free fall.So our apparent weight reduces a lot, when we MOVE down a slope with a large acceleration.

here A and B MEANS A and B VECTORA B cosФ=A B sinФsinФ/cosФ=1tanФ=1tanФ=tanπ/4Ф=π/4

?? I camp help EXACTLY TOMORROW I mean on THURSDAY

y is defined as CHANGE of DISPLACEMENT per unit  time.  Usually velocity is measured with respect to an inertial frame of reference, ie., a coordinate system which does not move.  So we choose the distant star in the universe which is stationary as the reference, and measure all velocities WRT to it.   In our world we measure velocities wrt objects that are stationary on Earth.  So velocity is measured as a relative quantity.Acceleration is defined as change of velocity per unit time.  That is           a = (v - u) / Δtlet  v and u  be measured relative to a frame of velocity    v₀.   Then the absolute velocities of an OBJECT at two points of time are:  v + v₀  and  u  + v₀  respectively.     Then acceleration a = [v + v₀  - u - v₀ ] / Δt                     a = (v - u) / ΔtThus acceleration is absolute, whether we TAKE absolute velocities or relative velocities of a particle to compute it.

rm motion - when ANYTHING covers EQUAL distances in equal intervals of time and travels in a straight line it is said to be in UNIFORM motion. like 3 km in 1 hr and 6km in 2hrs 2)non-uniform motion-when an object doesn 't travels in straight line and covers unequal distance it is called non-uniform motion.like 3KM in 1 hr and 7 km in 2 hrs Press thank you!!!!

nce CHEMICAL equation there r 4 steps 1)first write chemical formula in words2)then write the FORMULA3)check the formula WHETHER the coefficients are correct4)finnaly verify the formula

strange question to ask... the height of Buddha statue : 18 meters and 350 tons heavy.  it is made of rock of WHITE granite. The PEDESTAL is the rock (rock of GIBRALTAR) on which the statue is standing.  A concrete platform of about 4 and half meters wide 4 and 1/2 meters LONG is constructed under the pedestal.

's  left hand RULE principle is used in DC motor where the three FINGERS are perpendicular to each other where the thumb indicates direction of  force ,index finger indicates induced current & middle finger indicates the magnetic INDUCTION

see diagram.The pendulums are each displaced by an amount x.  The string is stretched by 2x totally.  The restoration force in the spring = F2 = k * 2x = 2k x.  (Hooke's law).             Bob's displacement X in the tangential direction to the string = X   x = displacement in the horizontal direction = X Cos Ф   x  = L sin Ф   mg CosФ = Tension T in the string    F1 = COMPONENT of weight  along the tangential direction     F1 = mg SinФ = m g x / L   Net force on the bob = F  = m a = - m * d²X / dt²           (-ve sign is because the force is in the decreasing x direction)   X = x / cos Ф = L tan Ф  ≈ L Sin Ф ≈ L Ф,   approximation for small Ф.   Net force on pendulum in the tangential direction =     F1 and F2 are at an angle Ф, and their vector sum is to be calculated for the resultant force and acceleration.  But since Ф is small, we approximate the resultant force to be along horizontal direction and F1 and F2 along this direction.  So we take the linear sum for simplicity.    a =  - d² X / dt²  ≈  - d² x / dt²      =>   - m d² x / dt²  =  2k x + m g x/L       Resultant force on the bob =  mg x/L + 2 k x = F = m a = - m d^2 x/dt^2             d² x/dt²  = - (g/L + 2k/m) x  The equation of motion above indicates that for small amplitudes x and displacements, the  pendulum oscillates in a SHM and  the corresponding       Angular VELOCITY = ω = √(g/L + 2k/m)=       TIME period = ======================================If displacements are large then, the resultant force F will be:..

3CWork done=18JAs we KNOW that  V=W/q=18/3=6V

ade MOTION is the motion in the DIRECTION OPPOSITE to the movement of something else and the CONTRARY or DIRECT motion

displacement HORIZONTALLY:    If the charge q is displaced slightly to the right side along the line joining the two CHARGES -q, then the force of attraction of the right side -q is more than the force of attraction of the -q on the left side.   So +q moves towards right side. As it moves closer to the -q (right side), the force increases.  Then it will ACCELERATE more and collide with it.================case 2)   displacement of +q along the perpendicular bisector of line joining two charges -q and -q.Let the direction of x be towards the TOP upwards from the center.  Let the origin be at the center of two charges -q.  Let the charge q be displaced by x , where x << d.   The distance between the two charges is 2 d.  Mass of charge q  is m.   The distance between -q and +q = √(d² + x²)Let K = 1/[ 4 πε ].see the diagram.  Let F1 be the force due to -q on the right and F2 be the force due to -q on the left side.   They are equal in magnitude and directions are as shown.   magnitude = F1 = F2 = K q² / (d² + x²)  Components of F1 and F2 along the perpendicular bisector are:            F1 Sin Ф = K q²  x / (x²+d²)³/² The components of these forces parallel to the line joining -q charges are = F1 Cos Ф and will cancel as they are in OPPOSITE directions.Let F and a be the instantaneous resultant force and net acceleration of +q.  Then the resultant force on q due to the two charges:       F =  2 K q² x / (d²+x²)³/²  towards the origin in the direction of - x.     m d² x/ dt²   = - 2 K q² x / [d³ (1 + x²/d²)³/² ]                       =  - 2 K q² x / d³   if  x << d, then we ignore x² as it is << d²   m  d² x / d t² =  - [ 2 q² / 4 π ε d³ ]  x                       = ⁻ω² x  This is the equation of motion of the charge q.  SO q oscillates in a simple harmonic motion with an amplitude of  x₀ that is the displacement at t = 0.     The angular frequency of oscillation =  ω  = q / √ (2π ε d³)

in WATER there is some palticles of SALT and salt is PASS ELECTRICITY so water is ALSO pass electricity from it.