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2 years ago

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(need help) " You make the following measurements of an object: 42kg and 22m^3. What would be the objects density be? Show work

for credit and include final units."

1 answer:

dlinn [17]2 years ago

6 0

Density + mass / volume = 42 / 22 = 1.909 kg / m^3 ( to the nearest thousandth)

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2 years ago

Ryan runs 50 m north and then 50 m east. What is his displacement?

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Answer:

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3 years ago

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3 years ago

You have a pendulum clock made from a uniform rod of mass M and length L pivoting around one end of the rod. Its frequency is 1

drek231 [11]

The new oscillation frequency of the pendulum clock is 1.14 rad/s.

The given parameters;

<em>Mass of the pendulum, = M </em>
<em>Length of the pendulum, = L</em>
<em>Initial angular speed, </em> $\omega _i$ <em> = 1 rad/s</em>

The moment of inertia of the rod about the end is given as;

$I_i = \frac{1}{3} ML^2$

The moment of inertia of the rod between the middle and the end is calculated as;

$I_f = \int\limits^L_{L/2} {r^2\frac{M}{L} } \, dr = \frac{M}{3L} [r^3]^L_{L/2} = \frac{M}{3L} [L^3 - \frac{L^3}{8} ] = \frac{M}{3L} [\frac{7L^3}{8} ]= \frac{7ML^2}{24}$

Apply the principle of conservation of angular momentum as shown below;

$I _i \omega _i = I _f \omega _f\\\\\frac{ML^2}{3} (1 \ rad/s)= \frac{7ML^2}{24} \times \omega _f\\\\\frac{24 \times ML^2}{3 \times 7 ML^2} (1 \ rad/s)= \omega _f\\\\1.14 \ rad/s = \omega _f$

Thus, the new oscillation frequency of the pendulum clock is 1.14 rad/s.

Learn more about moment of inertia of uniform rod here: brainly.com/question/15648129

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2 years ago