Answer:
Temperature : 92.9 F
Internal Energy change: -2.53 Btu/lbm
Explanation:
As
mh1=mh2
h1=h2
In table A-11 through 13E
p2=120Psi, h1= 41.79 Btu/lbm,
u1=41.49
So T1=90.49 F
P2=20Psi
h2=h1= 41.79 Btu/lbm
T2= -2.43F
u2= 38.96 Btu/lbm
T2-T1 = 92.9 F
u2-u1 = -2.53 Btu/lbm
Main sequence stars are characterised by the source of their energy.They are all undergoing fusion of hydrogen into helium within their cores. The mass of the star is the main element for such process or phenomenon to take place for it is a determinant of both the rate at which they perform the said activity and the amount of fuel available.
To answer the question, the lower mass limit for a main sequence star is about 0.08. If the mass of a main sequence star is lower than the above-mentioned value, there would be a deficit or insufficiency of gravitational force to generate a standard temperature for hydrogen core fusion to take place and the underdeveloped star would form into a brown dwarf instead.
Answer:
Moment of the force is 20 N-m.
Explanation:
Given:
Force exerted by the person is, 
Distance of application of force from the point about which moment is needed is, 
Now, we know that, moment of a force 'F' about a point at a perpendicular distance of 'd' from the same point is given as the product of the force and the perpendicular distance.
Therefore, the moment of the force about the end of the claw hammer is given as:

Hence, the moment of the force exerted by the person about the end of the claw hammer is 20 N-m.
Answer:
v₂ = 70 m / s
Explanation:
For this exercise let's use Bernoulli's equation
where subscript 1 is for the top of the mountain and subscript 2 is for Tuesday's level
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ +1/2 ρ v₂² + ρ g y₂
indicate that the pressure in the two points is the same, y₁ = 250 m, y₂ = 0 m, the water in the upper part, because it is a reservoir, is very large for which the velocity is very small, we will approximate it to 0 (v₁ = 0), we substitute
ρ g y₁ = ½ ρ v₂²
v₂ =
let's calculate
v₂ = √( 2 9.8 250)
v₂ = 70 m / s