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A thin rod of length 1 m is fixed in a vertical position inside a train, which is moving horizontally withconstant acceleration 4 m/s2. A bead can slide on the rod, and friction coefficient between them is 1/2.If the bead is released from rest at the top of the rod, find the time when it will reach at the bottom. Solution) If mass of bead is m;mg-µN=maNow, N=pseudo force due to acceleration of train, which is equal to 4m Newton. => mg-µ(4m)=ma=> g-4µ=a=> a=8ms-2 Using, second equation of motion:1=0.5(8)t2T=0.5sec
Negative feedback principle: The idea that in a system where there are self-propagating conditions, those new conditions tend to act against earlier existing circumstances. Su
Some wire of cross-sectional area 1mm 2 has a resistance of 20 ?. Determine (a) the resistance of a wire of the same length and material if the cross-sectional area is 4mm 2 ,
It is a parameter of an insulator that defInes the maximum voltage difference that can be applied across the material before the insulator collapse and conducts.
It describes the temperature above which the material loses its spontaneous polarization and piezoelectric characteristics.
CO has same electrons as or the ion that is isoelectronic with CO is: (1) N + 2 (b) CN - (3) O + 2 (4) O - 2
Rules to verify how many significant numbers has in a calculated quantity: 1. All nonzero digits are significant. As like 457 cm has three significant values; 0.25 g has two si
two insulated charged copper spheres A and B of identical size have charge q and -3q respectably .when they are brought in contact with each other and then separated,what are the n
The cross-sectional areas of the pistons in the system shown below have a ratio of 25 to 1. If the maximum force that can be applied to the small piston is 12 N. What is the
The current–voltage characteristic curve for a semiconductor diode as a function of temperature T is given by the equation i = i0(eev/kt - 1) Here the ?rst symbol e represen
mass=volume * density.By using this formula we can find the mass of sphere. Volume of sphere=(4/3)pi*r^3 V=(4/3)*3.14*(1*10^-2)^3=4.187*10^-6 Mass=(4.187*10^-6)*900=3768*10^-6kg
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