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a_sh-v [17]
2 years ago
13

What is the total momentum of a 30 kg object traveling left at 3 m/s and a 50 kg object traveling at 2 m/s to the right?

Physics
1 answer:
madam [21]2 years ago
4 0

Answer:

there are 25 kg objective travelling at 2m/s to the right.

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State the law of conservation of momentum<br>​
trapecia [35]

Answer:

Law of conservation of momentum states that. For two or more bodies in an isolated system acting upon each other, their total momentum remains constant unless an external force is applied. Therefore, momentum can neither be created nor destroyed.

Explanation:

Hope it helps

8 0
3 years ago
Read 2 more answers
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 car changes speed from 25 m/s to 10 m/s in 240 seconds. Describe its acceleration.​
Dmitry_Shevchenko [17]

Explanation:

Acceleration is change in velocity over change in time:

a = Δv / Δt

a = (10 m/s - 25 m/s) / (240 s - 0 s)

a = -0.0625 m/s²

So the car decelerates at 0.0625 m/s².

7 0
3 years ago
X = v xo t. x = (10.0 m/s)(3.53 s) x = ????
LenaWriter [7]

x = V<em>x</em> * t

given V<em>x</em> = 10m/s n t = 3.53s

x = 10 * 3.53

= 35.3m


7 0
3 years ago
Read 2 more answers
14. The average speed of a car was 60 m/s by the time it reached the finish line. The car moved in a straight line and traveled
Leni [432]
7.5m. Have a good day :)
4 0
3 years ago
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