Deduce a value for the latent heat of evaporation Lv of water. State clearly any simplifying assumptions that you make. Estimate
the pressure at which ice and water are in equilibrium at −2◦C given that ice cubes float with 4/5 of their volume submerged in water at the triple point (0.01◦C, 612Pa). [Latent heat of sublimation of ice at the triple point, Ls = 2776 °´ 103 Jkg−1.]
2 answers:
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Answer
given,
L(t) = 10 - 3.5 t
mass of particle = 2 Kg
radius of the circle = 3.1 m
a) torque
τ =
τ =
τ = -3.5 N.m
Particle rotates clockwise as i look down the plane. Hence, its angular velocity is downward.
L decreases the angular acceleration upward. so, net torque is upward.
b) Moment of inertia of the particle
I = m R^2
I = 2 x 3.1²
I = 19.22 kg.m²
L = I ω
ω =
ω =
ω =
A = 0.52 rad/s B = -0.182 rad/s²
Answer:
<em>The coefficient of static friction between the crate and the floor is 0.41</em>
Explanation:
<u>Friction Force</u>
When an object is moving and encounters friction in the air or rough surfaces, it loses acceleration and velocity because the friction force opposes motion.
The friction force when an object is moving on a horizontal surface is calculated by:
[1]
Where is the coefficient of static or kinetics friction and N is the normal force.
If no forces other then the weight and the normal are acting upon the y-direction, then the weight and the normal are equal in magnitude:
N = W = m.g
The crate of m=20 Kg has a weight of:
W = 20*9.8
W = 196 N
The normal force is also N=196 N
We can find the coefficient of static friction by solving [1] for :
The friction force is equal to the minimum force required to start moving the object on the floor, thus Fr=80 N and:
The coefficient of static friction between the crate and the floor is 0.41