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2 years ago

Voltage drop is a better measurement of resistance than testing static resistance because ________.

1 answer:

7 0

Voltage drop is a better measurement of resistance than testing static resistance because: current flow creates heat which adds to circuit resistance

The current flow or electric current is the measure of the electrons flow around a circuit. It is measured in amperes or "amps". If the current is higher the electrons flow is greater.

<h3>What is electricity?</h3>

Electricity or electrical energy is the energy that occurs due to the movement of electrons in the materials that are considered conductors.

Learn more about electricity at: brainly.com/question/23063355

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The sun's energy is probably produced by _____.

WARRIOR [948]

The suns energy is probably produced by nuclear fusion

i hope i helped

5 0

3 years ago

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The strength of the gravitational attraction depends on the mass of the objects involved and on the distance between them. The g

wlad13 [49]

Answer:

The gravitational force between two objects depends only on their masses.

i think sorry if its wrong!!

Explanation:

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3 years ago

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Three students attempt to define what it means for lines lll and mmm to be perpendicular. Can you match the teacher's comments t

Ber [7]

Incompletevquestion. However, I inferred from a general perspective about perpendicular lines.

Explanation:

Put simply, perpendicular lines are lines that are at right angles (90°) to each other. Thus, we could say based on this definition that for lines lll and mmm to be perpendicular they intersect and be at right angles (90°) to each other as found on the attached image.

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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;

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

Two Mississippi companies, Howard Industries and Kuhlman Electric Company, are responsible for most of the transformers and powe

yanalaym [24]

I think b but im not sure just helping

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3 years ago

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