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A spider is hanging by means of its own silk thread directly above a transparent fixed sphere of r = 20cm .... the refractive index of material of sphere is root(2).... and height of the spider is 2r = 40cm .... an insect initially sitting at the bottom directly below the spider starts crawling along vertical circular path with constant speed pi/4 cm/s.. for how long will the insct be invisible to the spider. Assume that it crawls once around the circle .Solution) Spider is able to see only half of the sphere which is topmost hemispherical surface. when insct enters in bottom hemispherical surface it becomes invisible for the spider. a/c question insct is travelling vertical circular path it travels pi/4cm in 1sec.then it will travel 2pi r dist i.e. 40picm in 160sec. hence it will travel pi r i.e20pi (half circle) in 80s.
Electric field (intensity) is a vector number. Electric field because of a positive particle is always away from the charge and that because of a negative charge is always toward
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ohmmeters: The use of ohmmeters is somewhat more involved. Most importantly when measuring resistance the circuit power must be switched off, power is derived from within the i
Average acceleration is described as the change in velocity over a given time interval Δt. Hence, The instantaneous acceleration of given particle is the rat
Radio Waves: Radio waves are electromagnetic waves with large range of wavelength from a few millimetres to many meters.
explain the construction and working of compound microscope
(i) The size examines cannot be operated to derive relations other than product of power functions, for example, s = ut + 1/2 at 2 or y = y 0 cosωt and so on, cannot be derived
There are six properties of surface tension:- Normal 0 false false false EN-IN X-NONE X-NONE MicrosoftInternetExplorer4
Finding the electric potential because of a continuous distribution of charge involves doing an integral. An integral is an infinite total of terms. In computing the electric poten
I need help solving an equation or two to find the elastic potential energy of a spring when compressed. Spring starts at .75 m, and is compressed to .25 m by a force of 12 newtons
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