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Doss [256]
3 years ago
7

Jon's bathtub is rectangular and its base is 18 ft2. How fast is the water level rising if Jon is filling the tub at a rate of 0

.7 ft3/min?
Physics
1 answer:
Dvinal [7]3 years ago
4 0

To solve this problem we will use the concepts related to the volumetric flow rate, which describes the amount of volume per unit of time, through the Area of the section and the speed of the fluid over it. Mathematically this is:

Q = A*V \rightarrow V = \frac{Q}{A}

Where,

Q = Discharge

A= Cross sectional Area

V = Velocity

Our values are given as,

A = 18ft^2

Q = 0.7ft^3/min

Replacing,

V = \frac{0.7}{18}

V = 0.038ft/min

Therefore the water level rises to a velocity of 0.038ft/min

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Answer:

The  the intensity at an 11° angle to the axis in terms of the intensity of the central maximum is  

   I_c = \frac{I}{I_o}  =8.48 *10^{-8}

Explanation:

From the question we are told that

   The  width of the slit is  D  =  0.3 \ mm =  0.3 *10^{-3} \ m

    The  wavelength is  \lambda =  254 \ nm =  254 *10^{-9} \ m

     The angle is  \theta  =  11^o

The intensity of at 11^o to the axis in terms of the intensity of the central maximum. is mathematically represented as

        I_c = \frac{I}{I_o}  = [ \frac{sin \beta  }{\beta }] ^2

Where \beta is mathematically represented as

        \beta  =  \frac{D sin (\theta ) *  \pi}{\lambda }

substituting values

      \beta  =  \frac{0.3 *10^{-3} sin (11 ) * 3.142}{254 *10^{-9} }

     \beta  =  708.1 \ rad

So

  I_c = \frac{I}{I_o}  = [ \frac{sin (708.1)  }{(708.1)}] ^2

   I_c = \frac{I}{I_o}  =8.48 *10^{-8}

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4 years ago
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