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3 years ago

9. You drive a car from Milwaukee to Chicago, which is a distance of 150km and it takes you 95

Physics

1 answer:

Katena32 [7]3 years ago

8 0

For this case we have to, by definition:

$v = \frac {x} {t}$

Where:

v: It's the velocity

x: It is the distance traveled

t: It is the time spent

1 hour equals 60 minutes and 1 minute equals 60 seconds.

On the other hand:

1 kilometer equals 1000 meters

So: $\frac {150} {95} \frac{km} {min} * \frac {1} {60} \frac {min} {sec} * \frac {1000} {1} \frac {m} {km} = \frac {150 * 1000} {95 * 60} \frac {m} {sec} = \frac {15000} {5700} \frac {m} {sec} = 2.63 \frac {m} {sec}$

On the other hand: $\frac {150} {95} \frac{km} {min} * \frac {60} {1} \frac {min} {hr} = \frac {150 * 60} {95} \frac {km} {h} = 94.74 \frac {km} {h}$

ANswer:

$2.63 \frac {m} {sec}\\94.74 \frac {km} {h}$

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Answer:

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Explanation:

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3 years ago

Of the superalloys, the cobalt-base types are the easiest to braze. True or False?

Ivahew [28]

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4 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;

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

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

Q = mcΔt

For copper metal, the quantity of heat transferred is calculated as;

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

The change in heat energy for copper metal;

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

For aluminum metal, the quantity of heat transferred is calculated as;

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

The change in heat energy for aluminum metal ;

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

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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3 years ago

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abruzzese [7]

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solniwko [45]

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Explanation:

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