1. Vertically oriented circular disks have strings wrapped around them. The other ends of the strings are attached to

Physics

1 answer:

Ymorist [56]2 years ago

5 0

Answer:

See the answers and explanation below

Explanation:

To solve this problem, we must have the full description of this problem, by doing an internet search we can find a problem with the same description and with the respective question.

Description of the problem

"Vertically oriented circular disks have strings wrapped around them. The other ends of the strings are attached to hanging masses. The diameters of the disks, the masses of the disks, and the masses of the hanging masses are

given. The disks are fixed and are not free to rotate. Specific values of the variables are given in the figures. Rank these situations, from greatest to least, on the basis of the magnitude of the torque on the disks. That is, put first the situation where the disk has the greatest torque acting on it and put last the situation where the disk has the least torque acting on it."

For case D

T = (20/2)*800 = 8000 [g-cm]

For case A

T = (20/2)*500 = 5000 [g-cm]

For case C

T = (10/2)*500 = 2500 [g-cm]

For case B

T = (10/2)*200 = 1000 [g-cm]

In this way it has been organized from the largest to the smallest torque present in each of the cases.

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vA 61.2-kg circus performer is fired from a cannon that is elevated at an angle of 57.8 ° above the horizontal. The cannon uses

dsp73

Answer:

The effective spring constant of the firing mechanism is 1808N/m.

Explanation:

First, we can use kinematics to obtain the initial velocity of the performer. Since we know the angle at which he was launched, the horizontal distance and the time in which it's traveled, we can calculate the speed by:

$v_0_x=\frac{x}{t}\\ \\v_0\cos\theta=\frac{x}{t}\\\\v_0=\frac{x}{t\cos\theta}$

(This is correct because the horizontal motion has acceleration zero). Then:

$v_0=\frac{20.8m}{(2.60s)\cos57.8\°}\\\\v_0=15.0m/s$

Now, we can use energy to obtain the spring constant of the firing mechanism. By the conservation of mechanical energy, considering the instant in which the elastic band is at its maximum stretch as t=0, and the instant in which the performer flies free of the bands as final time, we have:

$E_0=E_f\\\\U_e=K\\\\\frac{1}{2}kx^2=\frac{1}{2}mv^2\\\\\implies k=\frac{mv^2}{x^2}$