From an energy balance, we can use this formula to solve for the angular speed of the chimney
ω^2 = 3g / h sin θ
Substituting the given values:
ω^2 = 3 (9.81) / 53.2 sin 34.1
ω^2 = 0.987 /s
The formula for radial acceleration is:
a = rω^2
So,
a = 53.2 (0.987) = 52.494 /s^2
The linear velocity is:
v^2 = ar
v^2 = 52.949 (53.2) = 2816.887
The tangential acceleration is:
a = r v^2
a = 53.2 (2816.887)
a = 149858.378 m/s^2
If the tangential acceleration is equal to g:
g = r^2 3g / sin θ
Solving for θ
θ = 67°
I think their distance is a measurement of : B. space in two dimension
In two-dimensional space, both directions located in the same plane , and the distance in locations only separated by width and length (there is no volume in this model)
The water was heavier since it was more concentrated
The net force on the box parallel to the plane is
∑ F[para] = mg sin(24°) = ma
where mg is the weight of the box, so mg sin(24°) is the magnitude of the component of its weight acting parallel to the surface, and a is the box's acceleration.
Solve for a :
g sin(24°) = a ≈ 3.99 m/s²
The box starts at rest, so after 7.0 s it attains a speed of
(3.99 m/s²) (7.0 s) ≈ 28 m/s
Answer:
Rotational frequency=ω/2
Explanation:
Hence rotational frequency will be = 13.55 kHz
