How does Mr. Anderson define solubility in the video?

Một nguồn O phát sóng cơ dao động theo phương trình u = 4cos(4pi t - pi/4). Tốc độ truyền sóng là 1m/s.

brilliants [131]

Answer:

ytrxrddyoxswsdyxgxghfx

jdjdu3jthh

hhhujusbrnog

hhjfjtinrny

ykrjrhrnirjtjjtt

tkrjthr74uu3jt

hri4urjjrjtjjtjtjy

y

Explanation:

uueuhhhwuejroskanficndui39wn

jebfufkr

5 0

3 years ago

The most recently discovered system close to Earth is a pair of brown dwarfs known as Luhman 16. It has a distance of 6.5 light-

just olya [345]

Answer:

1.99 parsecs.

Explanation:

We have been given that the most recently discovered system close to Earth is a pair of brown dwarfs known as Luhman 16. It has a distance of 6.5 light-years.

We know that one light year equals to 0.306601 parsecs. To convert 6.5 light-years to parsecs, we will multiply 0.306601 by 6.5.

$6.5\text{ Light-years}=6.5\times 0.306601\text{ Parsecs}$

$6.5\text{ Light-years}=1.9929065\text{ Parsecs}$

$6.5\text{ Light-years}\approx 1.99\text{ Parsecs}$

Therefore, Luhman 16 is approximately 1.99 parsecs away from the Earth.

5 0

3 years ago

PLEASE HELP Scientists have observed that the average sunspots have increased in the last 50 years. Which of these statements be

Serga [27]

Answer:

option c is correct, earth temp increase

6 0

3 years ago

Air as an ideal gas enters a diffuser operating at steady state at 5 bar, 280 K with a velocity of 510 m/s. The exit velocity is

Nataly [62]

Answer:

Explanation:

Calculating the exit temperature for K = 1.4

The value of $c_p$ is determined via the expression:

$c_p = \frac{KR}{K_1}$

where ;

R = universal gas constant = $\frac{8.314 \ J}{28.97 \ kg.K}$

k = constant = 1.4

$c_p = \frac{1.4(\frac{8.314}{28.97} )}{1.4 -1}$

$c_p= 1.004 \ kJ/kg.K$

The derived expression from mass and energy rate balances reduce for the isothermal process of ideal gas is :

$0=(h_1-h_2)+\frac{(v_1^2-v_2^2)}{2}$ ------ equation(1)

we can rewrite the above equation as :

$0 = c_p(T_1-T_2)+ \frac{(v_1^2-v_2^2)}{2}$

$T_2 =T_1+ \frac{(v_1^2-v_2^2)}{2 c_p}$

where:

$T_1 = 280 K \\ \\ v_1 = 510 m/s \\ \\ v_2 = 120 m/s \\ \\c_p = 1.0004 \ kJ/kg.K$

$T_2= 280+\frac{((510)^2-(120)^2)}{2(1.004)} *\frac{1}{10^3}$

$T_2 = 402.36 \ K$

Thus, the exit temperature = 402.36 K

The exit pressure is determined by using the relation: $\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{k}{k-1}$

$P_2=P_1(\frac{T_2}{T_1})^\frac{k}{k-1}$

$P_2 = 5 (\frac{402.36}{280} )^\frac{1.4}{1.4-1}$

$P_2 = 17.79 \ bar$

Therefore, the exit pressure is 17.79 bar

7 0

3 years ago

Which is an example of an unstructured activity that promotes resistance training?

aleksandrvk [35]

One possible unstructured activity that promotes resistance training would be climbing playground equimpent - A.

This is by nature a unstructured ctivity. Furthermore, it promotes resistance training because you're forced to move and pull and push yourself.

3 0

3 years ago