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Svetradugi [14.3K]
3 years ago
13

Hufflepuff and Ravenclaw are having their annual tug of war contest. Each team can have 20 students. What is the most important

variable to have an edge up in the competition?
Physics
2 answers:
Leviafan [203]3 years ago
6 0

    I'm sorry if my answer is too late but they didn't provide the letter S as a variable to represent the students. If that were the answer they would have specifically provided that information in the question. I'm a student of K12 as well so I can confirm that the answer is

B)The overall mass of the team.

Have a nice day!

Troyanec [42]3 years ago
3 0

It should be S since S stands for students. Giving the team with more members the upper hand. Hope this helps.

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Examine the following equation.
alex41 [277]

Answer:

E)brain decay

Explanation:

Looking at the question causes it.

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
NaOH + FeCl3* Na Cl + Fe 10H)3<br> balanced
zvonat [6]
3NaOH + FeCl3 → 3NaCl + Fe(OH)3
8 0
3 years ago
Ron fills a beaker with glycerin (n = 1.473) to a depth of 5.0 cm. if he looks straight down through the glycerin surface, he wi
Tju [1.3M]

By law of refraction we know that image position and object positions are related to each other by following relation

\frac{\mu_1}{h_o} = \frac{\mu_2}{h_i}

here we know that

\mu_1 = 1.473

h_o = 5 cm

\mu_2 = 1

now by above formula

\frac{1.473}{5} = \frac{1}{h_i}

h_i = 3.39 cm

so apparent depth of the bottom is seen by the observer as h = 3.39 cm

7 0
3 years ago
28. Sound can be heard around a corner because of
Lesechka [4]

Answer:

Diffraction of sound wavelengths.

Explanation:

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5 0
3 years ago
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