All categories

3 years ago

21. For this table of data, how should the y-axis be labeled (with units)?

Physics

1 answer:

Snowcat [4.5K]3 years ago

8 0

Answer:

The y-axis would be labeled with displacement (m)

Explanation:

The displacement is dependent on the force applied.

Labels should have what you measure and how your units of measure.

The x-axis would be labeleed with force applied (N)

The y-axis would be labeled with displacement (m)

You might be interested in

Which of these is an example of potential energy?

Mama L [17]

Answer:

A person on top of a hill

Explanation:

7 0

2 years ago

Which part of the eye controls whether or not your pupil dilates or restricts?

valentinak56 [21]

The answer would be, the iris. Hope this helps:)

5 0

4 years ago

Explain how you think global winds can impact weather in an area.

weeeeeb [17]

Answer:

As the wind travels across the earth, air masses and global winds do not travel in straight lines. in the Northern hemisphere air veers to the right and in southern hemisphere to the left. this motion can result in large circulating weather systems, as air blows away from or into a high or low pressure areas.

on earth the main differences in air pressure are caused by differences in temperatures. cool air produces high air pressure and warm air produces low air pressure.

5 0

3 years ago

If a running system has a total change in heat of 295 joules, and it's running at a temperature of 402 kelvin, what is the entro

Liula [17]

The answer is either 75 or 55

7 0

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;

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

6 0

3 years ago