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erastovalidia [21]
3 years ago
10

During a testing process, a worker in a factory mounts a bicycle wheel on a stationary stand and applies a tangential resistive

force of 115 N to the tire's rim. The mass of the wheel is 1.70 kg and, for the purpose of this problem, assume that all of this mass is concentrated on the outside radius of the wheel. The diameter of the wheel is 50.0 cm. A chain passes over a sprocket that has a diameter of 9.50 cm. In order for the wheel to have an angular acceleration of 4.90 rad/s2, what force, in Newtons, must be applied to the chain? (Enter the magnitude only.) N
Physics
1 answer:
Alex3 years ago
5 0

Answer:

616.223684211 N

Explanation:

F_r = Resistive force on the wheel = 115 N

F = Force acting on sprocket

r_2 = Radius of sprocket = 4.75 cm

r_1 = Radius of wheel = 25 cm

Moment of inertia is given by

I=mr_1^2\\\Rightarrow I=1.7\times 0.25^2\\\Rightarrow I=0.10625\ kgm^2

Torque

\tau=I\alpha\\\Rightarrow \tau=0.10625\times 4.9\\\Rightarrow \tau=0.520625\ Nm

Torque is given by

\tau=Fr_2-F_rr_1\\\Rightarrow F=\dfrac{\tau+F_rr_1}{r_2}\\\Rightarrow F=\dfrac{0.520625+115\times 0.25}{0.0475}\\\Rightarrow F=616.223684211\ N

The force on the chain is 616.223684211 N

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A BODY STARTING FROM REST MOVES WITH CONSTANT ACCELERATON. what is the ratio of distance covered by the body during the fifth se
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Answer:

9/25

Explanation:

Distance covered in the first 5 seconds:

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Δx₀₋₅ = 25a/2

Distance covered in the first 4 seconds:

Δx = v₀ t + ½ at²

Δx₀₋₄ = (0) (4) + ½ a (4)²

Δx₀₋₄ = 8a

So the distance covered during the 5th second is:

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Δx₅ = 9a/2

So the ratio of the distance covered during the 5th second to the distance covered in the first 5 seconds is:

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8 0
3 years ago
An airplane traveling at half the speed of sound emits a sound of frequency 5.84 kHz. (a) At what frequency does a stationary li
sergejj [24]

Answer:

Stationary listener frequency = 77.68 kHz

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f0 = 5.84[(344+0)/(344-172)]

f0 = 5.84[(344)/172]

Stationary listener frequency = 77.68 kHz

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