List of priceless your bodies from largest to smallest in terms of their distance from earth
1 answer:
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Refer to the diagram shown below.
Let I = the moment of inertia of the wheel.
α = 0.81 rad/s², the angular acceleration
r = 0.33 m, the radius of the weel
F = 260 N, the applied tangential force
The applied torque is
T = F*r
= (260 N)*(0.33 m)
= 85.8 N-m
By definition,
T = I*α
Therefore,
I = T/α
= (85.8 N-m)/(0.81 rad/s²)
= 105.93 kg-m²
Answer: 105.93 kg-m²
Let I = the moment of inertia of the wheel.
α = 0.81 rad/s², the angular acceleration
r = 0.33 m, the radius of the weel
F = 260 N, the applied tangential force
The applied torque is
T = F*r
= (260 N)*(0.33 m)
= 85.8 N-m
By definition,
T = I*α
Therefore,
I = T/α
= (85.8 N-m)/(0.81 rad/s²)
= 105.93 kg-m²
Answer: 105.93 kg-m²
Answer:
The the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s
Explanation:
We are given that
Angular acceleration,
Diameter of the wheel, d=21 cm
Radius of wheel, cm
Radius of wheel,
1m=100 cm
Magnitude of total linear acceleration, a=
We have to find the linear speed of a at an instant when that point has a total linear acceleration with a magnitude of 1.7 m/s2.
Tangential acceleration,
Radial acceleration,
We know that
Using the formula
Squaring on both sides
we get
Hence, the the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s