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3 years ago

CAN SOMEONE DO THIS TO ME ITS OK IF YOUR DRAWING IS NOT THAT COOL OR GOOD AND DONT GET A PHOTO FROM SOCIAL MEDIA

Physics

1 answer:

Svet_ta [14]3 years ago

7 0

bc i cant draw good sorry but hope it helps!

and can i have brainliest? ight plz can i have brainliest?

ill give another picture of the volley ball thing

is that better?

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How long does a lunar eclipse last? A)seconds B)Minutes C)Days D) Weeks

bonufazy [111]

Well, it happens a few weeks ahead, then for a total of 3 hours and 40 minutes.

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3 years ago

Difference between on pitch and frequency

shepuryov [24]

Answer:

A high pitch sound corresponds to a high frequency sound wave and a low pitch sound corresponds to a low frequency sound wave. I hope I got it correct !!

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2 years ago

A pitcher exerts a force on a baseball that is 30 times the balls weight. How fast is the pitcher accelerating the ball?

iVinArrow [24]

Answer:

Pitcher is accelerating the ball at 30 times of acceleration due to gravity = 294 m/s²

Explanation:

Force applied on baseball = 30 times weight of the ball.

Weight of ball = mg, where m is the mass of ball and g is acceleration due to gravity value.

We have force applied is also equal to product of mass and acceleration.

F = ma = 30 x mg

a = 30g

So, pitcher is accelerating the ball at 30 times of acceleration due to gravity = 294 m/s²

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3 years ago

Argon gas enters steadily an adiabatic turbine at 900 kPa and 450C with a velocity of 80 m/s and leaves at 150 kPa with a veloc

Crazy boy [7]

Answer:

Temperature at the exit = $267.3 C$

Explanation:

For the steady energy flow through a control volume, the power output is given as

$W_{out}= -m_{f}(h_{2}-h_{1} + \frac{v_{2}^{2}}{2} - \frac{v_{1}^{2}}{2})$

Inlet area of the turbine = $60cm^{2}= 0.006m^{2}$

To find the mass flow rate, we can apply the ideal gas laws to estimate the specific volume, from there we can get the mass flow rate.

Assuming Argon behaves as an Ideal gas, we have the specific volume $v_{1}$

$v_{1}=\frac{RT_{1}}{P_{1}}=\frac{0.2081\times723}{900}=0.1672m^{3}/kg$

$m_{f}=\frac{1}{v_{1}}\times A_{1}V_{1} = \frac{1}{0.1672}\times(0.006)(80)=2.871kg/sec$

for Ideal gasses, the enthalpy change can be calculated using the formula

$h_{2}-h_{1}=C_{p}(T_{2}-T_{1})$

hence we have

$W_{out}= -m_{f}((C_{p}(T_{2}-T_{1}) + \frac{v_{2}^{2}}{2} - \frac{v_{1}^{2}}{2})$

$250= -2.871((0.5203(T_{2}-450) + \frac{150^{2}}{2\times 1000} - \frac{80^{2}}{2\times 1000})$

<em>Note: to convert the Kinetic energy term to kilojoules, it was multiplied by 1000</em>

evaluating the above equation, we have $T_{2}=267.3C$

Hence, the temperature at the exit = $267.3 C$

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3 years ago

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mote1985 [20]

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3 years ago