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3 years ago

9

In which of the following situation is light most likely to be a reflection

1 answer:

maxonik [38]3 years ago

7 0

Answer:

when the reflecting surface is plain and without even small hurdles that are not the visible by our naked eyes. Eg : plain mirror

Explanation:

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What is The most basic unit of an Element on the Periodic Table

lesya [120]

Answer:

atom(which contains protons, neutrons and electrons)

8 0

3 years ago

An air-standard Diesel cycle has a compression ratio of 16 and a cutoff ratio of 2. At the beginning of the compression process,

Sedbober [7]

Answer:

$a.T_3=1723.8kPa\\b.n=0.563\\c.MEP=674.95kPa$

Explanation:

a. Internal energy and the relative specific volume at $s_1$ are determined from A-17: $u_1=214.07kJ/kg, \ \alpha_r_1=621.2$ .

The relative specific volume at $s_2$ is calculated from the compression ratio:

$\alpha_r_2=\frac{\alpha_r_1}{r}\\=\frac{621.2}{16}\\=38.825$

#from this, the temperature and enthalpy at state 2, $s_2$ can be determined using interpolations $T_2=862K$ and $h_2=890.9kJ/kg$ . The specific volume at $s_1$ can then be determined as:

$\alpha_1=\frac{RT_1}{P_1}\\\\=\frac{0.287\times 300}{95} m^3/kg\\0.906316m^3/kg$

Specific volume, $s_2$ :

$\alpha_2=\frac{\alpha_1}{r}\\=\frac{0.906316}{16}m^3/kg\\=0.05664m^3/kg$

The pressures at $s_2 \ and\ s_3$ is:

$P_2=P_3=\frac{RT_2}{\alpha_2}\\\\=\frac{0.287\times862}{0.05664}\\=4367.06kPa$

.The thermal efficiency=> maximum temperature at $s_3$ can be obtained from the expansion work at constant pressure during $s_2-s_3$

$\bigtriangleup \omega_2_-_3=P(\alpha_3-\alpha_2)\\R(T_3-T_2)=P\alpha(r_c-1)\\T_3=T_2+\frac{P\alpha_2}{R}(r_c-1)\\\\=(862+\frac{4367\times 0.05664}{0.287}(2-1))K\\=1723.84K$

b.Relative SV and enthalpy at $s_3$ are obtained for the given temperature with interpolation with data from A-17 : $a_r_3=4.553 \ and\ h_3=1909.62kJ/kg$

Relative SV at $s_4$ is

$a_r_4=\frac{r}{r_c}\alpha _r_3$

= $=\frac{16}{2}\times4.533\\=36.424$

Thermal efficiency occurs when the heat loss is equal to the internal energy decrease and heat gain equal to enthalpy increase;

$n=1-\frac{q_o}{q_i}\\=1-\frac{u_4-u_1}{h_3-h_2}\\=1-\frac{65903-214.07}{1909.62-890.9}\\=0.563$

Hence, the thermal efficiency is 0.563

c. The mean relative pressure is calculated from its standard definition:

$MEP=\frac{\omega}{\alpa_1-\alpa_2}\\=\frac{q_i-q_o}{\alpha_1(1-1/r)}\\=\frac{1909.62-890.9-(65903-214.7)}{0.90632(1-1/16)}\\=674.95kPa$

Hence, the mean effective relative pressure is 674.95kPa

3 0

3 years ago

A wheel that was initially spinning is accelerated at a constant angular acceleration of 5.0 rad/s^2. After 8.0 s, the wheel is

notka56 [123]

Answer:

a) Initial angular speed = 30 rad/s

b) Final angular speed = 70 rad/s

Explanation:

a) We have equation of motion s = ut + 0.5at²

Here s = 400 radians

t = 8 s

a = 5 rad/s²

Substituting

400 = u x 8 + 0.5 x 5 x 8²

u = 30 rad/s

Initial angular speed = 30 rad/s

b) We have equation of motion v = u + at

Here u = 30 rad/s

t = 8 s

a = 5 rad/s²

Substituting

v = 30 + 5 x 8 = 70 rad/s

Final angular speed = 70 rad/s

8 0

3 years ago

If the ball shown in the figure lands in 0.5 s, about what height was it thrown from?

Ede4ka [16]

Answer: 1.22 m

Explanation:

The equation of motion in this situation is:

$y=y_{o}+V_{oy}t-\frac{g}{2}t^{2}$ (1)

Where:

$y=0$ is the final height of the ball

$y_{o}=h$ is the initial height of the ball

$V_{oy}=V_{o}sin(0\°)=0$ is the vertical component of the initial velocity (assuming the ball was thrown vertically and there is no horizontal velocity)

$t=0.5 s$ is the time at which the ball lands

$g=9.8 m/s^{2}$ is the acceleration due gravity

So, with these conditions the equation is rewritten as:

$h=\frac{g}{2}t^{2}$ (2)

$h=\frac{9.8 m/s^{2}}{2}(0.5 s)^{2}$ (3)

Finally:

$h=1.22 m$

4 0

3 years ago

A thunderstorm loses energy in the _______ stage.

Irina-Kira [14]

dissipation is the answer ;(

3 0

3 years ago

Read 2 more answers