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Fynjy0 [20]
3 years ago
12

Water flows over a section of Niagara Falls at the rate of 1.1 × 106 kg/s and falls 50.0 m. How much power is generated by the f

alling water? The acceleration of gravity if 9.81 m/s 2 . Answer in units of W.
Physics
1 answer:
Kryger [21]3 years ago
7 0
<h3>Answer:</h3>

5.395 × 10^8 Watts

<h3>Explanation:</h3>

<u>We are given;</u>

  • Rate of flow is 1.1 × 10^6 kg/s
  • Distance is 50.0 m
  • Gravitational acceleration is 9.8 m/s²

We are required to calculate the power that is generated by the falling water

  • Power is the rate of work done
  • It is given by dividing the energy or work done by time
  • Power = Work done ÷ time

But; work done = Force × distance

Therefore;

Power = (F × d) ÷ time

The rate is 1.1 × 10^ 6 Kg/s

But, 1 kg = 9.81 N

Therefore, the rate is equivalent to 1.079 × 10^7 N/s

Thus,

Power = Rate (N/s) × distance

           = 1.079 × 10^7 N/s × 50.0 m

           = 5.395 × 10^8 Watts

The power generated from the falling water is 5.395 × 10^8 Watts

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Marat540 [252]

Answer:

option D

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distance between the sound level = ?

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log (\dfrac{r_2}{r_1}) = \dfrac{L_1 -L_2}{20}

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7 0
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The top of the pool table is 0.810 m from the floor. the placement of the tape is such that 0 m is aligned with the edge of the
8090 [49]
Compute first for the vertical motion, the formula is:

y = gt²/2 

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t = 0.4064 s 


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x = (vx)t 

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4.65 m/ 0.4064s = (vx)

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(vy) = gt 

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6 0
3 years ago
A coil of wire containing N turns is in an external magnetic field that is perpendicular to the plane of the coil and it steadil
krok68 [10]

Answer:

The Resultant Induced Emf in coil is 4∈.

Explanation:

Given that,

A coil of wire containing having N turns in an External magnetic Field that is perpendicular to the plane of the coil which is steadily changing. An Emf (∈) is induced in the coil.

To find :-

find the induced Emf if rate of change of the magnetic field and the number of turns in the coil are Doubled (but nothing else changes).

So,

   Emf induced in the coil represented by formula

                          ∈  =   -N\frac{d\phi}{dt}                                  ...................(1)

                                          Where:

                                                    .   \phi = BAcos\theta     { B is magnetic field }

                                                                                 {A is cross-sectional area}

                                                    .  N = No. of turns in coil.

                                                    .  \frac{d\phi}{dt} = Rate change of induced Emf.

Here,

Considering the case :-

                                    N1 = 2N  &      \frac{d\phi1}{dt} = 2\frac{d\phi}{dt}

Putting these value in the equation (1) and finding the  new emf induced (∈1)

                           

                                      ∈1 =-N1\times\frac{d\phi1}{dt}

                                      ∈1 =-2N\times2\frac{d\phi}{dt}

                                       ∈1 =4 [-N\times\frac{d\phi}{dt}]

                                        ∈1 = 4∈             ...............{from Equation (1)}      

Hence,

The Resultant Induced Emf in coil is 4∈.        

                           

8 0
3 years ago
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