Answer:
The weight of the landing craft in the vicinity of Callisto's surface is 3480 N.
Explanation:
The engine of the craft provides an upward thrust of 
so that the space craft descends at a constant speed.
This implies that the net force on the space craft is zero.
The upward thrust will be equal to the downward gravitational pull by Callisto.
So the weight of the craft near the vicinity will be 3480 N.
The six commonly recognised metalloids are boron, silicon, germanium, arsenic, antimony, and tellurium....
Answer:
Explanation:
Orbital radius of satellite A , Ra = 6370 + 6370 = 12740 km
Orbital radius of satellite B , Rb = 6370 + 19110 = 25480 km
Orbital potential energy of a satellite = - GMm / r where G is gravitational constant , M is mass of the earth and m is mass of the satellite
Orbital potential energy of a satellite A = - GMm / Ra
Orbital potential energy of a satellite B = - GMm / Rb
PE of satellite B /PE of satellite A
= Ra / Rb
= 12740 / 25480
= 1 / 2
b ) Kinetic energy of a satellite is half the potential energy with positive value , so ratio of their kinetic energy will also be same
KE of satellite B /KE of satellite A
= 1 / 2
c ) Total energy will be as follows
Total energy = - PE + KE
- P E + PE/2
= - PE /2
Total energy of satellite B / Total energy of A
= 1 / 2
Satellite B will have greater total energy because its negative value is less.
Answer:
Check the explanation
Explanation:
In calculate kinetic energy, write out a formula whereby the kinetic energy will be equal to 0.5 times mass times velocity squared. include in the value for the object mass, then the velocity with which it is moving.
Kindly check the attached image below to get the step by step explanation to the question above.
Explanation:
Below is an attachment containing the solution
