you need G, grav constant, here.
F=GMmoonMEllen/R^2
all other numbers given ... calculate ...
Momentum = Mass × Velocity
According to this formula,
Momentum of deer = 176 × 19 = 3344 kg•m/s.
Since you are heading north and the deer is running towards you, the direction of the deer' s momentum is north as well.
<h2>
Hey There!</h2><h2>
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Answer:</h2><h2 /><h2>

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<h2>DATA:</h2>
mass = m = 2kg
Distance = x = 6m
Force = 30N
TO FIND:
Work = W = ?
Velocity = V = ?
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SOLUTION:</h2>
According to the object of mass 2 kg travels a distance when the force was exerted on it. The graph between the Force and position was plotted which shows that 30 N of force was used to push the object till the distance of 6.0m.
To find the work, I will use the method of determining the area of the plotted graph. As the graph is plotted in the straight line between the Force and work, THE PICTURE ATTCHED SHOWS THE AREA COVERED IN BLUE AS WORK DONE AND HEIGHT AS 30m AND DISTANCE COVERED AS 6m To solve for the area(work) of triangle is given as,

Base is the x-axis of the graph which is Position i.e. 6m
Height is the y-axis of the graph which is Force i.e. 30N
So,

W = 90 J
The work done is 90 J.
According to the principle of work and kinetic energy (also known as the work-energy theorem) states that the work done by the sum of all forces acting on a particle equals the change in the kinetic energy of the particle.



<h2>_____________________________________</h2><h2>Best Regards,</h2><h2>'Borz'</h2>
This question is in complete.The question is
A coin with a diameter 3.00 cm rolls up a 30.0° inclined plane. The coin starts out with an initial angular speed of 60.0 rad/s and rolls in a straight line without slipping. If the moment of inertia of the coin is(1/2) MR² , how far will the coin roll up the inclined plane (length along the ramp)? Hint: Conservation of mechanical energy.
Answer:
distance=0.124 m
Explanation:
