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m_a_m_a [10]
3 years ago
5

In an experiment what are all the parts of an experiment that remain unchanged are called​

Physics
1 answer:
stira [4]3 years ago
3 0
The answer for this question is Control Variable because it doesn’t change throughout the experiment.
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The conversion of thermal energy into mechanical energy requires
emmainna [20.7K]
Temperature difference is required, so i’m guessing - a. thermometer - would be required to check that temperature.
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3 years ago
Motion is always compared to something called
HACTEHA [7]

Movement can be without change in positions. As only a part of it can be made to move. If you move there is a motion and if there is a motion there is a movement so they're same thing. In motion whole of body moves, but in movements only the parts of body moves.

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;

  • <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

6 0
3 years ago
Set local ground level to 700 ft, and record the inches of mercury.
irakobra [83]

The main way in which an altimeter measures the altitude of an object is by calculating the location's air pressure.

<h3>What is an Altimeter?</h3>

This refers to the instrument that is used to measure the altitude of an object when it is at a fixed level.

Hence, we can see that an altimeter should NOT be confused with a barometer, because although they both measure pressure, a barometer calculates the change in air pressure and elevation based on available weather.

Please note that your question is incomplete so I gave you a general overview to help you get a better understanding of the concept.

Read more about altimeters here:

brainly.com/question/17633815

#SPJ1

6 0
1 year ago
An object is located 51 millimeter from a diverging lens. te object has a hight of 13 millimeters and the image height is 3.5 mi
Sati [7]
<span>An object is located 51 millimeters from a diverging lens the object has a height og 13 millimeters and the image height is 3.5 millimeters how far in front of the lens is image located?</span>
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3 years ago
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