At its peak, a tornado is 60.0 m in diameter and carries 500 km/h winds. What is its angular velocity in revolutions per second?
v=500000/3600 =138.9 m/s
ω=v/R=138.9/30=4.63 rad/s
n= ω/2π=4.63/2 π=0.74 rev/s
To find the angular velocity of the tornado in revolutions per second, we need to use the formula:
Angular velocity = Linear velocity / Radius
First, let's convert the linear velocity from km/h to m/s.
Linear velocity = 500 km/h * (1/3.6) m/s = 138.8889 m/s
Next, we need to find the radius of the tornado. The diameter is given as 60.0 m, so the radius will be half of that.
Radius = 60.0 m / 2 = 30.0 m
Now we can calculate the angular velocity.
Angular velocity = 138.8889 m/s / 30.0 m = 4.6296 rad/s
Since 1 revolution is equal to 2π radians, we can convert the angular velocity to revolutions per second.
Angular velocity in revolutions per second = 4.6296 rad/s / (2π rad/rev) ≈ 0.7354 rev/s
Therefore, the angular velocity of the tornado is approximately 0.7354 revolutions per second.
To find the angular velocity of the tornado in revolutions per second, we first need to convert the velocity from kilometers per hour to meters per second.
1 kilometer = 1000 meters
1 hour = 3600 seconds
So, the tornado's velocity in meters per second is calculated as follows:
500 km/h * (1000 m / 1 km) * (1 h / 3600 s) = 500,000 / 3600 m/s
= 138.89 m/s (rounded to two decimal places)
Now, let's calculate the tornado's angular velocity. The angular velocity (ω) is given by the equation:
ω = v / r
Where:
ω = Angular velocity (in radians per second)
v = Linear velocity (in meters per second)
r = Radius (diameter / 2)
Given that the diameter of the tornado is 60.0 m, the radius would be:
r = 60.0 m / 2 = 30.0 m
Now we can substitute the values into the equation:
ω = 138.89 m/s / 30.0 m
= 4.63 radians per second (rounded to two decimal places)
Since the question asks for angular velocity in revolutions per second, we need to convert radians to revolutions. There are 2π radians in one revolution:
1 revolution = 2π radians
So let's convert:
4.63 radians / (2π radians / 1 revolution)
≈ 0.736 revolutions per second (rounded to three decimal places)
Therefore, at its peak, the tornado has an angular velocity of approximately 0.736 revolutions per second.