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cle of mass 4 m at REST explodes into 3 fragments. Two fragments of mass m each move atspeed V each PERPENDICULAR to one ANOTHER. Then energy of the fragment of mass 2 m will be

tion:Average speed does not have to equal the magnitude of the average velocity. People may think that average speed and average velocity are just different names for the same QUANTITY, but average speed depends on DISTANCE and average velocity depends on DISPLACEMENT. If an object changes direction in its journey, then the average speed will be greater than the magnitude of the average velocity.Speed is a SCALAR, and average velocity is a vector. Average velocity indicates direction and can be represented as a negative number when the displacement is in the negative direction. Average speed does not indicate direction and can only be POSITIVE or zero.

Resistive FORCE of 80 N will be offered by the ground., in the direction opposite to the MOTION of cannon after firing the BALL.Explanation:Given,MASS of cannon ball, m = 2 kgMass of cannon, M = half tonne = 500 kg Final velocity of cannon ball, v = 100 m/sLet, final velocity of cannon = VDistance covered by cannon after firing the ball, s = 0.5 m To find,Resistive force offered by the ground to the motion of the cannon, F =?Solution,Initially both Cannon and ball would be at rest, therefore Initial velocity of Cannon ball = Initial velocity of cannon = u = 0Using LAW of conservation of momentum → m u + M u = m v + M V→ m (0) + M (0) = ( 2 ) ( 100 ) + ( 500 ) V→ 0 = 200 + 500 V→ V = -200/500→ V = -0.4 m/s Now, Let acceleration of cannon after firing the ball be a then, After firing the ball cannon will have U = V = -0.4 m/s as initial velocity and it will cover a distance of s = 0.5 m to come to rest hence, final velocity will be V' = 0Using third equation of motion for cannon→ 2 a s = V'² - U²→ 2 a ( 0.5 ) = ( 0 )² - ( -0.4 )²→ a = -0.16  m/s²Now, Using newton's second law of motion for cannon→ F = M a → F = ( 500 ) · ( -0.16 )→ F = -80 NTherefore, A resistive force of 80 Newtons will be offered by the Ground, in the direction opposite to the motion of Cannon after firing the ball.

The Force of Gravity Any two objects that have mass attract each other with a force we call gravity. You probably never noticed this for small objects, because the force is so weak. But the Earth has lots of mass, and so it exerts a big gravitational force on you. We call that force your weight. The fact that gravity is actually a force of attraction is not obvious. Prior to the work of Isaac NEWTON, it was assumed that gravity was simply the natural tendency of objects to move downward. If you weigh 150 lb, and are sitting about 1 meter (3.3 feet) from another person of similar weight, then the gravitational force of attraction between the two of you is 10-7 lb. This seems small, but such forces can be measured; it is about the same as the weight of a flea. You weigh less when you stand on the Moon, because the force of attraction is less. If you weigh 150 lb on the Earth, you would weigh only 25 lb on the Moon. You haven’t changed (you are made up of the same atoms), but the force exerted on you is different. Physicists like to say that your mass hasn’t changed, only your weight. Think of mass as the AMOUNT of material, and weight as the force of attraction of the Earth (or whatever other planet or SATELLITE you are standing on). Mass is commonly measured in kilograms. If you put a kilogram of material on the surface of the Earth, the pull of gravity will be a force of 2.2 lbs. So a good definition of a kilogram is an amount of materialBut if you are so small, how can you exert such a large force on the Earth? The answer is that, even though you are small, your mass exerts a force on EVERY piece of the Earth, simultaneously. When you add all those forces together, the sum is 150 lb. So you are pulling up on the Earth exactly as much as the Earth is pulling down on you. Think of it in the following way. If you push on some else’s hand, they feel your force. But you feel the force too. You push on them; they push back on you. The same thing WORKS with gravity. The Earth pulls on you; you pull on the Earth.

ेखर वेंकटरामन का जन्म ७ नवम्बर सन् १८८८ ई. में तमिलनाडु के तिरुचिरापल्‍ली नामक स्थान में हुआ था। आपके पिता चन्द्रशेखर अय्यर एस. पी. जी. कॉलेज में भौतिकी के प्राध्यापक थे। आपकी माता पार्वती अम्मल एक सुसंस्कृत परिवार की महिला थीं। सन् १८९२ ई. मे आपके पिता चन्द्रशेखर अय्यर विशाखापतनम के श्रीमती ए. वी.एन. कॉलेज में भौतिकी और गणित के प्राध्यापक होकर चले गए। उस समय आपकी अवस्था चार वर्ष की थी। आपकी प्रारम्भिक शिक्षा विशाखापत्तनम में ही हुई। वहाँ के प्राकृतिक सौंदर्य और विद्वानों की संगति ने आपको विशेष रूप से प्रभावित किया।cv RAMAN

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best trick to LEARN reactivity SERIES is ...Kashinath ka mali ALOO zara phika pakata HAI...kon hai age aaoExplanation:kaNaCaMgAlZn FePbHgCuHAgAu..HOPE u understood

knowledge of multivariable calculus (at LEAST MATH 212) and linear algebra (at least Math 221). A BASIC knowledge of classical MECHANICS and electromagnetism is DESIRABLE too.

t d s is a DEVICE USED to MEASURE the PURITY of WATER

Gifford SAD sad XA DD sad dp SEND ZAL fud zal as

it's ANSWER is 2Explanation:as SIGNIFICANT PLACE is 2

800.4 K is CORRECT ANSWER HAI

of WEIGHT 100 gf and volume 35 CM immersed in WATER. What is the apparent weight of stone in water ?⠀⠀⠀⠀⠀and⠀⠀⠀⠀⠀We have to find apparent weight of stone when it is immersed in water.Volume of stone is 35cm and volume of water displaced = (V) = 35cm Density of water (d) = 1 cm Mass of water displaced = V × d = 35 (cm) = 35 g Weight of water displaced = 35 gf.Upthrust = Weight of fluid (water) displaced = 35 gf Apparent weight of stone in water = (100 – 35) gf= 65 gf.

116640 jouleExplanation:mass =45 kgvelocity = 72 km/h KINETIC energy = 1/2 mv^2 K.E=1/2×45×72×72 =116640 JOULE

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free fall is defined as the motion of an object where gravity is the only force acting upon it. ... A skydiver may be PULLED towards earth by gravity, but they are ALSO affected by AIR resistance, a force opposing their DOWNWARD movement.

: ➥ The recoil VELOCITY of the pistol = 1 m/s Given : ➤ MASS of the bullet (m₁) = 10 g ➤ Velocity of the bullet (v₁) = 300 m/s ➤ Mass of the pistol (m₂) = 3 kg To FIND : ➤ Recoil velocity of the pistol (v₂) = ? Solution : • Mass of the bullet (m₁₁) = 10 g = 0.01 kgHence, the recoil velocity of the pistol is 1 m/s.               LOTS OF LOVE !

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2gfigure SHOWS a bob moving in horizontal circular motion at ANGULAR speed 0=2 Vhwhere AB=h. the ratio of tension in SUPPORTING steingsTTINis hiven by3value of N. strings are of EQUAL LENGTHS and intextensibalT

hat:Percentage error in radius is given as 2%or we can SAY thatSolutionwe KNOW that,volume → Percentage error in volume: is the required ANSWER.

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:Mass of BODY = 10g = 0.01kgInitial velocity = zeroFinal velocity = 10m/sTome interval = 10sWe have to find magnitude of applied force★ As per newton's second law of MOTION force is defined as the rate of change of linear momentum.F = dp/dtForce is a polar vector quantity as it has a point of application.Momentum is measured as the product of mass and velocity.F = m (V - u) / tF denotes forcem denotes massv denotes FINAL VELOCITYU denotes initial velocityt denotes timeBy substituting the given values;➝ F = m (v - u) / t➝ F = 0.01 (10 - 0) / 10➝ F = 0.01 × 10/10➝ F = 0.01 N

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can define the work done as a definite integral of the FORCE over the total displacement as Now, we ALREADY KNOW that KINETIC energy of an object can be expressed as: We can express the change in the kinetic energy with TIME as, Hence, the work-energy theorem is proved for the variable force as well.

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ρ= 3Gg/4πR IE RADIUS is inversely PROPORTIONAL to DENSITY and density decrease by factor 5

A MONOMIAL, also called power product, is a product of powers of variables with NONNEGATIVE integer EXPONENTS, or, in other words, a product of variables, possibly with repetitions. For example, is a monomial. The CONSTANT 1 is a monomial, being equal to the empty product and to x0 for any VARIABLE x

tion:1.for MAKING the TEXT look coloured refer the below syntax and IGNORE the dots [.tex]\red{the\:text\:appears\:in\:red}[./tex]result \red{the\:text\:appears\:in\:red}thetextappearsinred 2.for making the text marquee[.tex]{the\:text\:appears\:LIKE\:this}[./tex]result = < marquee > {the\:text\:appears\:like\:this} < /marquee > thetextappearslikethis 3. for making the text look bold [.tex]\bold{the\:text\: appears\: like\: this }[./tex]result = \bold{the\:text\: appears\: like\: this }thetextappearslikethis 4. for showing the exponential value [.tex]{no.}^{power}[./tex]result = {no.}^{power}[ /tex] 5. for showing the square root value [.tex]\sqrt{no.}[./tex] result = 5.forshowingthesquarerootvalue[.tex] no. [./tex]result=[tex] no. \bold{\boxed{thanx}} thanx follow me next

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