<span>To answer this question with specific values, you would need a phase diagram for the substance in question. In more general terms, the pressure needs to be lower than the pressure of the triple point, the substance must be present in its solid state, and the temperature must rise high enough to produce a gas.</span>
Answer:
Explanation:
This problem is based on conservation of rotational momentum.
Moment of inertia of rod about its center
= 1/12 m l² , m is mass of the rod and l is its length .
= 1 / 12 x 4.6 x .11²
I = .004638 kg m²
The angular momentum of the bullet about the center of rod = mvr
where m is mass , v is perpendicular component of velocity of bullet and r is distance of point of impact of bullet fro center .
5 x 10⁻³ x v sin60 x .11 x .5 where v is velocity of bullet
According to law of conservation of angular momentum
5 x 10⁻³ x v sin60 x .11 x .5 = ( I + mr²)ω , where ω is angular velocity of bullet rod system and ( I + mr²) is moment of inertia of bullet rod system .
.238 x 10⁻³ v = ( .004638 + 5 x 10⁻³ x .11² x .5² ) x 12
.238 x 10⁻³ v = ( .004638 + .000015125 ) x 12
.238 x 10⁻³ v = 55.8375 x 10⁻³
.238 v = 55.8375
v = 234.6 m /s
Answer:
μsmín = 0.1
Explanation:
- There are three external forces acting on the riders, two in the vertical direction that oppose each other, the force due to gravity (which we call weight) and the friction force.
- This friction force has a maximum value, that can be written as follows:
where μs is the coefficient of static friction, and Fn is the normal force,
perpendicular to the wall and aiming to the center of rotation.
- This force is the only force acting in the horizontal direction, but, at the same time, is the force that keeps the riders rotating, which is the centripetal force.
- This force has the following general expression:
where ω is the angular velocity of the riders, and r the distance to the
center of rotation (the radius of the circle), and m the mass of the
riders.
Since Fc is actually Fn, we can replace the right side of (2) in (1), as
follows:
- When the riders are on the verge of sliding down, this force must be equal to the weight Fg, so we can write the following equation:
- (The coefficient of static friction is the minimum possible, due to any value less than it would cause the riders to slide down)
- Cancelling the masses on both sides of (4), we get:
- Prior to solve (5) we need to convert ω from rev/min to rad/sec, as follows:
- Replacing by the givens in (5), we can solve for μsmín, as follows:
Answer:
the acceleration due to gravity g at the surface is proportional to the planet radius R (g ∝ R)
Explanation:
according to newton's law of universal gravitation ( we will neglect relativistic effects)
F= G*m*M/d² , G= constant , M= planet mass , m= mass of an object , d=distance between the object and the centre of mass of the planet
if we assume that the planet has a spherical shape, the object mass at the surface is at a distance d=R (radius) from the centre of mass and the planet volume is V=4/3πR³ ,
since M= ρ* V = ρ* 4/3πR³ , ρ= density
F = G*m*M/R² = G*m*ρ* 4/3πR³/R²= G*ρ* 4/3πR
from Newton's second law
F= m*g = G*ρ*m* 4/3πR
thus
g = G*ρ* 4/3π*R = (4/3π*G*ρ)*R
g ∝ R
