Temperature of 273K is the temperature at which water
1 answer:
You might be interested in
Answer:
0.767
Explanation:
The work done on the truck by the frictional drag force is given by
where
F is the magnitude of the frictional force
d = 38.0 m is the maximum displacement allowed for the truck
The negative sign is due to the fact that the force of friction is opposite to the motion of the truck
The force of friction can also be written as:
where
is the coefficient of kinetic friction between the truck and the lane
m is the mass of the truck
g is the acceleration of gravity
So we can rewrite the work done as
(1)
According to the work-energy theorem, the work done by friction is equal to the change in kinetic energy of the truck:
(2)
where
v = 0 is the final velocity of the truck
u = 23.9 m/s is the initial velocity of the truck
By combining (1) and (2) we get
And solving for , we find the minimum coefficient of kinetic friction able to stop the truck in a distance d:
Answer:
w = 4,786 rad / s , f = 0.76176 Hz
Explanation:
For this problem let's use the concept of angular momentum
L = I w
The system is formed by the two discs, during the impact the system remains isolated, we have the forces are internal, this implies that the external torque is zero and the angular momentum is conserved
Initial Before sticking
L₀ = 0 + I₂ w₂
Final after coupling
= (I₁ + I₂) w
The moments of inertia of a disk with an axis of rotation in its center are
I = ½ M R²
How the moment is preserved
L₀ =
I₂ w₂ = (I₁ + I₂) w
w = w₂ I₂ / (I₁ + I₂)
Let's reduce the units to the SI System
d₁ = 60 cm = 0.60 m
d₂ = 40 cm = 0.40 m
f₂ = 200 min-1 (1 min / 60 s) = 3.33 Hz
Angular velocity and frequency are related.
w₂ = 2 π f₂
w₂ = 2π 3.33
w₂ = 20.94 rad / s
Let's replace
w = w₂ (½ M₂ R₂²) / (½ M₁ R₁² + ½ M₂ R₂²)
w = w₂ M₂ R₂² / (M₁ R₁² + M₂ R₂²)
Let's calculate
w = 20.94 8 0.40² / (12 0.60² + 8 0.40²)
w = 20.94 1.28 / 5.6
w = 4,786 rad / s
Angular velocity and frequency are related.
w = 2π f
f = w / 2π
f = 4.786 / 2π
f = 0.76176 Hz