<span>Let F be the force of gravity, G be the gravitational constant, M be the mass of the earth, m your mass and r the radius of the earth, then:
F = G(Mm / (4(pi)*r^2))
The above expression gives the force that you feel on the earth's surface, as it is today!
Let us now double the mass of the earth and decrease its diameter to half its original size.
This is the same as replacing M with 2M and r with r/2.
Now the gravitational force (F' ) on the new earth's surface is given by:
F' = G(2Mm / (4(pi)(r/2)^2)) = 2G(Mm / ((1/4)*4(pi)*r^2)) = 8G(Mm / (4(pi)*r^2)) = 8F
So:
F' = 8F
This implies that the force that you would feel pulling you down (your weight) would increase by 800%!
You would be 8 times heavier on this "new" earth!</span>
Explanation:
The newton's laws of motion are:
First law:
"A body will remain in its state of rest or of uniform motion along a path unless it is acted upon by an external force. ".
This is popularly called the law of inertia.
Second law:
"the acceleration of a body is produced by a net force that is inversely proportional to the mass of the body".
Third law:
"action and reaction forces are equal and opposite in direction".
learn more:
Newton's laws brainly.com/question/11411375
#learnwithBrainly
The rotational speed of the person is 0.4 rad/s.
<h3>
Rotational speed (rad/s)</h3>
The rotational speed of the person in radian per second is calculated as follows;
v = ωr
where;
- v is linear speed in m/s
- r is radius in meters
- ω is speed in rad/s
ω = v/r
ω = 2/5
ω = 0.4 rad/s
Thus, the rotational speed of the person is 0.4 rad/s.
Learn more about rotational speed here: brainly.com/question/6860269
Answer:
(a) The equivalent spring constant is 598.485 N/m
(b) The work done is 46.926 J
Explanation:
From Hooke's law of elasticity
K (spring constant) = F/e
F is the range of force exerted = 237 - 0 = 237 N
e is the extension of bowstring = 0.396 m
K = F/e = 237/0.396 = 598.485 N/m
Work done = 1/2 Fe = 1/2 × 237 × 0.396 = 46.926 J
