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masya89 [10]
3 years ago
11

A ladybug falls into small hole. it takes 8.6s for the ladybug to hit the bottom. how deep is the hole? (take g=9.81 m/s²)

Physics
1 answer:
Jobisdone [24]3 years ago
4 0

Answer:

Length of hole (s) = 362.77 m (Approx)

Explanation:

Given:

Time taken to hit the bottom (t) = 8.6 s

Acceleration due to gravity (g) = 9.81 m/s²

Find:

Length of hole (s) = ?

Computation:

Initial velocity (u) = 0 m/s

S = ut +1/2(gt²)

S = (0)(8.6) +1/2(9.81)(8.6)(8.6)

S = 1/2(9.81)(8.6)(8.6)

S = 1/2(9.81)(8.6)(8.6)

S = 362.7738

Length of hole (s) = 362.77 m (Approx)

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A long, thin solenoid has 450 turns per meter and a radius of 1.17 cm. The current in the solenoid is increasing at a uniform ra
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Answer:

\frac{di}{dt}  = 7.31 \  A/s

Explanation:

From the question we are told that  

     The  number of turns is  N =  450 \  turns

      The  radius is  r =  1.17 \ cm =  0.0117 \ m

       The  position from the center consider is  x =  3.45 cm  =  0.0345 m

       The  induced emf is  e  =  8.20 *10^{-6} \  V/m

Generally according to Gauss law

        \int\limits { e } \, dl  =  \mu_o *  N  *  \frac{di}{dt }  *  A

=>    e *  2\pi x  =  \mu_o  *  N  *  \frac{d i }{dt }  *  A

Where A is the  cross-sectional area of the solenoid which is mathematically represented as

                A =  \pi r ^2

=>      e *  2\pi x  =  \mu_o  *  N  *  \frac{d i }{dt }  *  \pi r^2

=>       \frac{di}{dt}  =  \frac{2e * x  }{\mu_o * N  * r^2}ggl;

Here  \mu_o is the permeability of free space with value

          \mu_o  =  4\pi * 10^{-7} \  N/A^2

=>     \frac{di}{dt}  =  \frac{2 *  8.20*10^{-6} *  0.0345  }{ 4\pi * 10^{-7} * 450  * (0.0117)^2}

=>      \frac{di}{dt}  = 7.31 \  A/s

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4 years ago
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