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Nataliya [291]
3 years ago
13

If two proton approach each other with speed of 400m\s and 450m\s respectively what is the total momentum of the two particle sy

stem
Physics
1 answer:
lilavasa [31]3 years ago
7 0

Answer: 83.65\times 10^{-27}\ kg.m/s

Explanation:

Given

The velocities of the protons are 400 m/s and 450 m/s

Mass of the Proton is m=1.673\times 10^{-27}\ kg

Since they travel in the opposite direction, therefore, their momentum cancels out each other.

Momentum is the product of mass and velocity

\therefore m(450-400)\\\Rightarrow 1.673\times 10^{-27}\times 50\\\Rightarrow 83.65\times 10^{-27}\ kg.m/s

The direction of momentum is in the direction of proton having 450 m/s velocity.

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To solve this problem we will use the related concepts in Newtonian laws that describe the force of gravitational attraction. We will use the given value and then we will obtain the proportion of the new force depending on the Radius. From there we will observe how much the force of attraction increases in the new distance.

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Applying the new distance,

F = \frac{GMm}{(\frac{R}{2})^2}

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Replacing with the previous force,

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F = 36*10^{22}N

Therefore the magnitude of the force on the star due to the planet is  36*10^{22}N

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\lambda=\frac{v_{air}}{f} \\\lambda=\frac{344}{298.25} \\\lambda=1.15 m

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