With a 30 mph head wind it takes the plane 18.52 hours to fly 5000 miles. ANSWER 2: With a 30 mph tail wind it takes the plane 15.15 hours to fly 5000 miles.
According to the Bernoulli's equation,the pressure difference between the wide and narrow ends of the pipe is given by
![\Delta P= \frac{1}{2} \rho ( v^2_{2} - v^2_{1} )](https://tex.z-dn.net/?f=%5CDelta%20P%3D%20%5Cfrac%7B1%7D%7B2%7D%20%5Crho%20%28%20v%5E2_%7B2%7D%20-%20v%5E2_%7B1%7D%20%29)
Here,
is the velocity of water through wide ends of cylindrical pipe and
is the velocity of water through narrow ends of cylindrical pipe.
Given, ![v_{1} =1.4 m/s](https://tex.z-dn.net/?f=v_%7B1%7D%20%3D1.4%20m%2Fs)
Now from equation continuity,
.
Here,
and
are cross- sectional areas of wide and narrow ends of cylindrical pipe.
As pipe is circular, so
.
At the second point, the diameter is halved, which means the radius is also halved. Therefore,
![v_{1} r^2_{1} = v_{2}(\frac{1}{2} r_{1})^2 \\\\ v_{2} = 4 v_{1}](https://tex.z-dn.net/?f=v_%7B1%7D%20r%5E2_%7B1%7D%20%3D%20v_%7B2%7D%28%5Cfrac%7B1%7D%7B2%7D%20r_%7B1%7D%29%5E2%20%5C%5C%5C%5C%20v_%7B2%7D%20%3D%204%20v_%7B1%7D)
![v_{2} = 4 \times 1.4 = 5.6 m/s](https://tex.z-dn.net/?f=v_%7B2%7D%20%3D%204%20%5Ctimes%201.4%20%3D%205.6%20m%2Fs)
Substituting these values with the density of water is
in pressure difference formula we get.
![\Delta P= \frac{1}{2} \rho ( v^2_{2} - v^2_{1} )=\frac{1}{2}\times 1000 kg/m^3(5.6^2-1.4^2)\\\\ \Delta P = 14700\ Pa](https://tex.z-dn.net/?f=%5CDelta%20P%3D%20%5Cfrac%7B1%7D%7B2%7D%20%5Crho%20%28%20v%5E2_%7B2%7D%20-%20v%5E2_%7B1%7D%20%29%3D%5Cfrac%7B1%7D%7B2%7D%5Ctimes%201000%20kg%2Fm%5E3%285.6%5E2-1.4%5E2%29%5C%5C%5C%5C%20%5CDelta%20P%20%3D%2014700%5C%20Pa)
Answer:
b. 14
Explanation:
= Initial temperature = 27 °C = 27 + 273 = 300 K
= Final temperature = 37 °C = 37 + 273 = 310 K
= Initial Power radiated by the object
= Final Power radiated by the object
We know that the power radiated is directly proportional to fourth power of the temperature. hence
![\frac{P_{f}}{P_{i}} = \frac{T_{f}^{4} }{T_{i}^{4} }\\\frac{P_{f}}{P_{i}} = \frac{(310)^{4} }{(300)^{4} }\\\frac{P_{f}}{P_{i}} = 1.14\\P_{f} = (1.14) P_{i}](https://tex.z-dn.net/?f=%5Cfrac%7BP_%7Bf%7D%7D%7BP_%7Bi%7D%7D%20%3D%20%5Cfrac%7BT_%7Bf%7D%5E%7B4%7D%20%7D%7BT_%7Bi%7D%5E%7B4%7D%20%7D%5C%5C%5Cfrac%7BP_%7Bf%7D%7D%7BP_%7Bi%7D%7D%20%3D%20%5Cfrac%7B%28310%29%5E%7B4%7D%20%7D%7B%28300%29%5E%7B4%7D%20%7D%5C%5C%5Cfrac%7BP_%7Bf%7D%7D%7BP_%7Bi%7D%7D%20%3D%201.14%5C%5CP_%7Bf%7D%20%3D%20%281.14%29%20P_%7Bi%7D)
Percentage increase in power is given as
![\frac{(P_{f} - P_{i})\times100}{P_{i}} \\\frac{((1.14) P_{i} - P_{i})\times100}{P_{i}} \\14](https://tex.z-dn.net/?f=%5Cfrac%7B%28P_%7Bf%7D%20-%20P_%7Bi%7D%29%5Ctimes100%7D%7BP_%7Bi%7D%7D%20%5C%5C%5Cfrac%7B%28%281.14%29%20P_%7Bi%7D%20-%20P_%7Bi%7D%29%5Ctimes100%7D%7BP_%7Bi%7D%7D%20%5C%5C14)
C. meter per second
Velocity and speed share the same SI unit.
Work is (force applied) x (distance through which the force moves).
Since the suitcase doesn't move up or down during the 15 minutes,
no work is done ... zero, zip, nada ... according to the real Physics
definition of 'work'.