Hey
SI unit for measure the amount of material contained in a solid sphere metre cube(m^3).
mean it have Volume .
Answer: 0.9264 kg
Explanation: [I'll use "cc" for cubic centimeter, instead of cm^3.
The volume is 6cm*4cm*2cm = 48 cm^3 (cc).
Density of Au is 19.3 g/cc
Mass of gold = (48 cc)*(9.3 g/cc) = 926.4 grams Au
1 kg = 1,000 g
(926.4 grams Au)*(1 kg/1,000 g) = 0.9264 kg, 0.93 kg to 2 sig figs
At gold's current price of $57,500/kg, this bar is worth $53,268. Keep it hidden from your lab partner (and instructor).
it is just a matter of integration and using initial conditions since in general dv/dt = a it implies v = integral a dt
v(t)_x = integral a_{x}(t) dt = alpha t^3/3 + c the integration constant c can be found out since we know v(t)_x at t =0 is v_{0x} so substitute this in the equation to get v(t)_x = alpha t^3 / 3 + v_{0x}
similarly v(t)_y = integral a_{y}(t) dt = integral beta - gamma t dt = beta t - gamma t^2 / 2 + c this constant c use at t = 0 v(t)_y = v_{0y} v(t)_y = beta t - gamma t^2 / 2 + v_{0y}
so the velocity vector as a function of time vec{v}(t) in terms of components as[ alpha t^3 / 3 + v_{0x} , beta t - gamma t^2 / 2 + v_{0y} ]
similarly you should integrate to find position vector since dr/dt = v r = integral of v dt
r(t)_x = alpha t^4 / 12 + + v_{0x}t + c let us assume the initial position vector is at origin so x and y initial position vector is zero and hence c = 0 in both cases
r(t)_y = beta t^2/2 - gamma t^3/6 + v_{0y} t + c here c = 0 since it is at 0 when t = 0 we assume
r(t)_vec = [ r(t)_x , r(t)_y ] = [ alpha t^4 / 12 + + v_{0x}t , beta t^2/2 - gamma t^3/6 + v_{0y} t ]
To solve this problem we will apply the theorem given in the conservation of energy, by which we have that it is conserved and that in terms of potential and kinetic energy, in their initial moment they must be equal to the final potential and kinetic energy. This is,
Replacing the 5100MJ for satellite as initial potential energy, 4200MJ for initial kinetic energy and 5700MJ for final potential energy we have that
Therefore the final kinetic energy is 3600MJ
