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Tom [10]
3 years ago
11

A jetliner is moving at a speed of 245m/s. The vertical component of the plane's velocity is 40.6 m/s. Determine the magnitude o

f the horizontal component of the plane velocity.
Physics
1 answer:
Harman [31]3 years ago
8 0

Answer:

Horizontal component of velocity = 241.61 m/s

Explanation:

Given that,

Velocity of a jetliner is 245 m/s

The vertical component of the plane's velocity is 40.6 m/s, v_y=40.6\ m/s

We need to find the magnitude of the horizontal component of the plane velocity.

The resultant velocity is given by :

v=\sqrt{v_x^2+v_y^2}

Where, v_x is the horizontal component of the plane velocity

v^2=v_x^2+v_y^2\\\\v_x^2=v^2-v_y^2\\\\v_x^2=(245)^2-(40.6)^2\\\\v_x=\sqrt{58376.64}\\\\v_x=241.61\ m/s

So, the magnitude of the horizontal component of the plane velocity is 241.61 m/s.

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According to the law of multiple proportions, if 12 g of carbon combine with 16 g of oxygen to form co, the number of grams of c
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The number of grams of carbon that combine with 16 g of oxygen in the formation of CO₂ is 6g.

When two elements combine to make more than one compound, the masses of one element combined with a fixed amount of another element are in the ratio of whole numbers, according to the law of multiple proportions.

When combined with oxygen, carbon can produce two different compounds. They are referred to as carbon dioxide (CO₂) and carbon monoxide (CO).

Carbon monoxide is formed by combining 12 g of carbon with 16 g of oxygen whereas Carbon dioxide is formed when 12 g of carbon reacts with 32 g of oxygen. The amount of carbon is fixed at 12 g in each case. The mass ratio of carbon monoxide to carbon dioxide is 16: 32, or 1: 2.

But in the given case, 16g of oxygen is reacting instead of 32g. Therefore, the number of grams of carbon reacting will be:

\frac{12}{2}=6g

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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.

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Answer:

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Explanation:

D=18^{\circ}

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