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Dmitry [639]
3 years ago
14

A machine accelerates a 5 kg missile from rest to a speed of 5 km/s. The net force accelerating the missile 500,000 N. How long

does it take to arrive at the speed of 5 km/s?
Physics
1 answer:
Gnoma [55]3 years ago
3 0
Final velocity = 5km/sec = 5000 m/sec
Initial velocity = 0, mass = 5kg and force = 500,000 N
F = m(Vf-Vi)/time
Time = m(Vf-Vi)/force = 5*5000/500000 = 0.05 seconds
So, it takes 0.05 sec to arrive at the speed of 5km/s.
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An electron and a proton are held on an x axis, with the electron at x = + 1.000 m
mixas84 [53]

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}

for the electron at x = 1 m

          U₂ = k \frac{e^2 }{r-1}

starting point.

        Em₀ = K + U₁ + U₂

        Em₀ = \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1}

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

4 0
3 years ago
A hypothetical planet has a mass of 1.66 times that of Earth, but the same radius. What is gravitiy near its surface?
baherus [9]

Answer: 16.22 m/s^2

Explanation: g= GM/r^2 G= (6.67x 10^-11) M= 1.66(6x 10^24) r=(6400x 10^3) so

((6.67x10^-11)(1.66x 6x 10^24))/ (6400x10^3)^2 = 16.22 m/s^2

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3 years ago
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Answer:

a luminous ball of plasma

6 0
2 years ago
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3 years ago
H. A truck starts to move from rest. If it gains the acceleration of 3 m/s2 in 5 sec,
SashulF [63]

Answer:

1. The final velocity of the truck is 15 m/s

2. The distance travelled by the truck is 37.5 m

Explanation:

1. Determination of the final velocity

Initial velocity (u) = 0 m/s

Acceleration (a) = 3 m/s²

Time (t) = 5 s

Final velocity (v) =?

The final velocity of the truck can be obtained as follow:

v = u + at

v = 0 + (3 × 5)

v = 0 + 15

v = 15 m/s

Therefore, the final velocity of the truck is 15 m/s

2. Determination of the distance travelled

Initial velocity (u) = 0 m/s

Acceleration (a) = 3 m/s²

Time (t) = 5 s

Distance (s) =?

The distance travelled by the truck can be obtained as follow:

s = ut + ½at²

s = (0 × 5) + (½ × 3 × 5²)

s = 0 + (½ × 3 × 25)

s = 0 + 37.5

s = 37.5 m

Therefore, the distance travelled by the truck is 37.5 m

6 0
2 years ago
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