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Irina-Kira [14]
1 year ago
10

Type the correct answer in the box.use numerals instead of words. if necessary, use / for the fraction bar. δabc is a right tria

ngle in which ∠b is a right angle, ab = 1, ac = 2, and bc = . cos c × sin a =
Physics
1 answer:
NNADVOKAT [17]1 year ago
8 0

From the diagram The value of cos C × sin A = \frac{3}{4}

<h3>Determine the numerical value of cos C × sin A</h3>

First step : determine the values of cos C and sin A

cos C = adjacent / hypotenuse

          = a / b

          = \sqrt{3}  / 2

          = √3/2

sin A = sin 60⁰

         = √3/2

Therefore the numerical value of cos C * sin A = \frac{3}{4}

In conclusion From the diagram The value of cos C × sin A = \frac{3}{4}

Learn more about right angle : brainly.com/question/24323420

#SPJ1

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If the range of a projectile's trajectory is six times larger than the height of the trajectory, then what was the angle of laun
zvonat [6]

Answer:

H = 1/2 g t^2    where t is time to fall a height H

H = 1/8 g T^2   where T is total time in air  (2 t  = T)

R = V T cos θ       horizontal range

3/4 g T^2 = V T cos θ       6 H = R    given in problem

cos θ = 3 g T / (4 V)           (I)

Now t = V sin θ / g     time for projectile to fall from max height

T = 2 V sin θ / g

T / V = 2 sin θ / g

cos θ = 3 g / 4 (T / V)     from (I)

cos θ = 3 g / 4 * 2 sin V / g = 6 / 4 sin θ

tan θ = 2/3      

θ = 33.7 deg

As a check- let V = 100 m/s

Vx = 100 cos 33.7 = 83,2

Vy = 100 sin 33,7 = 55.5

T = 2 * 55.5 / 9.8 = 11.3 sec

H = 1/2 * 9.8 * (11.3 / 2)^2 = 156

R = 83.2 * 11.3 = 932

R / H = 932 / 156 = 5.97        6 within rounding

3 0
2 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

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Answer: Go ask your teacher and im sure if you search it on good it will answer faster than this.

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