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3 years ago

5

When is the net force on an object equal to zero? A. When the object is in motion B. When the object is changing direction C. Wh

en the forces acting on the object are balanced D. When inertia is not present pls i need the answers quick

1 answer:

spin [16.1K]3 years ago

5 0

Answer: C. when the forces acting on the object are balanced.

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3 years ago

The sound source of a ship's sonar system operates at a frequency of 18.0 kHzkHz . The speed of sound in water (assumed to be at

Flura [38]

Answer:

Explanation:

a ) wave length of waves in water

= velocity / frequency

= 1482 / (18 x 1000)

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= 18000 [ 1482 / (1482 - 4.95 ) ]

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3 years ago

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

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pantera1 [17]

Answer: an steppe climate

Explanation:

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3 years ago

We know that charged objects with the same sign repel each other and objects charged with opposite signs attract. What happens w

choli [55]

solution:

When an uncharged conducting object brought near to a charged insulating object there is a force on the conducting object to move the electrons within it to opposite sides of the conductor. That means there is a separation of charges in the conducting object in the presence of the charged insulating object near to it but the charge on the conducting object is neutral.

Thus, the conducting object is uncharged.

There is a force of attraction between the uncharged conducting object and the insulating object when it brought near to the insulating object.

Thus, there is a force on the conducting object.

The conductor remains uncharged and a force is exerted on it.

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3 years ago