Answer:
671.76 kg or 6590 N
Explanation:
So the buoyant force generated by the floating ice is equals to the mass of water displaced by the submerged ice. We also need to account for gravity of ice. The resulting additional mass that the ice sheet can support is the difference between the mass of water displaced by ice and the mass of ice submerged totally in water.
So the ice piece can support an additional 671.76 kg of bear, or 671.76 * 9.81 = 6590 N
Answer:
16.6 ms or 0.0166 s
Explanation:
If Q is the final charge, Q' is the initial charge, C the capacitance ,R is the resistance , t the time taken and τ the time constant,
[tex]Q = Q'( 1- e^{-t\div \tau })
τ = R C = (1.20×10³) (20×10⁻⁶) = 0.024 s
15 = 30 ( 1- e^{-t\div \ 0.024 })
( 1- e^{-t\div \ 0.024 }) = 15 ÷ 30
⇒ - e^{-t\div \ 0.024 }) = 0.5 -1
⇒ e^{-t\div \ 0.024 }) = 0.5
Taking logarithm to the base e on both sides of this equation,
⇒ t = 0.0166 seconds = 16.6 milli seconds
Convert cm into meters,
50/100=0.5m
work done= 10x0.5
=5J
we dont consider the weight of the spring as it acts downwards.
