The added weight of the sand puts more downward pressure on the wheels contacting the rails, which would cause the trains speed to decrease.
Sponges have appendages jointed
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
Answer:
75.84%
Explanation:
We were given Speed of the sports car, v as 80 mph , we can convert to m/s for unit consistency.
v=80mph= 35.76 m/s
The radius of curvature is given as , r = 540 m
✓ the normal weight can be denoted as Wn
✓ the apparent weight of the person can be denoted as Wa
Wn= normal weight= mg
Wa=apparent weight = (mg - mv^2/r)
g= acceleration due to gravity= 9.8m/s^2
The apparent weightand normal weight has a ratio of
Mn/Ma= [mg - mv^2/r]/mg ........eqn(1)
If we simplify eqn(1) we have
Mn/Ma=[g - vr^2/g].............eqn(2)
Then substitute the given values
Mn/Ma=9.8 - [(35.76^2)/540]/ 9.8
=0.758×100%
Mn/Ma=75.84%
Hence, the required fraction is 75.84%