Answer:
the maximum is I₁ axis of rotation at the end
the minimum moment is I₂ axis of rotation at the center of mass
Explanation:
For this exercise we use the definition moment of inertia
I = ∫ r² dm
for bodies of high symmetry it is tabulated; In this case we can approximate a broomstick to a thin rod, the moment of inertia with respect to a perpendicular axis when varying are
at one end
I₁ = ⅓ mL²
in in center
I₂ = m L²
There is another possible axis of rotation around the axis of the broom, in this case we have a solid cylinder
I₃ = m r²
remember that the diameter of the broom is much smaller than its length, therefore this moment of inertia is very small
when examining the different moments of inertia:
the maximum is I₁ axis of rotation at the end
the minimum moment is I₂ axis of rotation at the center of mass
The equation that describes the graph is y = 16.7x + 150.
<h3>What is the equation?</h3>
We have been given the graph of a certain function data set in physics. We know that the graph is the representation of data on cartesian coordinates. In this case, we are asked to find the equation of the graph in the form; y=mx+b
m = slope of the graph
b = intercept of the graph.
To obtain the slope;
m = y2 - y1/ x2 - x1
m = 400 - 200/ 16 - 4
m = 200/ 12
m = 16.7
Then we can see from the graph that the y - intercept is 150. Having these data, the equation that could describe the graph is now;
y = 16.7x + 150
Learn more about graph:brainly.com/question/13301664
#SPJ1
The fastest speed before the string breaks is 9.5 m/s
Explanation:
The motion of the block is a uniform circular motion, which is a circular motion with constant speed. The force that keeps the block in circular motion is called centripetal force; its direction is towards the centre of the circle and its magnitude is given by:
where
m is the mass of the block
v is its speed
r is the radius of the circle
In this problem, the centripetal force is provided by the tension in the string, T, so we can write:
The string breaks when the centripetal force becomes larger than the maximum tension in the string:
Re-arranging the equation for v,
and here we have:
T = 450 N
m = 10 kg
r = 2 m
Substituting,
So, the fastest speed before the string breaks is 9.5 m/s.
Learn more about circular motion:
brainly.com/question/2562955
brainly.com/question/6372960
#LearnwithBrainly
Explanation:
light travel slower in daimond
The vector force on the unit positive charge placed at any location in the field defines the strength of the electric field at that point. The charge used to determine field intensity (field strength) is known as the test charge. Now, a field line is defined as a line to which the previously mentioned field strength vectors are tangents at the relevant places. When we study positive charge field lines, the field strength vectors point away from the positive charge. If there is a negative charge anywhere in the vicinity, the field lines that began from the positive charge will all terminate at the negative charge if the value of the negative charge is the same as the value of the positive charge. Remember that the number of field lines originating from positive charge is proportional to the charge's value, and similarly, the number of field lines terminating at negative charge is proportionate to the charge's value. As a result, if all charges are equivalent, all lines originating from the positive charge terminate at the negative charge. If the value of the positive charge is greater than the value of the negative charge, the number of lines ending at the negative charge will be proportionally less than the number of lines beginning at the positive charge. The remaining lines that do not end at the negative charge will go to infinity. If the positive charge is less, all lines from it terminate at a negative charge, and any other reasonable number of ines terminate at a negative charge from infinity. We should also keep in mind that the number of lines that run perpendicular to the field direction across a surface of unit area is proportional to the field strength at that location. As a result, lines are dense in the strong field zone and sparse in the low intensity region.