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scoundrel [369]
3 years ago
8

During the launch from a board, a diver's angular speed about her center of mass changes from zero to 8.10 rad/s in 240 ms. Her

rotational inertia about her center of mass is 11.6 kg·m2. During the launch, what are the magnitudes of (a) her average angular acceleration and (b) the average external torque on her from the board?
Physics
1 answer:
Artemon [7]3 years ago
6 0

Answer:

Part a)

\alpha = 33.75 rad/s^2

Part b)

\tau = 391.5 Nm

Explanation:

Part a)

Average angular acceleration is given as rate of change in angular speed

so it is given as

\alpha = \frac{\omega_f - \omega_i}{\Delta t}

\alpha = \frac{8.10 - 0}{240\times 10^{-3}}

\alpha = 33.75 rad/s^2

Part b)

average external torque is given as

\tau = I\alpha

here we know that

I = 11.6 kg m^2

\tau = 11.6 \times 33.75

\tau = 391.5 Nm

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If 61.4 cm of copper wire (diameter = 1.08 mm, resistivity = 1.69 × 10-8Ω·m) is formed into a circular loop and placed perpendic
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Answer:

the thermal energy generated in the loop = 6.64*10^{-4}  \ W

Explanation:

Given that;

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r = \frac{0.614}{2 \pi}

r = 0.0977 m

However , the area of the loop is :

A_L = \pi r^2

A_L = \pi (0.0977)^2

A_L = 0.02999 \ m^2

Change in the magnetic field is \frac{dB}{dt}= 0.0914 \ T/s

Then the induced emf e = A_L \frac{dB}{dt}

e = 0.02999 * 0.0914

e = 2.74 × 10⁻³ V

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diameter of the wire = 1.08 mm

radius of the wire = 0.54 mm = 0.54 × 10⁻³  m

Thus, the resistance of the wire  R = \frac {\rho L}{\pi r^2}

R =  \frac{(1.69*10^{-8})(0.614)}{   \pi (0.54*10^{-3})^2}

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Finally,  the thermal energy generated in the loop (i.e the power) = \frac{e^2}{R}

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Answer:

P = 800 Pa

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The pressure of water at the bottom of the pot can be given by the following formula:

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