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3 years ago

12

In establishing standing waves with a Slinky®, your wrist oscillates with f frequency of 8.0 Hz, and the velocity of the wave is

40 m/s. What is the wavelength of this wave?

1 answer:

emmainna [20.7K]3 years ago

6 0

The answer is 5.0, as velocity = frequency * wavelength.

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A piece of aluminum (bulk modulus 7.1 x 1010 N/m2) is placed in a vacuum chamber where the air pressure is 0.781 x 105 Pa. The v

Jet001 [13]

Answer:

${\dfrac{\Delta V}{V}}=11\times 10^{-7}$

Explanation:

Given that

Bulk modulus ,K = 7.1 x 10¹⁰ Pa

The pressure P₁ = 0.781 x 10⁵ Pa

The final pressure P ₂ = 0 Pa

We know that bulk modulus given as

$K=\dfrac{\Delta P}{\dfrac{\Delta V}{V}}$

Now by putting the values

$K=\dfrac{\Delta P}{\dfrac{\Delta V}{V}}$

$7.1\times 10^{10}=\dfrac{0.781\times 10^5}{\dfrac{\Delta V}{V}}$

${\dfrac{\Delta V}{V}}=\dfrac{0.781\times 10^5}{7.1\times 10^{10}}$

${\dfrac{\Delta V}{V}}=11\times 10^{-7}$

Therefore fractional change will be 11 x 10⁻⁷.

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4 years ago

How can positive feedback influence the behavior of a system

never [62]

Answer

It can cause the system to increase its output more and more.

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3 years ago

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Identifying the guilty party was mainly based on eyewitness accounts during what time period?

kondaur [170]

The time period for guilty party was between 1900-1988.

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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.

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3 years ago

2nd grade work. Anyone?

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Answer:

A. shadow

Explanation:

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