A trough is 10 feet long and the ends have the shape of isosceles triangles that measure 3 feet across at the top and have a height of 1 foot. If the trough is being filled with water at a rate of 12 ft3 per minute, how fast is the water level rising when the water is 6-inches deep

0.8 ft³/min

Step-by-step explanation:

Assuming that the cross section formed by the water at a depth of x is b. Then, if we use the property of similar triangles we have...

See the attachment for the workings.

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Continuation of the workings

From the question, we're told that the water is 6 inches deep. Converting this inch to ft, we have

6 ft = 0.5 ft.

So we use 0.5 as our x, and then use 12 ft³/min as our dV/dt. Doing this, we have

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12 = 30 * 0.5 dx/dt

12 = 15 dx/dt

dx/dt = 12/15

dx/dt= 0.8 ft³/min

(w – 2.5)(w + 2.5)

(–4v – 9)(–4v + 9)

step-by-step explanation:

step-by-step explanation:

we can use this expression to calculate the energy gain as a particle moves with a field or the energy loss as it moves against it. (note that it is also used to define the electron-volt, a unit of energy equal to that gained by an electron as it falls through a pd of 1 v, i.e. w = 1 ev = 1.6 10-19 c 1 v = 1.6 10-19j).in fact this expression is really nothing more than the statement that potential is the potential energy per unit charge. in this case, v= w/qis simply stating that the change in potential energy(or work done) per unit charge is equal to the change in potential (or potential difference), v.student questions: uniform electric fields2electron beamcathode+5 kvanode0 v