Answer: 1.76 Nm
Explanation:
If the force pulls horizontally, this means that the force is tangent to the disk at any point of the string unwinding process, so the distance d is irrelevant.
In this case, the torque is directly given by the product of the force times the distance perpendicular to the center of the disk, which is just the radius, as follows:
τ = F * r = 16 N. (0.11) m = 1.76 Nm
Answer:
ω = 630.2663 = 630[rad/s]
Explanation:
Solution:
- We can tackle this question by simple direct proportion relation between angular speed for the disk to rotate a cycle that constitutes 20 holes. We will use direct relation with number of holes per cycle to compute the revolution per seconds i.e frequency of speed f.
1rev(20 hole) -> 20(cycle)/rev
2006.2(cycle) -> f ?
f = 2006.2/20 = 100.31rev at second
- The relation between angular frequency and angular speed is given by:
ω = 2πf
ω = 2*3.14*100.31
ω = 630.2663 = 630[rad/s]
The goal of scientific method is to produce public knowledge. That is knowledge available to any person with sufficient education, all such knowledge can be verified by or deduced from experiments and observations.
Answer:
4.545454 seconds
Explanation:
1500/330=4.545454 seconds
Answer: An archer pulls her bowstring back 0.400m by exerting a force that increases uniformly from zero to 230N. The equivalent spring constant of the bow will be 575N/m
Explanation: To find the answer, we have to know about the Simple Harmonic Motion.
<h3>What is Simple Harmonic Motion?</h3>
- A particle is said to execute simple harmonic motion, if it moves to and fro about mean position under the action of restoring force.
- The restoring force is directly proportional to its displacement from the mean position and always directed towards the mean position.
- If x is the displacement from the mean position, and F is the restoring force, then
∝
where, k is called the spring constant.
<h3>How to approach the problem?</h3>
- Given that, an archer pulls her bowstring back 0.400m by exerting a force that increasing from zero to 230N.
- Here, from the question given, we can write,
and 
. - Thus, our spring constant k will be,
Thus, we can conclude that, the equivalent spring constant of the bow will be 575N/m.
Learn more about the Simple Harmonic Motion here:
brainly.com/question/28019840
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