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avanturin [10]
3 years ago
11

A motorist driving at 25 meters/second decelerates to

Physics
2 answers:
Ilia_Sergeevich [38]3 years ago
8 0
Work done = 0.5*m*[(v2)^2 - (v1)^2]
where m is mass,
v2 and v1 are the velocities.

Given that m = 1.50 x 10^3 kg, v2 = -15 m/s (decelerates), v1 = 25 kg,

Work done = 0.5 * 1.50 x 10^3 * ((-15)^2 - 25^2) = 3 x 10^5 joules

Just ignore the negative value for the final result because work is a scalar quantity.
andrew11 [14]3 years ago
3 0

Answer:

C.

3.0 × 10^5 joules

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Two antennas located at points A and B are broadcasting radio waves of frequency 96.0 MHz, perfectly in phase with each other. T
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a)   Δφ = 1.51 rad , b)  x = 21.17 m

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This is an interference problem, as they indicate that the distance AP is on the x-axis the antennas must be on the y-axis, the phase difference is

          Δr /λ = Δfi / 2π

          Δfi = Δr /λ 2π

          Δr = r₂-r₁

let's look the distances

         r₁ = 57.0 m

We use Pythagoras' theorem for the other distance

         r₂ = √ (x² + y²)

         r₂ = √(57² + 9.3²)

         r₂ = 57.75 m

The  difference is

         Δr = 57.75 - 57.0

         Δr = 0.75 m

Let's look for the wavelength

        c =  λ f

          λ = c / f

          λ = 3 10⁸ / 96.0 10⁶

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           Δr = (2n + 1) λ / 2

           

For the first interference n = 0

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           Δr = r₂ - r₁

We substitute the values

        √ (x² + y²) - x = 3.12 / 2

Let's solve for distance x

          √ (x² + y²) = 1.56 + x

          x² + y² = (1.56 + x)²

          x² + y² = 1.56² + 2 1.56 x + x²

          y2 = 20.4336 +3.12 x

          x = (y² -20.4336) /3.12

          x = (9.3² -20.4336) /3.12

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This is the distance for the first minimum

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3 years ago
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