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meababy2009ow9ewa
16.01.2020 •
Physics
The 2 kg 30 cm diameter disk is spinning at 300 rpm. how much friction force must the brake appply to the rim to bring the disk to a halt?
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Ответ:
Force of Friction= 1.57 N
Explanation:
Given Data
Mass=2 kg
Diameter=30 cm
To Find
Friction Force=?
Solution
First We need to find angular deceleration rate
a = (300×2π/60 rad/s)/3.0 s
a= (10/3)×π = 10.47 rad/s²
The required torque T is given by
T = I×a
where
I is the moment of inertia of the disc, (1/2) M×R² (M is mass and R is radius)
a is angular deceleration
Therefore
T= (1/2) M×R² ×10.47
first we need to find radius in meter so
R=diameter/(2×100)
R=(30 cm)/(2×100) m
R=0.15 m
T=(1/2)(2 kg)×(0.15 m)²(10.47 rad/s² )
T=0.2355 N.m
Once you have determined torque T, use the below formula to find friction
Force of Friction= T/R
Force of Friction= 0.2355 N.m/0.15 m
Force of Friction= 1.57 N
Ответ:
A qualitative data is something that is not described by numbers. The pH of water, oxygen concentration, and changed in the size of fishes can all be described by a number.
Only the occurrences of usual odors cannot be described by a number.