A 6.00-kg block and an 8.00-kg block are connected as shown. When released, the 6.00-kg block accelerates downward and the 8.00-kg block accelerates to the right. After each block has moved 2.00 cm, the total work done on the 8.00-kg block: A. is greater than the total work done on the 6.00-kg block. B. is the same as the total work done on the 6.00-kg block. C. is less than the total work done on the 6.00-kg block. D. not enough information given to decide
Since both masses are connected by a single string both have equal acceleration = a Both traveled with same distance = d Work done on mass m1(8.00kg)= W1= Net force on m1*distance traveled = F1*d = m1*a*d -----------(1) Work done on mass m2(6.00kg)= W2 = Net force on m2*distance traveled = F2*d = m2*a*d -----------(2) m1>m2, Comparing eqn (1) and (2), W1>W2 Thus total work done on the 8.00-kg block is greater than the total work done on the 6.00-kg block.
An underground telephone cable, consisting of a pair of wires, has suffered a short somewhere along its length (at point P in the Figure). But, where to dig to fix it? The telephone cable is 6.00 km long, and there is a clever trick which takes advantage of the fact that real wires do have a finite resistivity (so their resistance, while small, is not zero, and depends on length!) In order to determine where the short is, a technician first measures the resistance between terminals AB; then she measures the resistance across the terminals CD. The first measurement yields 105.00 Ohm; the second 5.00 Ohm. Where is the short? Give your answer as a distance from point A.
Let us use eqn, R=ρL/A Hence, R α L So we can write relation, R(AB)/R(CD) = L(AB)/L(CD) 105.00/5.00 = L(AB)/[12.00-L(AB)] L(AB)=11.45km Distance between point A and short is 11.45km
Two spherical drops of mercury each have a charge of 0.10 nC and a potential of 300 V at the surface. The two drops merge to form a single drop. What is the potential at the surface of the new drop? __________ V E
Let us first find the relation between radius of small drop(r) and radius of large drop(R) Volume of large drop = 2*volume of small drop 4/3*π*R^3 = 2*4/3*π*r^3 R= (cube root2)*r -----------(1) Potential at the surface of small drop=v = kq/r -------(2) Potential at the surface of large drop =V= k*2q/R---------(3) Plug (1) in (3), V=k*2q/R = k*2q/((cube root2)*r) = sqrt2*kq/r = [2/(cube root2)]*v Plugging values, V= [2/(cube root2)]*300 = 476.22 V