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n200080 [17]
1 year ago
8

16) The depth of pond is ....... if it seems to be 10m a) 10m b) 7.7m c) 13.3m d) 20m​

Physics
1 answer:
Alexxandr [17]1 year ago
7 0

The answer is d) 20 m.

The depth of pond is 20 m if it seems to be 10 m.

The formula relating real depth and app. depth :

\boxed {A = \frac{R}{n}}}

In this case, refractive index has to be an whole number for real depth to give a whole number for app. depth.

  • Then, real depth must be a multiple of 10, but refractive index cannot be 1
  • The option is 20 m
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ioda

Answer:

Explanation:

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distance of the nearest side from long wire, r = 2 cm = 0.02 m

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(a) The magnetic field due to the current carrying wire at a distance r is given by

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(b)

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\phi=\int B\times a dr

\phi=\int \frac{\mu_{0}\times 9t}{2\pi r}\times a dr

\phi=\frac{\mu_{0}\times 9t\times a}{2\pi}\times ln\left ( \frac{2 + 7}{2} \right )

\phi=\frac{\mu_{0}\times 9t\times 0.07}{2\pi}\times ln(4.5)

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(c)

R = 3 ohm

e = -\frac{d\phi}{dt}

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e = 1.89 x 10^-7 V

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i = 6.3 x 10^-8 A

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3 years ago
On a violin, the highest note Mandy can play is an A-note, which produces a sound wave with a high frequency. The lowest note sh
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6 0
3 years ago
The maximum speed with which you can throw a stone is about 20 m/s. Can you hit a window 45 m away horizontally and 10 m up from
Allushta [10]

Answer:

 y = 17 m

Explanation:

For this projectile launch exercise, let's write the equation of position

          x = v₀ₓ t

          y = v_{oy} t - ½ g t²

let's substitute

          45 = v₀ cos θ t

          10 = v₀ sin θ t - ½ 9.8 t²

the maximum height the ball can reach where the vertical velocity is zero

 

           v_{y} = v_{oy} - gt

           0 = v₀ sin θ - gt

           0 = v₀ sin θ - 9.8 t

Let's write our system of equations

         45 = v₀ cos θ  t

         10 = v₀ sin θ t - ½ 9.8 t²

         0 = v₀ sin θ - 9.8 t

We have a system of three equations with three unknowns for which it can be solved.

Let's use the last two

        v₀ sin θ = 9.8 t

we substitute

        10 = (9.8 t) t - ½ 9.8 t2

        10 = ½ 9.8 t2

        10 = 4.9 t2

        t = √ (10 / 4.9)

        t = 1,429 s

Now let's use the first equation and the last one

         45 = v₀ cos θ t

         0  = v₀ sin θ - 9.8 t

         9.8 t = v₀  sin θ

         45 / t = v₀ cos θ

we divide

         9.8t / (45 / t) = tan θ

          tan θ = 9.8 t² / 45

          θ = tan⁻¹ ( 9.8 t² / 45 )

          θ = tan⁻¹ (0.4447)

          θ = 24º

Now we can calculate the maximum height

         v_y² = v_{oy}^2 - 2 g y

         vy = 0

          y = v_{oy}^2 / 2g

          y = (20 sin 24)²/2 9.8

          y = 3,376 m

the other angle that gives the same result is

       θ‘= 90 - θ

       θ' = 90 -24

       θ'= 66'

for this angle the maximum height is

 

          y = v_{oy}^2 / 2g

          y = (20 sin 66)²/2 9.8

           y = 17 m

thisis the correct

6 0
3 years ago
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