A lead weight with volume of .82 x10 -5 m 3 is lowered on a fishing line into a

Question

A lead weight with volume of .82 x10-5m3is lowered on a fishing line into a lake to a depth of 1m.(a) whatis the tension required in the fishing line to give the weight anupward acceleration of 2.1m/s2.(b) if the initial depthof the weiight is increased to 2.0m,does the tension found in part(A)increase, decrease or remains the same?explain(c)whatacceleration will the weight have if the tension in the fishingline is 1.2N?Give both direction and magnitude? A lead weight with volume of .82 x10-5m3is lowered on a fishing line into a lake to a depth of 1m.(a) whatis the tension required in the fishing line to give the weight anupward acceleration of 2.1m/s2.(b) if the initial depthof the weiight is increased to 2.0m,does the tension found in part(A)increase, decrease or remains the same?explain(c)whatacceleration will the weight have if the tension in the fishingline is 1.2N?Give both direction and magnitude?

Explanation / Answer

volume of lead = V= .82*10^-5 m^3 mass of lead = m= .82*10^-5 * 11350 = .093 kg weight = mg = .912 N buoyancy = B = weight of displace water = V(1000)(g) = .080 N a) if tension (upward) is T, T + B - mg = ma T = m(a+g) - B = .093*(2.1+9.8) - .080 = 1.027 N b) no change, since B does not change so long the mass iscompletely inside the fluid. c) if the tension is 1.2 N, 1.2 + .08 - .912 = .093*a a = 3.96 m/s^2

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