The downward acceleration of the solid cylinder at the given tension in the string is determined as 2Tr/MR.
<h3>
Downward acceleration of the cylinder</h3>
The downward acceleration of the solid cylinder is determined from the principle of conservation of angular momentum as shown below;
Iα = Tr
where;
- I is moment of inertia of the solid cylinder
- α is angular acceleration of the cylinder
- T is tension in the string
- r is length of the string
α = Tr/I

where;
- a is the downward acceleration of the solid cylinder
- R is radius of the cylinder
Thus, the downward acceleration of the solid cylinder at the given tension in the string is determined as 2Tr/MR.
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Answer:
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Answer:
v=1.295
Explanation:
What we are given:
a=5÷(3s^(1/3)+s^(5/2)) m/s^2
Start by using equation a ds = v dv
This problem requires a numeric method of solving. Therefore, you can integrate v ds normally, but you must use a different method for a ds The problem should look like this:

<em>a=2</em>
<em>b=1</em>
<em>x=5÷(3s^(1/3)+s^(5/2)) </em><em>m/s^2</em>
<em>dx=dv</em>
Integrate the left side the standard method.

<em>a=v</em>
<em>b=0</em>
<em>dx=dv</em>
<em>Integrating</em>
=v^2/2
Use Simpson's rule for the right site.

<em>a=b</em>
<em>b=a</em>
<em>x=f(x)</em>
f(x)=b-a/6*(f(a)+4f(a+b/2)+f(b)
If properly applied. you should now have the following equation:
v^2/2=5[(1/6*(0.25+4(0.162)+(0.106)]
=0.8376
Solve for v.
v=1.295
Refer to the diagram shown below.
The component of the applied force perpendicular to the door is
F * sin(60°) = 0.866F N
Because the moment arm is 0.40 m, the torque is
(0.866F N)*(0.4 m) = 0.3464F N-m
This torque is equal to 1.4 N-m, therefore
0.3464F = 1.4
F = 4.04 N
Answer: 4.04 N