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tatyana61 [14]
2 years ago
10

A housekeeper is cleaning the kitchen and decides to to use a ladder to reach a top shelf. As he/she is cleaning he/she is start

ed by a mouse and falls off the ladder to the floor. As the he/she is falling, what is true about the magnitude of the Earth's gravitational force exerted on the housekeeper compared to the magnitude of gravitational force that the housekeeper exerts on the Earth?
Physics
1 answer:
pshichka [43]2 years ago
3 0

The amount of gravitational force between both objects will be the same.

The magnitude of the Earth's gravitational force exerted on the housekeeper is calculated by applying Newton's second law of motion;

F = mg

where;

<em>m </em><em>is the mass of the housekeeper</em>

<em>g </em><em>is acceleration due to gravity</em>

According to Newton's third law of motion, action and reaction are equal and opposite.

The force exerted on the housekeeper by the Earth is equal in magnitude to the force exerted on the Earth by the housekeeper.

F_{H} = - F_{E}

The two forces are equal in magnitude but opposite in direction.

Thus, the correct option is " the amount of gravitational force between both objects will be the same"

<em>The</em><em> missing part</em><em> of the </em><em>question </em><em>is below:</em>

a. the Earth exerts the largest amount of gravitational force

b. the housekeeper exerts the largest amount of gravitational force

c. the amount of gravitational force between both objects will be the same

Learn more about Newton's third law of motion here: brainly.com/question/15507

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One electron collides elastically with a second electron initially at rest. After the collision, the radii of their trajectories
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Answer:

114.92749 keV

Explanation:

r = Radius of trajectory

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The centripetal force and the magnetic forces are conserved

m\frac{v^2}{r}=Bqv\\\Rightarrow v=\frac{Bqr}{m}

Velocity of first electron

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Velocity of second electron

v=\frac{Bqr_2}{m}\\\Rightarrow v_2=\frac{0.044\times 1.6\times 10^{-19}\times 0.024}{9.11\times 10^{-31}}\\\Rightarrow v_2=185466520.30735\ m/s

Total kinetic energy is given by

K=K_1+K_2\\\Rightarrow K=\frac{1}{2}mv_1^2+\frac{1}{2}mv_2^2\\\Rightarrow K=\frac{1}{2}m(v_1^2+v_2^2)\\\Rightarrow K=\frac{1}{2}\times 9.11\times 10^{-31}(77277716.79473^2+185466520.30735^2)\\\Rightarrow K=1.83884\times 10^{-14}\ J

Converting to eV

1\ J=\frac{1}{1.6\times 10^{-19}}\ eV

1.83884\times 10^{-14}\ J=1.83884\times 10^{-14}\times \frac{1}{1.6\times 10^{-19}}\ eV\\ =114927.49\ ev=114.92749\ keV

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