(5 mi/hr) x (1hr/60min) x (10min) = 5 x 10 / 60 = <em>5/6 mile</em>
(5/6 mile) x (1,760 yd/mile) = <em>1,466 and 2/3 yards</em>
To solve this problem we will use the work theorem, for which we have that the Force applied on the object multiplied by the distance traveled by it, is equivalent to the total work. From the measurements obtained we have that the width and the top are 14ft and 7ft respectively. In turn, the bottom of the tank is 15ft. Although the weight of the liquid is not given we will assume this value of
(Whose variable will remain modifiable until the end of the equations subsequently presented to facilitate the change of this, in case be different). Now the general expression for the integral of work would be given as

Basically under this expression we are making it difficult for the weight of the liquid multiplied by the area (Top and widht) under the integral of the liquid path to be equivalent to the total work done, then replacing

![W = (14*7*62)\big [15y-\frac{y^2}{2}\big ]^{15}_0](https://tex.z-dn.net/?f=W%20%3D%20%2814%2A7%2A62%29%5Cbig%20%5B15y-%5Cfrac%7By%5E2%7D%7B2%7D%5Cbig%20%5D%5E%7B15%7D_0)
![W = (14*7*62)[15(15)-\frac{(15)^2}{2}]](https://tex.z-dn.net/?f=W%20%3D%20%2814%2A7%2A62%29%5B15%2815%29-%5Cfrac%7B%2815%29%5E2%7D%7B2%7D%5D)

Therefore the total work in the system is 
Answer:
5p
Explanation:
We are given that a tin has 50 electrons.
We have to find in which subshell electrons experience the lowest effective nuclear charge.
We know that the electron in outermost shell experience the lowest effective nuclear charge.
Electronic configuration is given by

Outer most sub-shell is 5p. Therefore, 5p subshell experience the lowest effective nuclear charge because the distance of 5p sub-shell is large from nucleus.
Answer: 5p
Answer:a goal is scored by throwing the ball into the goal post and is confirmed by deciding that’s if the ball has hit the tape mark or not
Explanation:
A volt-ampere is the product of the voltage and the current (measured in amperes) of the electricity on a line. Voltage is electric potential difference in charge between two points in an electrical field. it is measured in volts. while the Current is the flow of electrical charge which is often carried by moving electrons in a wire or a conductor. Current is measured in Amperes.