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lara [203]
3 years ago
9

Gravitational attraction depends on the mass of the objects as well as their distance. The gravitational force between objects i

ncreases as the masses of the objects increase. The gravitational force decreases as the distance between objects increases. The impact of mass and distance are not equal. If the mass of one object doubles, the gravitational force also doubles. If the distance between objects doubles, the gravitational force decreases by 1/4. Using the model predict what will happen to the gravitational force if the masses both double and the distance also doubles.
Physics
2 answers:
shepuryov [24]3 years ago
7 0
<h2>Answer: Gravitational attraction will be the same</h2>

According to the law of universal gravitation, which is a classical physical law that describes the gravitational interaction between different bodies with mass:

F=G\frac{m_{1}m_{2}}{r^2}    (1)

Where:

F is the module of the force exerted between both bodies

G is the universal gravitation constant.

m_{1} and m_{2} are the masses of both bodies.

r is the distance between both bodies

Now, if we double both masses and the distance also doubles, this means:

m_{1} and m_{2} will be now 2m_{1} and 2m_{2}

r will be now 2r

Let's rewrite the equation (1) with this new values:

F=G\frac{(2m_{1})(2m_{2})}{(2r)^2}    (2)

Solving and simplifying:

F=4G\frac{m_{1}2m_{2}}{4r^2}    

F=G\frac{m_{1}m_{2}}{r^2}     (3)

As we can see, equation (3) is the same as equation (1).

So, if the masses both double and the distance also doubles the <u>Gravitational attraction between both masses will remain the same.</u>

Dmitry [639]3 years ago
7 0

Answer:

C) The gravitational attraction remains constant.

even though the distance is double The gravitational attraction remains constant no matter what.

I hope this helps!!!

Your welcome,

                         Shadow Lolbit

Can we be friends!!!!

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A piece of wire 10 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral tria
aleksandrvk [35]

Answer:

a) in order to be maximum the length of the square wire should be 10m and no cut for making the triangle

b) in order to be minimum the length of the square wire should be 3.84 m and the triangle wire should be 6.16 m

Explanation:

for a square with a side length of a and a equilateral triangle of length b

area square = a²

area triangle = base* height/2

for an equilateral triangle height = base * sin 60 = (√2 / 2 )* base

therefore

area triangle = base* height/2 =  base² (√2 /4) =(√2 /4) b²

the total length of the wire is = square length + triangle length = 4*a + 3*b

therefore

A=a²+(√2 /4) b²

4*a + 3*b=L→ b = (L-4*a)/3

A=  a²+(√2 /4) (L-4*a)²/ 9

the maximum and minimum amount can be found taking the derivative of the area respect with a:

dA/da= 2*a + (√2 /36) 2*(L-4*a)*(-4) = 0

a -(√2 /9) (L-4*a) = 0

a - √2 /9 * L + √2 /9*4*a =0

(√2 /9*4+1)*a = √2 /9 * L

a= √2 /9 * L / (√2 /9*4+1) = √2 /9 * L / (√2 /9*4+1) = √2 /9 * 10 m / (√2 /9*4+1)

a= 0.96 m

therefore

b = (L-4*a)/3 = (10 m - 4*0.96m)/3= 2.053 m

A=a²+(√2 /4) b² = (0.96 m)²+ (√2 /4) (2.053m)² = 2.411 m²

in the extreme cases

a= 10 m/4=2.5 m and b=0

thus A= (2.5 m)² = 6.25 m²

b=10 m/3= 3.33 m and a=0

thus A= (√2 /4) (3.33 m)² = 3.92 m²

therefore minimum area A=3.92 m² with Length 1=4*0.96 m=3.84 m , Length 2 =2.053 m*3 = 6.16 m

the maximum area is A=6.25 m² with Length 1=10 m and Length 2=0 m

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