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2 years ago

All heat engines involve _____. the refrigeration cycle evaporation and condensation combustion

Physics

2 answers:

AURORKA [14]2 years ago

7 0

The answer is Combustion took the test and made a 93

pishuonlain [190]2 years ago

4 0

Combustion is the answer i just took the test

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Which characteristic should a good scientific question have?

Brilliant_brown [7]

The best answer to this question is D. a god scientific question should be based on observations instead of just imagination.

Hope this helps.

7 0

2 years ago

Read 2 more answers

water has an index of refraction of 1.33. What is the critical angle for light leaving a pool of water into air? 0 37 90 49

vesna_86 [32]

When light travels from a medium with higher refractive index into a medium with lower refractive index, there is a maximum angle (called critical angle) for which all the light is reflected, so there is no refraction.

The value of the critical angle is given by:
$\theta = \arcsin ( \frac{n_2}{n_1} )$
when n1 is the refractive index of the first medium, while n2 is the refractive index of the second medium. In our case, n1=1.33 (the water) and n1=1.00 (the air). Putting numbers in, we get
$\theta = \arcsin ( \frac{1.00}{1.33} )=49^{\circ}$

6 0

3 years ago

The volume of an ideal gas is adiabatically reduced from 151 L to 80.6 L. The initial pressure and temperature are 1.50 atm and

Zolol [24]

Answer:

gas is dioatomic

T_f = 330.0 K

$\eta = 7.07 mole$

Explanation:

Part 1

below equation is used to determine the type Gas by determining $\gamma$ value

$\frac{V_{1}}{V_{F}}\gamma=\frac{P_{i}}{P_{f}}$

where V_i and V_f is initial and final volume respectively

and P_i and P_f are initial and final pressure

$\gamma = \frac{ln(P_f/P_i)}{ln(V_i/V_f)}$

$\gamma = \frac{ln(3.61/1.50)}{ln(151/80.6}$

\gamma = 1.38

therefore gas is dioatomic

Part 2

final temperature in adiabatic process is given as

$T_f = T_i*[\frac{v_i}{V_f}](^\gamma-1)$

substituing value to get final temperature

$T_f = 260*[\frac{151}{80.6}]^ {(1.38-1)}$

T_f = 330.0 K

Part 3

determine number of moles by using following formula

$\eta =\frac{PV}{RT}$

$\eta =\frac{1.013*10^{5}*0.151}{8.314*260}$

$\eta = 7.07 mole$

4 0

3 years ago

1. An EM wave representing 650 nm laser light is traveling in the +zˆ direction. At one point in space and time, the electric fi

ivann1987 [24]

Answer:

Explanation:

a ) Direction of the magnetic field will be in positive x direction.

The direction of the vector E X B gives the direction of motion of wave.

b ) Magnitude of magnetic field is given by the relation

E₀ / B₀ = c , c is velocity of light

B₀ = E₀ / c

= 20 / (3 x 10⁸)

= 6.67 x 10⁻⁸ T

c ) Average power flowing per unit area by this wave is called Poynting vector

c ε₀E₀² , ε₀ = 8.85X10⁻¹²

= 3 X 10⁸ X 8.85 X 10⁻¹² X 20²

= 1.062 W m⁻²

5 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;

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

6 0

2 years ago