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yoojayypee1236
06.05.2020 •
Physics
A mass m = 3.9 kg hangs from a massless string wrapped around a uniform cylinder with mass Mp = 10.53 and radius R= 0.96 m. The cylinder has a frictionless pivot in the center. The mass is released from rest and falls downwards unwinding the string from the outside of the cylinder.
What is the rotational inertia of the cylinder, and how fast is the mass moving after it has fallen a distance of 7.56 m.?
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Ответ:
the rotational inertia of the cylinder = 4.85 kgm²
the mass moved 7.942 m/s
Explanation:
Formula for calculating Inertia can be expressed as:
For calculating the rotational inertia of the cylinder ; we have;
I ≅ 4.85 kgm²
mg - T ma and RT = I ∝
T =![\frac{Ia}{R^2}](/tpl/images/0645/0784/6e587.png)
a = 4.1713 m/s²
Using the equation of motion
Ответ:
In 1897, Thomson set out to prove that the cathode rays produced from the cathode were actually a stream of negatively charged particles called electrons. (See Figure 1.8 in the textbook for Thomson's experimental setup). From Maxwell's theory, he knew that charged particles could be deflected in a magnetic field.
Hopefully this helped and have a blessed day, hun!