Answer:
The gauge pressure in Pascals inside a honey droplet is 416 Pa
Explanation:
Given;
diameter of the honey droplet, D = 0.1 cm
radius of the honey droplet, R = 0.05 cm = 0.0005 m
surface tension of honey, γ = 0.052 N/m
Apply Laplace's law for a spherical membrane with two surfaces
Gauge pressure = P₁ - P₀ = 2 (2γ / r)
Where;
P₀ is the atmospheric pressure
Gauge pressure = 4γ / r
Gauge pressure = 4 (0.052) / (0.0005)
Gauge pressure = 416 Pa
Therefore, the gauge pressure in Pascals inside a honey droplet is 416 Pa
Answer:
heat pressure, electron degeneracy, neutron degeneracy, and nothing
Explanation:
Main Sequence Star: It is a star in which nuclear fusion is happening in the core of the star. Hydrogen molecules fuse together to generate Helium. This nuclear fusion generates outward gas pressure and radiation pressure which balances the inward gravity thus creating an equilibrium which keeps the stars in shape.
White dwarf: It is the end stage of a medium sized star like the Sun. Outer layers of the star are thrown in the form a shell/bubble leaving a small and dense core in the center called as white dwarf. This core consists of carbon and oxygen. Nuclear fusion doesn't occur in the core of white dwarfs. The inward gravity is balanced by the electron degeneracy pressure. Thus these stars will keep on radiating the remaining heat and will turn in to a black dwarf at the end.
Neutron Star: This is the end stage of a supermassive star (1-3 times the mass of the Sun). At the last stage of the life the core collapses. In these stars the inward gravity is so huge that the pressure overcomes the electron degeneracy pressure and crushes together the electron and proton to form neutron. The neutron then stops the collapse and balances the inward gravity.
Black Hole: This is the end stage of a hyper massive stars weighing more than 3 times the mass of the Sun. The inward gravitational force is so huge that even the neutrons are not able to stop the collapse the core. thus the mass of the star collapses into a very small area of immense gravity. There is nothing that can balance this inward gravity.
Answer:
Eastward, at 11 m/s^2
Explanation:
64N-31N=unbalanced force of 33N
F=ma
33N=(3kg)a
a=11m/s^2 to the East
Answer:
B. 7.07 m/s
Explanation:
The velocity of the stone when it leaves the circular path is its tangential velocity, 
, which is given by
where 
is the angular speed and 
is the radius of the circular path.
is given by
where 
is the frequency of revolution.
Thus
Using values from the question,
<em>Note the conversion of 75 cm to 0.75 m</em>
Answer:
Explanation:
The centripetal acceleration requirement must equal gravity at the top of the circle
mg = mv²/R
v = √Rg
v = √(1.0(9.8))
v = 3.1304951...
v = 3.1 m/s
