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Kazeer [188]
3 years ago
7

A car has a velocity of 10m/s.it now accelerates for 1m/s for 1/4 minutes. Find the distance travelled in this time and final sp

eed of the car​
Physics
1 answer:
Sonja [21]3 years ago
6 0

¡Hellow!

For this problem, lets recabe information:

v (Velocity) = 10 m/s

a (Aceleration) = 1 m/s²

t (Time) = 1/4 min = 25 s

d (Distance) = ?

v' (Final velocity) = ?

First, for calculate distance, lets applicate formula:

                                        \boxed{\boxed{\text{d = Vo * t + (a * t}^{2})\text{ * 0,5}   } }

Lets replace according we information and let's resolve it:

d = 10 m/s * 25 s + (1 m/s² * (25 s)²) * 0,5

d = 250 m + (625 m) * 0,5

d = 2,5 m + 312,5 m

d = 314 meters.

Now, for calculate final speed, lets applicate formula:

                                                 \boxed{\boxed{\text{v' = v + a * t}   } }

Lets replace according we information and let's resolve it:

v' = 10 m/s + 1 m/s² * 25 s

v' = 10 m/s + 25 m/s

v' = 35 m/s

¿Good Luck?

Att: That guy who use the "ñ".

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An electron and a proton are held on an x axis, with the electron at x = + 1.000 m
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Answer:

  r2 = 1 m

therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m

Explanation:

For this exercise we must use conservation of energy

the electric potential energy is

          U = k \frac{q_1q_2}{r_{12}}

for the proton at x = -1 m

          U₁ =- k \frac{e^2 }{r+1}

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          U₂ = k \frac{e^2 }{r-1}

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final point

         Em_f = k e^2 ( -\frac{1}{r_2 +1} + \frac{1}{r_2 -1})

   

energy is conserved

        Em₀ = Em_f

        \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e^2 (- \frac{1}{r_2 +1} + \frac{1}{r_2 -1})              

       

        \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e²(  \frac{2}{(r_2+1)(r_2-1)} )

we substitute the values

½ 9.1 10⁻³¹ 450 + 9 10⁹ (1.6 10⁻¹⁹)² [ - \frac{1}{20+1} + \frac{1}{20-1} ) = 9 109 (1.6 10-19) ²( \frac{2}{r_2^2 -1} )

          2.0475 10⁻²⁸ + 2.304 10⁻³⁷ (5.0125 10⁻³) = 4.608 10⁻³⁷ ( \frac{1}{r_2^2 -1} )

          2.0475 10⁻²⁸ + 1.1549 10⁻³⁹ = 4.608 10⁻³⁷     \frac{1}{r_2^2 -1}

          \frac{2.0475 \ 10^{-28} }{1.1549 \ 10^{-37} } = \frac{1}{r_2^2 -1}

          r₂² -1 = (4.443 10⁸)⁻¹

           

          r2 = \sqrt{1 + 2.25 10^{-9}}

          r2 = 1 m

therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m

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