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sergiy2304 [10]
1 year ago
14

Question 9 of 10

Physics
1 answer:
romanna [79]1 year ago
5 0

A atomic neutron particle that is attracting another atomic neutron is an example of the strong nuclear force acting in an atom (Option B).

<h3>What are nuclear forces?</h3>

The expression 'nuclear forces' makes reference to the attraction between different subatomic particles in a given nucleus of a particular atom, which represents the strongest force in the universe.

Subatomic particles include positively charged particles (i.e., protons), neutrally charged particles called neutrons, and electrically negative charged particles called electrons (e-).

In conclusion, a atomic neutron particle that is attracting another atomic neutron is an example of the strong nuclear force acting in an atom (Option B).

Learn more about nuclear forces here:

brainly.com/question/16959709

#SPJ1

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The magnitude of the net displacement is 95.3 m

Explanation:

To find the magnitude of the net displacement, we have to resolve each of the two displacements into the horizontal and vertical direction first.

1st displacement is:

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So its components are

d_{1x}=(79)(cos 16.9^{\circ})=75.6 m\\d_{1y}=(79)(sin 16.9^{\circ})=23.0 m

2nd displacement is:

d_2=16.7 m at 31.1^{\circ}

So its components are

d_{2x}=(16.7)(cos 31.1^{\circ})=14.3 m\\d_{2y}=(16.7)(sin 31.1^{\circ})=8.6 m

Therefore, the x- and y-components of the net displacement are:

d_x=d_{1x}+d_{2x}=75.6+14.3=89.9 m\\d_y=d_{1y}+d_{2y}=23.0+8.6=31.6 m

Therefore, the magnitude of the final displacement is:

d=\sqrt{d_x^2+d_y^2}=\sqrt{(89.9)^2+(31.6)^2}=95.3 m

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