Answer:
conservation of momentum, general law of physics according to which the quantity called momentum that characterizes motion never changes in an isolated collection of objects; that is, the to
Answer:
8.94*10^22 kg
Explanation:
Given that
Mass of Lo, M = ?
Radius of Lo, r = 1.82*10^6 m
Acceleration on Lo, g = 1.80 m/s²
Gravitational constant, G = 6.67*10^-11
Using the formula
g = GM/r²
Solution is attached below
Answer is 8.94*10^22 kg
Answer:
c
Explanation:
all the others r physical
Answer:
The boiling point temperature of this substance when its pressure is 60 psia is 480.275 R
Explanation:
Given the data in the question;
Using the Clapeyron equation
where is the change in enthalpy of saturated vapor to saturated liquid ( 250 Btu
T is the temperature ( 15 + 460 )R
m is the mass of water ( 0.5 Ibm )
is specific volume ( 1.5 ft³ )
we substitute
/
272.98 Ibf-ft²/R
Now,
where P₁ is the initial pressure ( 50 psia )
P₂ is the final pressure ( 60 psia )
T₁ is the initial temperature ( 15 + 460 )R
T₂ is the final temperature = ?
we substitute;
480.275 R
Therefore, boiling point temperature of this substance when its pressure is 60 psia is 480.275 R
P=4800kgm/s
As
p=mΔv
where p is momentum, m is mass and v is velocity
Given values is
m =1200kg
Δv= 17m/s-13m/s=4m/s
Now
p=mΔv
p=(1200kg)*(4m/s)
p=4800kgm/s