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3 years ago

9

Which change will always result in an increase in gravitational force between two objects

2 answers:

grin007 [14]3 years ago

7 0

Increasing the masses of the objects and decreasing the distance between the objects

jolli1 [7]3 years ago

5 0

Answer: Hello, the gravitational force has the form $F = G \frac{M1*M2}{r^{2} }$ , where M1 and M2 are the masses of the objects, G is a constant and r is the distance between the two objects.

first: increasing the mass of the objects, the masses are in the numerator, then if whe increase either M1 or M2 (or both of them) then the gravitational force will increase.

second: decreasing the distance between the objects ( r in this case). r is in the denominator, si if r is smaller, the result of the division will be bigger, so if r decrease, then the gravitational force increase.

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olga55 [171]

Answer:

4 m/ $s^{2}$

Explanation:

From Equilibrium of forces, The Tension in string is cancelled by the Weight (product of mass and acceleration due to gravity) of the body acting downwards.

The Net force = Mass * Acceleration.

Since Net Force = 20 Newton, Mass = 5kg, therefore;

20 = 5kg * acceleration. Dividing the RHS and LHS of the equation by 5, we have;

Acceleration = $\frac{20}{5}$ which gives 4.

Note: RHS means Right Hand Side.

LHS means Left Hand Side.

7 0

2 years ago

Two airplanes leave an airport at the same time. The velocity of the first airplane is 730 m/h at a heading of 65.3 ◦ . The velo

yanalaym [24]

Answer:

Plane will 741.6959 m apart after 1.7 hour

Explanation:

We have given time = 1.7 hr

So if we draw the vectors of a 2d graph we see that the difference in angles is = $102^{\circ}-65.3^{\circ}=36.7^{\circ}$

Speed of first plane = 730 m/h

So distance traveled by first plane = 730×1.7 = 1241 m

Speed of second plane = 590 m/hr

So distance traveled by second plane = 590×1.7 = 1003 m

We represent these distances as two sides of the triangle, and the distance between the planes as the side opposing the angle 58.6.

Using the law of cosine, $r^2$ representing the distance between the planes, we see that:

$r^2=1241^2+1003^2-2\times 1003\times 1241cos(36.7)=550112.8295$

r = 741.6959 m

3 0

3 years ago

Explain what major mistake producers made when editing the battle scenes for the movie Star Wars

Artyom0805 [142]

Answer: Air

Explanation:

Roffns dn b bdd

4 0

2 years ago

Calculating average speed

djverab [1.8K]

Average speed =

(distance covered during some period of time)
divided by
(length of time to cover that distance).

7 0

2 years ago

Suppose that a planet were discovered between the sun and Mercury, with a circular orbit of radius equal to 2/3 of the average o

Musya8 [376]

Explanation:

It is given that,

A planet were discovered between the sun and Mercury, with a circular orbit of radius equal to 2/3 of the average orbit radius of Mercury.

Mass of the Sun, $M=1.99\times 10^{30}\ kg$

Radius of Mercury's orbit, $r=5.79\times 10^{10}\ m$

Radius of discovered planet, $R=\dfrac{2}{3}r$

$R=\dfrac{2}{3}\times 5.79\times 10^{10}\ m=3.86\times 10^{10}\ m$

Let T is the orbital period of such a planet. Using Kepler's third law of planetary motion as :

$T^2\propto R^3$

$T^2=\dfrac{4\pi^2R^3}{GM}$

$T^2=\dfrac{4\pi^2\times (3.86\times 10^{10})^3}{6.67\times 10^{-11}\times 1.99\times 10^{30}}$

$T=\sqrt{1.71\times 10^{13}}$

T = 4135214.625 s

or

T = 47.86 days

So, the orbital period of such a planet is 47.86 days. Hence, this is the required solution.

6 0

2 years ago