Answer:
1072.86 Newtons.
Step-by-step explanation:
It seems like you are trying to find the tension force at the bottom of a rotating motion given some parameters. The information provided includes the mass (11g), the force (1.44N), the radius (11.5), and an "f" value (5). I assume "f" represents the angular velocity or frequency of rotation.
To find the tension force at the bottom of the motion, you can use the following equation for circular motion:
Tension force (T) = (m * v^2) / r
Where:
- T is the tension force (what you want to find).
- m is the mass in kilograms (convert 11g to kg: 0.011 kg).
- v is the tangential velocity.
- r is the radius.
First, let's find the tangential velocity (v). In the case of uniform circular motion, you can use the following equation:
v = r * ω
Where:
- v is the tangential velocity.
- r is the radius (11.5 meters).
- ω is the angular velocity in radians per second. You mentioned an "f" value of 5, which may represent the frequency (in Hz). You can find ω from f using the formula: ω = 2πf.
ω = 2π * 5 = 10π radians per second.
Now, calculate v:
v = 11.5 m * 10π rad/s = 115π m/s
Now, you have v and r, so you can calculate the tension force (T):
T = (0.011 kg * (115π m/s)^2) / 11.5 m
T ≈ 110π^2 N
T ≈ 1072.86 N (approximately)
So, the tension force at the bottom of the motion is approximately 1072.86 Newtons.