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Brrunno [24]
2 years ago
7

A gymnast dismounts off the uneven bars in a tuck position with a radius of 0.3m (assume she is a solid sphere) and an angular v

elocity of 2rev/s. During the dismount she stretches out into the straight position, with a length of 1.5m, (assume she is a uniform rod through the center) for her landing. The gymnast has a mass of 50kg. What is her angular velocity in the straight position?
Physics
1 answer:
kifflom [539]2 years ago
6 0

Here we will say that there is no external torque on the system so we will have

L_i = L_f

here we know that

L_i = I_1\omega_1

where we know that

I_1 = \frac{2}{5}mr^2

Also we know that

I_2 = \frac{1}{12}mL^2

initial angular speed will be

\omega_1 = 2\pi(2rev/s) = 4\pi rad/s

now from above equation

\frac{2}{5}mr^2 (4\pi) = \frac{1}{12}mL^2 \omega

0.4(0.3)^2(4\pi) = \frac{1}{12}(1.5)^2\omega

0.452 = 0.1875 \omega

now we have

\omega = 2.41 rad/s

so final speed will be 2.41 rad/s

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3 years ago
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For Part A, Sebastian decided to use the copper cylinder. How would the magnitude of his q and ∆H compare if he were to redo Par
Vitek1552 [10]

The magnitudes of his q and ∆H for the copper trial would be lower than the aluminum trial.

The given parameters;

  • <em>initial temperature of metals, =  </em>T_m<em />
  • <em>initial temperature of water, = </em>T_i<em> </em>
  • <em>specific heat capacity of copper, </em>C_p<em> = 0.385 J/g.K</em>
  • <em>specific heat capacity of aluminum, </em>C_A = 0.9 J/g.K
  • <em>both metals have equal mass = m</em>

The quantity of heat transferred by each metal is calculated as follows;

Q = mcΔt

<em>For</em><em> copper metal</em><em>, the quantity of heat transferred is calculated as</em>;

Q_p = (m_wc_w + m_pc_p)(T_m - T_i)\\\\Q_p = (T_m - T_i)(m_wc_w ) + (T_m - T_i)(m_pc_p)\\\\Q_p = (T_m - T_i)(m_wc_w ) + 0.385m_p(T_m - T_i)\\\\m_p = m\\\\Q_p = (T_m - T_i)(m_wc_w ) + 0.385m(T_m - T_i)\\\\let \ (T_m - T_i)(m_wc_w )  = Q_i, \ \ \ and \ let \ (T_m- T_i) = \Delta t\\\\Q_p = Q_i + 0.385m\Delta t

<em>The </em><em>change</em><em> in </em><em>heat </em><em>energy for </em><em>copper metal</em>;

\Delta H = Q_p - Q_i\\\\\Delta H = (Q_i + 0.385m \Delta t) - Q_i\\\\\Delta H = 0.385 m \Delta t

<em>For </em><em>aluminum metal</em><em>, the quantity of heat transferred is calculated as</em>;

Q_A = (m_wc_w + m_Ac_A)(T_m - T_i)\\\\Q_A = (T_m -T_i)(m_wc_w) + (T_m -T_i) (m_Ac_A)\\\\let \ (T_m -T_i)(m_wc_w)  = Q_i, \ and \ let (T_m - T_i) = \Delta t\\\\Q_A = Q_i \ + \ m_Ac_A\Delta t\\\\m_A = m\\\\Q_A = Q_i \ + \ 0.9m\Delta t

<em>The </em><em>change</em><em> in </em><em>heat </em><em>energy for </em><em>aluminum metal </em><em>;</em>

\Delta H = Q_A - Q_i\\\\\Delta H = (Q_i + 0.9m\Delta t) - Q_i\\\\\Delta H = 0.9m\Delta t

Thus, we can conclude that the magnitudes of his q and ∆H for the copper trial would be lower than the aluminum trial.

Learn more here:brainly.com/question/15345295

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It is found experimentally that the electric field in a certain region of Earth's atmosphere is directed vertically down. At an
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Answer:

q=1.7346×10⁻⁶C

Explanation:

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Ф(total)=E₃₀₀A Cos(180)+E₂₃₀ACos(0)

Ф(total)=A(E₃₀₀ - E₂₃₀)

The total flux is given by Gauss Law as:

Ф(total)=q/ε₀

q=ε₀Ф(total)

q=ε₀(A(E₃₀₀ - E₂₃₀))

Substitute the given values

q=(8.85×10⁻¹²){(70²)(100 - 60)}

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