Answer:
Yes, this is <em><u>true</u></em>. Hope this helps :)
if the distance between the objects is doubled the force is reduced by a factor of 4
<h3>Whats is gravitational force?</h3>
Gravitational force is the force of attraction between objects in the universe.
f = G * m1 * m2 / r^2
f = gravitational force
G = gravitational constant
m1 = mass of object 1
m2 = mass of object 2
r = distance between the objects
From the formular, the gravitational force and the distance is an inverse relationship so increasing the distance by a factor results to reduction of the force by the square of the factor. hence doubling the distance which is distance mutiplied by 2 will lead to reduction of the force by 2^2 = 4
Therefore: The force decreases by a factor of 4.
hope it helps
<span>3.834 m/s.
In this problem we need to have a centripetal force that is at least as great as the gravitational attraction the object has. The equation for centripetal force is
F = mv^2/r
and the equation for gravitational attraction is
F = ma
Since m is the same in both cases, we can cancel it out and then set the equations equal to each other, so
a = v^2/r
Substitute the known values (radius is diameter/2) and solve for v
9.8 m/s^2 <= v^2/1.5 m
14.7 m^2/s^2 <= v^2
3.834057903 m/s <= v
So the minimum velocity needed is 3.834 m/s.</span>
From Newton's law v^2 = u^2 + 2as where a is the acceleration and s is the distance.
But to go any further, we need to know how fast the vehicle is accelerating
From v = u +at
We have a = u/t where the final velocity v = 0
So in one minute acceleration = (35 / 60) / 60 = 0.0097 ms/2. The first
experession in bracket is the initial velocity, u, in metres per seconds.
Hence v^2 = (0.583)^2 + 2 (0.0097)(30)
v^2 = 0.3398 + 0.5826 = 0.9224
v = âš 0.9224 = 0.960m
