What change will always result in an increase in gravitational force between two objects
1 answer:
The gravitational force between two object depends on their masses and on their distance.
Since the formula is
If the masses grow, the force also grows. But I'm assuming the two objects are fixed, so you can't enlarge their mass.
So, the only option remaining is to lower their distance: since it sits at the denominator, a smaller value of d results in a bigger value for F.
So, if you reduce the distance between two objects, the gravitational force between them will always result in an increase
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Increasing the angle of inclination of the plane decreases the velocity of the block as it leaves the spring.
- The statement that indicates how the relationship between <em>v</em> and <em>x</em> changes is;<u> As </u><u><em>x</em></u><u> increases, </u><u><em>v</em></u><u> increases, but the relationship is no longer linear and the values of </u><u><em>v</em></u><u> will be less for the same value of </u><u><em>x</em></u><u>.</u>
Reasons:
The energy given to the block by the spring =
According to the principle of conservation of energy, we have;
On a flat plane, energy given to the block = = kinetic energy of
block =
Therefore;
0.5·k·x² = 0.5·m·v²
Which gives;
x² ∝ v²
x ∝ v
On a plane inclined at an angle θ, we have;
The energy of the spring =
- The force of the weight of the block on the string,
The energy given to the block = = The kinetic energy of block as it leaves the spring =
Which gives;
Which is of the form;
a·x² - b = c·v²
a·x² + c·v² = b
Where;
a, b, and <em>c</em> are constants
The graph of the equation a·x² + c·v² = b is an ellipse
Therefore;
- As <em>x</em> increases, <em>v</em> increases, however, the value of <em>v</em> obtained will be lesser than the same value of <em>x</em> as when the block is on a flat plane.
<em>Please find attached a drawing related to the question obtained from a similar question online</em>
<em>The possible question options are;</em>
- <em>As x increases, v increases, but the relationship is no longer linear and the values of v will be less for the same value of x</em>
- <em>The relationship is no longer linear and v will be more for the same value of x</em>
- <em>The relationship is still linear, with lesser value of v</em>
- <em>The relationship is still linear, with higher value of v</em>
- <em>The relationship is still linear, but vary inversely, such that as x increases, v decreases</em>
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