Answer:
a) 3-in. pipe
Explanation:
Given that
Fluid flow is in same amount in the same time it means that volume flow rate is same for the pipes
Volume flow rate
Q = A V
A=Area ,V=Velocity

If diameter d is more then the velocity will be less for same volume flow rate .We also Know that if pressure is more then the velocity will be less.
The second pipe 3 in diameter having more diameter then the velocity will be less but the pressure will be more.
That is why the 3 in diameter is having more pressure than 2 in diameter pipe.
Therefore the answer will be a.
a) 3-in diameter pipe
Neutrons have a zero charge but consist of mass.
The average power produced by the soccer player is 710 Watts.
Given the data in the question;
- Mass of the soccer player;

- Energy used by the soccer player;

- Time;

Power; 
Power is simply the amount of energy converted or transferred per unit time. It is expressed as:

We substitute our given values into the equation
![Power = \frac{5100000J}{7200s}\\\\Power = 708.33J/s \\\\Power = 710J/s \ \ \ \ \ [ 2\ Significant\ Figures]\\\\Power = 710W](https://tex.z-dn.net/?f=Power%20%3D%20%5Cfrac%7B5100000J%7D%7B7200s%7D%5C%5C%5C%5CPower%20%3D%20708.33J%2Fs%20%5C%5C%5C%5CPower%20%3D%20710J%2Fs%20%5C%20%5C%20%5C%20%5C%20%5C%20%5B%202%5C%20Significant%5C%20Figures%5D%5C%5C%5C%5CPower%20%3D%20710W)
Therefore, the average power produced by the soccer player is 710 Watts.
Learn more: brainly.com/question/20953664
Answer:
b.it depends on the distance it falls
<u>Given </u><u>:</u><u>-</u>
- An elevator is moving vertically up with an acceleration a.
<u>To </u><u>Find</u><u> </u><u>:</u><u>-</u>
- The force exerted on the floor by a passenger of mass m .
<u>Solution</u><u> </u><u>:</u><u>-</u>
As the man is in a accelerated frame that is <u>non </u><u>inertial</u><u> frame</u><u> </u>, we would have to think of a pseudo force .
- The direction of this force is opposite to the direction of acceleration the frame and its magnitude is equal to the product of mass of the concerned body with the acceleration of the frame .
For the FBD refer to the attachment . From that ,
<u>Hence</u><u> </u><u>option</u><u> </u><u>d </u><u>is </u><u>correct</u><u> </u><u>choice </u><u>.</u>
<em>I </em><em>hope</em><em> this</em><em> helps</em><em> </em><em>.</em>