To solve this problem we will apply the concepts related to Orbital Speed as a function of the universal gravitational constant, the mass of the planet and the orbital distance of the satellite. From finding the velocity it will be possible to calculate the period of the body and finally the gravitational force acting on the satellite.
PART A)
Here,
M = Mass of Earth
R = Distance from center to the satellite
Replacing with our values we have,
PART B) The period of satellite is given as,
PART C) The gravitational force on the satellite is given by,
If each side is 0.1 feet extra,
The volume will be 5.1*2.1*1.1= about 11.781.
Perhaps this helps.
No they have eukaryotic cells
Answer:
W = ½ m v²
Explanation:
In this exercise we must solve it in parts, in a first part we use the conservation of the moment to find the speed after the separation
We define the system formed by the two parts of the rocket, therefore the forces during internal separation and the moment are conserved
initial instant. before separation
p₀ = m v
final attempt. after separation
= m /2 0 + m /2 v_{f}
p₀ = p_{f}
m v = m /2 
v_{f}= 2 v
this is the speed of the second part of the ship
now we can use the relation of work and energy, which establishes that the work is initial to the variation of the kinetic energy of the body
initial energy
K₀ = ½ m v²
final energy
= ½ m/2 0 + ½ m/2 v_{f}²
K_{f} = ¼ m (2v)²
K_{f} = m v²
the expression for work is
W = ΔK = K_{f} - K₀
W = m v² - ½ m v²
W = ½ m v²
Answer:
The minimum work per unit heat transfer will be 0.15.
Explanation:
We know the for a heat pump the coefficient of performance (
) is given by
where, 
is the magnitude of heat transfer between cyclic device and high-temperature medium at temperature 
and 
is the required input and is given by 
, 
being magnitude of heat transfer between cyclic device and low-temperature 
. Therefore, from above equation we can write,
Given, 
and 
. So, the minimum work per unit heat transfer is given by
