V = I • R
V = (0.054 A) • (152 ohms)
V = 8.208 volts
None of the listed choices is correct.
When an object is slowing down, the acceleration is in the opposite direction as the velocity.
Answer:
m/min
Explanation:
You have to use the volume of a cone, which is:
![V=\frac{1}{3}\pi r^{2}h](https://tex.z-dn.net/?f=V%3D%5Cfrac%7B1%7D%7B3%7D%5Cpi%20r%5E%7B2%7Dh)
where r is the radius of the base and h is the height.
In this case, r=5 and h=10. The radius can be written as r=h/2
Replacing it in the equation:
(I)
The rate of the volume is the derivate of volume respect time, therefore you have to perform the implicit differentiation of the previous equation and equal the result to 3.14 m³/min
![\frac{dV}{dt}=\frac{\pi }{12}(3)h^{2}\frac{dh}{dt} =\frac{\pi }{4}h^{2}\frac{dh}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7BdV%7D%7Bdt%7D%3D%5Cfrac%7B%5Cpi%20%7D%7B12%7D%283%29h%5E%7B2%7D%5Cfrac%7Bdh%7D%7Bdt%7D%20%3D%5Cfrac%7B%5Cpi%20%7D%7B4%7Dh%5E%7B2%7D%5Cfrac%7Bdh%7D%7Bdt%7D)
Replacing dV/dt= 3.14, h=7.5 and solving for dh/dt, which represents how fast the level is rising:
![3.14=\frac{\pi }{4}(7.5)^{2}\frac{dh}{dt}\\3.14=\frac{225\pi }{16}\frac{dh}{dt}](https://tex.z-dn.net/?f=3.14%3D%5Cfrac%7B%5Cpi%20%7D%7B4%7D%287.5%29%5E%7B2%7D%5Cfrac%7Bdh%7D%7Bdt%7D%5C%5C3.14%3D%5Cfrac%7B225%5Cpi%20%7D%7B16%7D%5Cfrac%7Bdh%7D%7Bdt%7D)
Multiplying by 16/225π both sides:
m/min
Answer:
0.08 ft/min
Explanation:
To get the speed at witch the water raising at a given point we need to know the area it needs to fill at that point in the trough (the longitudinal section), which is given by the height at that point.
So we need to get the lenght of the sides for a height of 1 foot. Given the geometry of the trough, one side is the depth <em>d</em> and the other (lets call it <em>l</em>) is given by:
![l=\frac{3-2}{2}\,ft+2\,ft\\l=2.5\,ft](https://tex.z-dn.net/?f=l%3D%5Cfrac%7B3-2%7D%7B2%7D%5C%2Cft%2B2%5C%2Cft%5C%5Cl%3D2.5%5C%2Cft)
since the difference between the upper and lower base is the increase in the base and we are only at halft the height.
Now we can calculate the longitudinal section <em>A</em> at that point:
![A=d\times l\\A=5\,ft \times 2.5\, ft\\A=12.5\, ft^{2}](https://tex.z-dn.net/?f=A%3Dd%5Ctimes%20l%5C%5CA%3D5%5C%2Cft%20%5Ctimes%202.5%5C%2C%20ft%5C%5CA%3D12.5%5C%2C%20ft%5E%7B2%7D)
And the raising speed <em>v </em>of the water is given by:
![v=\frac{q}{A}\\v=\frac{1\, \frac{ft^3}{min}}{12.5\, ft^2}\\v=0.08\, \frac{ft}{min}](https://tex.z-dn.net/?f=v%3D%5Cfrac%7Bq%7D%7BA%7D%5C%5Cv%3D%5Cfrac%7B1%5C%2C%20%5Cfrac%7Bft%5E3%7D%7Bmin%7D%7D%7B12.5%5C%2C%20ft%5E2%7D%5C%5Cv%3D0.08%5C%2C%20%5Cfrac%7Bft%7D%7Bmin%7D)
where <em>q</em> is the water flow (1 cubic foot per minute).
Answer:
a) The resistance of the calf between the electrodes is ![85\Omega](https://tex.z-dn.net/?f=85%5COmega)
b) The average resistivity of this part of the leg is ![57.81m\Omega/cm^{3}](https://tex.z-dn.net/?f=57.81m%5COmega%2Fcm%5E%7B3%7D)
Explanation:
Hi
a) Using Ohm's law
, solving for
, we obtain ![R=\frac{V}{I}=\frac{17mV}{200uA}=85\Omega](https://tex.z-dn.net/?f=R%3D%5Cfrac%7BV%7D%7BI%7D%3D%5Cfrac%7B17mV%7D%7B200uA%7D%3D85%5COmega)
b) The volume of the calf is like a cylinder, so
, with
and
, therefore
. Then we can use
, this is the average resistivity of this part of the leg.