QUESTION 2 25 a) (5 p) Interpret the rocker equation dv(t)M(t)=-udMO (EQ.1) within the framework of...
a) (5 p) Interpret the rocker equation dv(t)M(t)=-udM(t) (EQ.1) within the framework of the law of momentum conservation, written in a closed system, here M(t) is the rocker mass, at time t, whereas M(t) is by definition, dM(t)-M(t+dt)-M(t): - dM(t)-dM(t), is the mass of the gas thrown by the rocket through the infinitely small period of time dt; on the other hand, dv(t) is, still by definition, dv(t)-v(t+dt)-v(t), i.e. the increase in the velocity of the rocker through the period...
QUESTION 2 a) (5 p) Interpret the rocket equation dv(OM(t)=-udMO [EQ.1) within the framework of the law of momentum conservation, written in a closed system, here Mt) is the rocket mass, time t, whereas M(t) is by definition, dM(t)=M(t+dt)-M(t): -dM(t)-dM(t), is the mass of the gas thrown by the rocket through the infinitely small period of time dt: on the other hand, dv(t) is, still by definition, dv(t)v(t+dt)-vít), i.e. the increase in the velocity of the rocket through the period...
QUESTION 2 a) (5 p) Interpret the rocket equation dv(OM(t)=-udMO (EQ.1) within framework of the law of momentum conservation, written in a closed system, here M(t) is the rocket mass, at time t, whereas M(t) is by definition, dM(t)=M(t+dt)-M(t): -dM(t)=dM(t), is the mass of the gas thrown by the rocket through the infinitely small period of time dt; on the other hand, dv(t) is, still by definition, dv(t)=v(t+dt)-vít).i.e. the increase in the velocity of the rocket through the period of...
please solve 2 QUESTION 2 a) (5 p) Interpret the rocket equation dv(t)M(1)=-udMO [EQ.1) within the framework of the law of momentan conservation, written in a closed system here M(t) is the rocket mass, at time t, whereas dMt) is by definition, dM(t)-M(t+dt)-M(t): -SM(t)-M(!), is the mass of the gas thrown by the rocket through the infinitely small period of time dt; on the other hand, dv(t) is still by definition, dy(t){t+dt)-v(t), i.e. the increase in the velocity of the...
Interpret the rocket equation dv(t)M(t)=-udM(t) [EQ.1] within the framework of the law of momentum conservation, written in a closed system; here M(t) is the rocket mass, at time t, whereas dM(t) isby definition, dM(t)=M(t+dt)-M(t); -dM(t)=|dM(t)|, is the mass of the gas thrown by the rocket through the infinitely small period of time dt; on the other hand, dv(t) is, still by definition, dv(t)=v(t+dt)–v(t), i.e. theincrease in the velocity of the rocket through the period of time dt; u is the relative...
of the a) (5 p) Interpret the rocket equation dv(t)M(t)=-udM(t) [EQ.1) within the framework law of momentum conservation, written in a closed system; here M(t) is the rocket mass, at time t, whereas dM(t) is by definition, dM(t)=M(t+dt)-M(t); dM(t)=|dM(t), is the mass of the gas thrown by the rocket through the infinitely small period of time dt; on the other hand, dv(t) is, still by definition, dv(t)=v(t+dt)-v(t), i.e. the increase in the velocity of the rocket through the period of...
a) (15 p) We consider a nuclear reactor of power output P=1000 Megawatt (1000 million watts) electric, functioning with Plutonium. It is fueled, initially, with 1000 kg of Plutonium. The nuclear material in question is made of Plutonium nuclei, each 239 consisting in 94 protons and 239-94-145 neutrons, which is denominated by the symbol 94Pu. For thermodynamical rd reasons, only 1/3 of the nuclear energy in the form of heat produced by the reactor, can be converted into electricity. How...
*) (15 p) We considera muclear reactor of power output P-1000 Megowott 1000 million wot elect, can with Plutonium. Bifueled, initially with 1000 kg of Plutonium. The nuclear teslim 239 question is made of Plutonium nuclei, each consisting in 94 protons and 239-94145 euro the symbol For thermodynamical reasons, only 1.3 in the form of heat produced by the reactor, can be converted into electricity. How much mass deficit should the nucle fuel of concer delicates if the reactor is...