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3 years ago

10

What does WNBA stand for

2 answers:

zvonat [6]3 years ago

7 0

Answer:

Women's national basketball association

Explanation:

Alexeev081 [22]3 years ago

6 0

Answer:The Women's National Basketball Association,

Explanation:Branliest:)

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A boxer hits punching bag and gives it a change in momentum of 12 kg multiplied by m divided by s over 7.0ms what is the magnitu

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Answer: 1700

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2 years ago

When light passes through an object unchanged, scientists call that process _____.

Nataly [62]

<u>Answer</u>:

When light passes through an object unchanged, scientists call that process Transmission.

<u>Explanation</u>:

Transmission is the process where all the light that is passed through the material moves via the material without being absorbed. The Transmission depends on the affected radiation.The Transmittance of the medium is defined as the ratio between transmitted radiant power and incident radiant power. The light that is passed through the medium and not reflected will be either scattered or reflected. The light can be transmitted only through transparent or translucent material. Opaque object does not allows transmission of light.

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3 years ago

A particle of mass 2kg resting on a smooth table attached to a fixed point on the table by a rope 1.0m making 300revolution per

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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

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2 years ago

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Answer:

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3 years ago

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