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Pie
4 years ago
13

Ted is a high school student. He is very good at math, but he is not a very good English student. He likes all of his teachers;

he thinks they are fair and want him to learn. On Tuesday, Ted was sitting in his English classroom, listening to his teacher’s lecture. The teacher asked the class a question. He called on Ted. Ted had not been paying attention. His palms began to sweat, and his heart rate increased. He felt embarrassed that he did not know the answer. On Wednesday, Ted was asked to answer another question in his math classroom. Even though this time he was listening and knew the answer, his palms began to sweat and his heart rate increased. Why did Ted’s palms sweat and heart rate increase on Wednesday? A. Ted associated his embarrassment with being in high school. B. Ted associated his math teacher with feelings of embarrassment. C. Ted was not completely sure that his answer was correct. D. Ted associated being asked a question with embarrassment.
Physics
2 answers:
oee [108]4 years ago
7 0

Answer:

I think its D...?

Explanation:

TAking the test rn

scoray [572]4 years ago
3 0
I think the correct answer is

D) Ted associated being asked a question with embarrassment.

Glad I could help, and good luck!

AnonymousGiantsFan
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12)A black body is heated from 27°C to 127° C. The ratio of their energies of radiations emitted will be
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Answer:

81:256.

Explanation:

Let T denote the absolute temperature of this object.

Calculate the value of T before and after heating:

T(\text{before}) = 27 + 273 = 300\; \rm K.

T(\text{after}) = 127 + 273 = 400\; \rm K.

By the Stefan-Boltzmann Law, the energy that this object emits (over all frequencies) would be proportional to T^4.

Ratio between the absolute temperature of this object before and after heating:

\displaystyle \frac{T(\text{before})}{T(\text{after})} = \frac{3}{4}.

Therefore, by the Stefan-Boltzmann Law, the ratio between the energy that this object emits before and after heating would be:

\displaystyle \left(\frac{T(\text{before})}{T(\text{after})}\right)^{4} = \left(\frac{3}{4}\right)^{4} = \frac{81}{256}.

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3 years ago
Which part of an object's rate of change best defines acceleration?
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3 years ago
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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
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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

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