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2 years ago

A 1500 kg car

1 answer:

5 0

<h2>We determine the final velocity, </h2><h2>v</h2><h2>f</h2><h2>, of the car. We do this by applying the kinematic equation,</h2><h2 /><h2>v</h2><h2>2</h2><h2>f</h2><h2>=</h2><h2>v</h2><h2>2</h2><h2>i</h2><h2>+</h2><h2>2</h2><h2>a</h2><h2>d</h2><h2 /><h2>where </h2><h2>v</h2><h2>i</h2><h2> is the initial velocity, a is the acceleration, and d is the distance covered across the movement. We can express the acceleration using the equation,</h2><h2 /><h2>a</h2><h2>=</h2><h2>F</h2><h2>m</h2><h2 /><h2>where F is the applied force and m is the mass of the car. We use the following values for the variables:</h2><h2 /><h2>v</h2><h2>i</h2><h2>=</h2><h2>5.25</h2><h2> </h2><h2>m</h2><h2>s</h2><h2>−</h2><h2>1</h2><h2>F</h2><h2>=</h2><h2>1250</h2><h2> </h2><h2>N</h2><h2>m</h2><h2>=</h2><h2>1500</h2><h2> </h2><h2>k</h2><h2>g</h2><h2>d</h2><h2>=</h2><h2>42.8</h2><h2> </h2><h2>m</h2><h2>We proceed with the solution.</h2><h2 /><h2>v</h2><h2>2</h2><h2>f</h2><h2>=</h2><h2>v</h2><h2>2</h2><h2>i</h2><h2>+</h2><h2>2</h2><h2>a</h2><h2>d</h2><h2>v</h2><h2>2</h2><h2>f</h2><h2>=</h2><h2>v</h2><h2>2</h2><h2>i</h2><h2>+</h2><h2>2</h2><h2>F</h2><h2>d</h2><h2>m</h2><h2>Take the positive root.</h2><h2>v</h2><h2>f</h2><h2>=</h2><h2>√</h2><h2>v</h2><h2>2</h2><h2>i</h2><h2>+</h2><h2>2</h2><h2>F</h2><h2>d</h2><h2>m</h2><h2>=</h2><h2>√</h2><h2>(</h2><h2>5.25</h2><h2> </h2><h2>m</h2><h2>s</h2><h2>−</h2><h2>1</h2><h2>)</h2><h2>2</h2><h2>+</h2><h2>2</h2><h2>(</h2><h2>1250</h2><h2> </h2><h2>N</h2><h2>)</h2><h2>(</h2><h2>42.8</h2><h2> </h2><h2>m</h2><h2>)</h2><h2>1500</h2><h2> </h2><h2>k</h2><h2>g</h2><h2>≈</h2><h2>9.9</h2><h2> </h2><h2>m</h2><h2>s</h2><h2>−</h2><h2>1</h2><h2>Help improve Study.com. Report an Error</h2>

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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.

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<em>Mass of the pendulum, = M </em>
<em>Length of the pendulum, = L</em>
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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

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Delvig [45]

Answer: Pedaling your bike : acceleration :: applying the brakes : inertia.

The reason I think this to be the answer to the analogy is because there is energy and work used in both processes (and the unit focuses on forces); gravity is constant and does not change whether one pedals or applies brakes. And I do not think it's deceleration, as deceleration tends to equate to acceleration within the physics perspective.

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2 years ago