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Answer:
Work done, W =1520 J
Explanation:
We have,
The brakes on a bicycle apply 95 N of force to the wheels. When the brakes are applied, the bicycle comes to a stop in 16 m.
It is required to find the work done by the brakes on the wheels. We know that the product of force and displacement is equal to the work done. It is given by :
So, the work done by the brakes is 1520 J.
Answer:
216.31 (the work done by gravity is -216.31) positive for going up.
Explanation:
We look at this question first by getting the right equation for <em>work</em>.
Which should be... W = F x D.
From this, we can do everything, we need the Force (F) first - the question tells us that Joe is lying on his back and moves his arms upward to raise the barbell. This means that he is countering the force of graving on the object.
What is the formula for the force of gravity on an object near the earth?
Right here --- = mg
m = the mass and...
g = the acceleration due to gravity which is <em>9.81 m/s2</em>
Before we plug things in though, we need to convert everything to SI units,
the weight is in kg - so we're good to go there, but the length of Joe's arms are in "cm" we need m or meters. Converting 70 cm to m = .7 m.
Now, we just put it all together - (31.5kg)(9.81m/s2)(.7m) = 216.31 J or 216.31 N m.
(a) The final angular velocity of the flywheel after 3 complete revolutions is 4.96 rad/s.
(b) The time taken for the flywheel to make 3 complete revolutions is 5.93 s.
<h3>
Final angular velocity</h3>
The final angular velocity of the flywheel after 3 complete revolutions is determined by applying third kinematic equation as shown below;
θ = 2π (rad/rev) x (3 rev) = 18.85 rad
ωf² = ωi² + 2αθ
ωf² = (1.4)² + 2(0.6)(18.85)
ωf² = 24.58
ωf = √24.58
ωf = 4.96 rad/s
<h3>Time of motion</h3>
The time taken for the flywheel to make 3 complete revolutions is calculated as follows;
ωf = ωi + αt
t = (ωf - ωi)/α
t = (4.96 - 1.4)/0.6
t = 5.93 s
Learn more about time of motion here: brainly.com/question/2364404
Answer:
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