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bearhunter [10]
3 years ago
15

A construction worker pushes a crate horizontally on a frictionless floor with a net force of 10\, \text 10N, start text, N, end

text for 4.0\,\text m4. 0m4, point, 0, start text, m, end text. How much kinetic energy does the crate gain?
Physics
1 answer:
Strike441 [17]3 years ago
7 0

Answer:

k_f = 40J

Explanation:

The work and energy theorem says that:

W_f =k_f-k_i

where W_f is the work of the force, k_f the final kinetic energy and k_i the initial kinetic energy.

Addittionally, the work of the force is calculate as force multiply by distance and if the crate inittialy is at rest, the initial kinetic energy is zero, so:

Fd = k_f

where F is the force and d the distance. Then, replacing values, we get:

(10N)(4m) = k_f

40J = k_f

it means that the system gain 40J of kinetic energy.

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Explanation:

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A stereo speaker is placed between two observers who are 35 m apart, along the line connecting them. If one observer records an
kompoz [17]

Answer:

   x = 2,864 m ,       Ra = 32.1 m                       

Explanation:

Let's solve this problem in parts, let's start by finding the intensity of the sound in each observer

observer A β = 64 db

             β = 10 log Iₐ / I₀

where I₀ = 1 10⁻¹² W / m²

              Iₐ = I₀ 10 (β/ 10)

let's calculate

              Iₐ = 1 10⁻¹² (64/10)

              Iₐ = 2.51 10⁻⁶ W / m²

Observer B β = 85 db

             I_b = 1 10-12 10 (85/10)

             I_b = 3.16 10⁻⁴ W / m²

now we use that the emitted power that is constant is the intensity over the area of ​​the sphere where the sound is distributed

              P = I A

therefore for the two observers

              P = Ia Aa = Ib Ab

the area of ​​a sphere is

               A = 4π R²

we substitute

               Ia 4pi Ra2 = Ib 4pi Rb2

               Ia Ra2 = Ib Rb2

Let us call the distance from the observer be to the haughty R = ax, so the distance from the observer A to the haughty is R = 35 ax; we substitute

             Ia (35 -x) 2 = Ib x2

we develop and solve

           35-x = Ra (Ib / Ia) x

           35 = [Ra (Ib / Ia) +1] x

           x (11.22 +1) = 35

           x = 35 / 12.22

            x = 2,864 m

This is the distance of observer B

The distance from observer A

            Ra = 35 - x

            Ra = 35 - 2,864

            Ra = 32.1 m

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