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1​= 2 and N2​ = 1.5 moles. The initial temperatures are T1​ = 175 K and T2​ = 400 K. The total volume is 0.025 m^3. 1/T1=(3/2)R*N1/U1 ,P1/T1=R*N1/V11/T2=(5/2)R*N2/U2 , P2/T2=R*N2/V2​To find:Determine the final temperature of the system (in Kelvin).Solution:Let the volumes of two systems be, V1 and V2From given, we have,V1 + V2 = V = 0.025 m³As given, that both systems are in a closed CONTAINER in ADIABATIC condition, we have,T1 = 175 K T2 = 400 Kw.k.t PV = nRTnRT1/P1 + nRT2/P2 = nRT/P2R×175/P + 1.5R×400/P = 3.5RT/P2 × 175 + 1.5 × 400 = 3.5T350 + 600 = 3.5T950 = 3.5TT = 950/3.5∴T = 271.42 KHence the final temeperature.

⬇⬇⬇⬇⬇⬇⬇⬇⬇⬇⬇⬇⬇⬇⬇⬇⬇⬇⬇⬇⬇⬇⬇QUESTION :An object of 30 cm is placed in front of convex lens of FOCAL length 20 cm find its positionSOLUTION :Given that, Object distance (u) = - 30 cm. Focal length (f) = 20 cm. IMAGE distance (v) = ? By USING " Lens Formula .. ⟹ 1/f = 1/v - 1/u➡ 1/20 = 1/v - 1/(-30) ➡ 1/v = 1/20 - 1/30➡ 1/v = 30 - 20/600➡ 1/v = 10/600➡ v = 600/10➡ v = 60 Therefore, it's image distance is 60 cm. ➢ Magnification (m) = -v/u➡ m = - 60/30➡ m = -2Hence, the image distance is 60 cm. ⬆⬆⬆⬆⬆⬆⬆⬆⬆⬆⬆⬆⬆⬆⬆⬆⬆⬆⬆⬆⬆⬆⬆Explanation:℘ℓḙᾰṧḙ ՊᾰԻк Պḙ ᾰṧ ♭Իᾰ!ℵℓ!ḙṧт ✌✌✌

the LINES around the MAGNET is CALLED MAGNETIC FIELD lines

Explanation:y = mx + cWe are using the same concept .BRING 5x + 2Y = 0 in y = mx+ C format :2y = -5x + 0y = -5/2 X + 0So , m denotes the SLOPE .That implies slope = -5/2.As simple as that :)

% reduction in the VALUE of electric field would go by this method :(E - 4/14 E) * 100 the WHOLE upon E = Which will GIVE you around 1000/14 % and thats = 70.7 %Option AThe question cannot be explained to more BASIC level than this by a student. I advise you to ask your educator for more better clarification.Keep going! Awesome !

sir please SHOW me the DISTANCE time GRAPH....

a is ANS...............

nitial velocity of bicycle, u= 0 m/sFinal velocity of bicycle, v= 10 m/sTime TAKEN, t= 30 sTo Find:Acceleration of bicycleSolution:We KNOW that,According to first equation of MOTION for constant acceleration, where, v is final velocity u is initial velocity a is accelerationt is TIME takenLet the acceleration of bicycle be aOn applying first equation of motion on bicycle, we GET Hence, the acceleration of the bicycle is 0.33 m/s².

The RESISTANCE WOULD be 660wats

total DISTANCE is 67metre Explanation:PLEASE MARK as BRAINLIEST

but what cai I do if technician has only 2 resistance it's not my MATTER EXPLANATION:its techicians matter

of the building (d) = 90mLet’s find the velocity of STONE thrown up:We know, v^2 – u^2= 2adVelocity of stone after reaching top of the building = 0Acceleration is opposite to acceleration due to gravity = -9.8m/s^20- u^2= 2*(-9.8)*90- u^2=-1764Therefore, velocity of stone MOVING up= 42m/sDistance at time t = h(t) =ut-1/2at^2h(t) = 42t-4.9t^2 -----(1)h1(T) = HEIGHT of building – 1/2aT^2T=t-2 (since second stone is thrown after two seconds)H1= 90 – 4.9(t-2)^2When the two stones meet, distance are same. So, Set h= h142t - 4.9t^2 = 90 -4.9t^2 +19.6t -19.642t-19.6t=90-19.622.4t = 70.4 => t = 3.14sSubstitute the VALUE of “t” in equation (1)                            h =83.6m

Can you take a picture of the QUESTION and UT it ... I will SURELY answer it ...EXPLANATION:

Ohm's law STATES that current flowing through a conductor is directly PROPORTIONAL to the potential difference maintained across the two ends of a conductor at constant temperature and pressure. ... As per the question, the resistance of a conductor is 1 ohm.. Pls follow me and have a NICE DAY.. ❤️

▪ Initial velocity of car = zero (i.e. rest)▪ Final velocity of car = 60kmph▪ Time interval = 10minTo Find :▪ Acceleration of car.▪ Distance covered by car in the given interval of time.Concept :☞ Acceleration is defined as the ratio of change in velocity to the time interval.☞ SINCE, acceleration has SAID to be uniform, we can easily apply EQUATION of kinematics to solve this type of questions.Conversion :✏ 1kmph = 5/18mps✏ 60kmph = 60×5/18 = 16.67mps✏ 10MIN = 10×60 = 600sCalculation :↗ Acceleration of car :→ a = (v - U) / t→ a = (16.67 - 0) / 600→ a = 16.67/600→ a = 0.028 m/s²↗ Distance covered by car :→ d = ut + (1/2)at²→ d = (0×600) + (0.5)(0.028)(600)²→ d = 0 + (0.014×360000)→ d = 5040m = 5.04km

none of theseExplanation:An artificial Satellite revolves ROUND the Earth in circular orbit, which QUANTITY remains CONSTANT?A. ANGULAR MomentumB. Linear VelocityC. Angular DisplacementD. None of these

no heat and TEMPERATURE are not the same conceptExplanation:this is because heat is ALWAYS emmited or RECEIVED from a source but temperature refers to the exact READING of a substance in degree celcius or Fahrenheit

tion:initial velocity(u)=0m/s as the BODY starts from the restfinal velocity(v)=4/2 m/s=2m/stime =10S acceleration= v-u /t =2-0/10 =2/10 =1/5m/s²distance (s)s= UT + 1/2at =0*10 + 1/2*1/5*10 =0 + 1/2*2 =0 + 1 =1M(ans) this May help you!!please MARK this answer as brainlist answer please please please please please please please please please please please please

0.18 metres Explanation:since the formula for refractive power of lens is given by,, D=1÷f that is f=1÷D , SUBSTITUTE the VALUE of d in the equation, we GET f=1÷-5.5=-0.18181=-0.18metrs since distance can't be in negative so by neglecting negative SIGN, we have the answer as 0.18 metres.

c) a = 1Explanation:Absorptive power is 1 because it ABSORBS all radiation INCIDENT upon it.(Technically r = 0 is ALSO CORRECT as nothing is reflected, everything is absorbed.)

light is a form of energy.Light is needed to see things around US.Light enables us to see objects from which it comes or from which it is reflected.➤ NATURE of light :- There are two theories of nature of light.☛ According to wave theory:- light consist of electormagnetic waves which do not require a material medium (LIKE solid, liquid,gas) for their proportion.☛ According to PARTICLE theory:-light is composed of particles which travel in a STRAIGHT line at very high speed.The elementary particles that defines light is the ‘photon'.

resistorsExplanation:Given :Many 150 ohms resistor are connected 2 AMPERES of CURRENT should flow in them Source shall have voltage of 50 volts.To find :No of RESISTORS to be connected.Using ohms law:V=IR50=2×Rr = R = 25 ohmsNow we NEED to find number of 150 ohms resistor to be connected in parallel to get equivalent resistance equal to 25 ohms.If resistance is same in all resistors, req = R/no.of resistors.150/no.of resistors = 25No.of resistors = No.of resistors = 6Six 150 ohms resistors must be connected in parallel to get an equivalent resistance of 25 ohms.

So the ANSWER is 1/200N/cm2.

20 cm.Explanation:If an object is PLACED at a DISTANCE of 10 cm in front of a plane MIRROR, it would be 20 cm away from it's IMAGE, SINCE the image formed is at the same distance from the mirror as the object is in front of it...

this TELL in ANOTHER WAY

4:3Explanation:See the image. HOPE it helps. PLEASE mark as BRAINLIEST PLEASEEEEE

It's easy.If ONE RESISTOR is of 2OHM then the other WOULD be 1\xNow,Since they are CONNECTED in parallel,1/2 + 1/X = 5/61\x = 5\6 - 1/21\x = 1/3Therefore X = 3ohmSo,the other resistor was of 3ohmhope it helps

fossils which DEAD organisms REMAINS in the earth crust for millions of YEARS

tion:SINCE the angle of the apex is 60, that means the other two angles are also 60 and this is an equilateral triangle. That means the length from the apex to the CROSS SIDE is the same as the cross side = 10CM which is 1ohm/cm * 10cm = 10 ohm.

We neglect air resistance when compared to the acceleration of the falling body.Explanation:A body is said to free fall when there is no other rigid object preventing the body from falling. A table restricts a book on it from falling toward the ground. In the same way ASTRONAUTS in the SPACE STATION are in the state of free fall since the space station is also EXPERIENCING an equal acceleration towards the earth. Here in your CASE air is not a body which can stop the movement of the body. Hence the resistance offered by air is negligible in this case.

a medium of constant refractive index, N, the OPL for a path of geometrical length s is just a medium of constant refractive index, n, the OPL for a path of geometrical length s is just{\displaystyle \mathrm {OPL} =ns.\,}{\displaystyle \mathrm {OPL} =ns.\,} a medium of constant refractive index, n, the OPL for a path of geometrical length s is just{\displaystyle \mathrm {OPL} =ns.\,}{\displaystyle \mathrm {OPL} =ns.\,}If the refractive index varies along the path, the OPL is given by a line INTEGRAL a medium of constant refractive index, n, the OPL for a path of geometrical length s is just{\displaystyle \mathrm {OPL} =ns.\,}{\displaystyle \mathrm {OPL} =ns.\,}If the refractive index varies along the path, the OPL is given by a line integral{\displaystyle \mathrm {OPL} =\INT _{C}n\mathrm {d} s,\quad }{\displaystyle \mathrm {OPL} =\int _{C}n\mathrm {d} s,\quad } a medium of constant refractive index, n, the OPL for a path of geometrical length s is just{\displaystyle \mathrm {OPL} =ns.\,}{\displaystyle \mathrm {OPL} =ns.\,}If the refractive index varies along the path, the OPL is given by a line integral{\displaystyle \mathrm {OPL} =\int _{C}n\mathrm {d} s,\quad }{\displaystyle \mathrm {OPL} =\int _{C}n\mathrm {d} s,\quad }where n is the local refractive index as a function of distance along the path C. a medium of constant refractive index, n, the OPL for a path of geometrical length s is just{\displaystyle \mathrm {OPL} =ns.\,}{\displaystyle \mathrm {OPL} =ns.\,}If the refractive index varies along the path, the OPL is given by a line integral{\displaystyle \mathrm {OPL} =\int _{C}n\mathrm {d} s,\quad }{\displaystyle \mathrm {OPL} =\int _{C}n\mathrm {d} s,\quad }where n is the local refractive index as a function of distance along the path C.An electromagnetic WAVE propagating along a path C has the phase shift over C as if it was propagating a path in a vacuum, length of which, is equal to the optical path length of C. Thus, if a wave is traveling through several different media, then the optical path length of each medium can be added to find the total optical path length. The optical path difference between the paths taken by two identical waves can then be used to find the phase change. a medium of constant refractive index, n, the OPL for a path of geometrical length s is just{\displaystyle \mathrm {OPL} =ns.\,}{\displaystyle \mathrm {OPL} =ns.\,}If the refractive index varies along the path, the OPL is given by a line integral{\displaystyle \mathrm {OPL} =\int _{C}n\mathrm {d} s,\quad }{\displaystyle \mathrm {OPL} =\int _{C}n\mathrm {d} s,\quad }where n is the local refractive index as a function of distance along the path C.An electromagnetic wave propagating along a path C has the phase shift over C as if it was propagating a path in a vacuum, length of which, is equal to the optical path length of C. Thus, if a wave is traveling through several different media, then the optical path length of each medium can be added to find the total optical path length. The optical path difference between the paths taken by two identical waves can then be used to find the phase change. Fermat's PRINCIPLE states that the path light takes between two points is the path that has the minimum optical path length.

Dry ice sombline. SNOW and ice stumbline during WINTER season without melting. MOTH BALLS sumbline. Room fresheners which are used in toilet sumbline.Frozen food will sumbline and you will find ice crystals INSIDE of the box.

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tion:1/mv²-mu² OK FRIEND this is URS ANSWER